JEE-Physics MAGNETIC EFFECT OF CURRENT The branch of physics which deals with the magnetism due to electric current or moving charge (i.e. electric current is equivalent to the charges or electrons in motion) is called electromagnetism. ORESTED'S DISCOVERY The relation between electricity and magnetism was discovered by Orested in 1820. Orested showed that the electric current through the conducting wire deflects the magnetic needle held below the wire. • When the direction of current in conductor is reversed then deflection of magnetic needle is also reversed • On increasing the current in conductor or bringing the needle closer to the conductor the deflection of magnetic needle increases. Oersted discovered a magnetic field around a conductor carrying electric current. Other related facts are as follows: (a) A magnet at rest produces a magnetic field around it while an electric charge at rest produce an electric field around it. (b) A current carrying conductor has a magnetic field and not an electric field around it. On the other hand, a charge moving w th a u iform velocity has an electric as well as a magnetic fie d around it. (c) An electric field cannot be produced without ch rge whereas a magnetic field can be produced without a m g et. (d) No poles are produced in a coil carrying cu et but such a coil shows north and south polarities. (e) All oscillating or an accelerated charge produces E.M. waves also in additions to electric and magnetic fields. • Current Element A very small element ab of length of a thin conductor carrying current I called current element. Current element is a vector quantity whose magnitude is equal to the product of current and length of small element having the direction of the flow of current. • Biot – Savart's Law Edujournal S N North – + S I N N Oersted's experiment. Current in the wire deflects the compass needle dB P r I I a d b d NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65 With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based on the following experimental observations for the magnetic field dB at a point P associated with a length element d of a wire carrying a steady current I. 1 Id sin 0 Id sin dB I, dB d , dB sinand dB dB dB = r 2 r 2 4r 2 Vector form of Biot-Savar t's law 0 Id sin ˆ n =unit dB n vector perpendicular to the plane of ( Id ) and ( r ) 4 r 2 0 Id r [ Id × r = (Id ) (r)sinn ˆ ] dB 4r 3 E 1
Transcript
JEE-Physics
MAGNETIC EFFECT OF CURRENTThe branch of physics which deals with the magnetism due to electric current or moving charge (i.e. electric
current is equivalent to the charges or electrons in motion) is called electromagnetism.
ORESTED'S DISCOVERY
The relation between electricity and magnetism was discovered by Orested in
1820. Orested showed that the electric current through the conducting wire
deflects the magnetic needle held below the wire.
• When the direction of current in conductor is reversed then deflection of
magnetic needle is also reversed
• On increasing the current in conductor or bringing the needle closer to
the conductor the deflection of magnetic needle increases.
Oersted discovered a magnetic field around a conductor carrying electric
current. Other related facts are as follows:
(a) A magnet at rest produces a magnetic field around it while an
electric charge at rest produce an electric field around it.(b) A current carrying conductor has a magnetic field and not an electric
field around it. On the other hand, a charge moving w th a u iform
velocity has an electric as well as a magnetic fie d around it.
(c) An electric field cannot be produced without ch rge whereas a
magnetic field can be produced without a m g et.
(d) No poles are produced in a coil carrying cu e t but such a coil
shows north and south polarities.
(e) All oscillating or an accelerated charge produces E.M. waves also
in additions to electric and magnetic fields.
• Current Element
A very small element ab of length of a thin conductor carrying
current I called current element. Current element is a vectorquantity whose magnitude is equal to the product of current and
length of small element having the direction of the flow of current.• Biot – Savart's Law Edujournal
S N
North
– +S
I
N
N
Oersted's experiment. Current in the wire deflects the compass needle
dB
Pr
II
a
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d
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With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based
on the following experimental observations for the magnetic field dB at a point P associated with a length
element d of a wire carrying a steady current I.
1 Id sin 0
Id sin
dB I,
dB d , dB sin and dB dB dB =r2 r2 4 r2
Vector form of Biot-Savar t's law
0Id sin ˆ
n =unit
dB n vector perpendicular to the plane of ( Id ) and ( r )4 r 2
0 Id r [ Id × r =(Id ) (r)sin n
ˆ ]dB 4 r 3
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JEE-Physics
GOLDEN KEY POINTS
0 Id r
• According to dB 4 r 3 , direction of magnetic field vector (dB) is always perpendicular to the plane of
vectors Id and ( r ), where plane of Id and ( r
) is the plane of wire.
• Magnetic field on the axis of current carrying conductor is always zero (=0° or = 180°)
• Magnetic field on the perimeter of circular loop or coil is always minimum.
MAGNETIC FIELD LINES (By Michal Faraday)
In order to visualise a magnetic field graphically, Michal faraday introduced the concept of field lines.
Field lines of magnetic field are imaginary lines which represents direction of magnetic field continuously.
GOLDEN KEY POINTS
• Magnetic field lines are closed curves.
• Tangent drawn at any point on field line represents direction of the field at that point.
• Field lines never intersects to each other.
• At any place crowded lines represent stronger field while distant l es represents weaker field.• Edujournal is uniform otherwise not.
In any region, if field lines are equidistant and straight the fie dNon-uniform Field Uniform Field
Both magnitude
Magnitude is Directi n is Both magnitude
not constant not c nstant and direction are and direction arenot constant constant
• Magnetic field lines emanate from or enters in the surface of magnetic material at any angle.
• Magnetic field lines exist inside every magnetised material.
• Magnetic field lines can be mapped by using iron dust or using compass needle.
RIGHT HAND THUMB RULE
This rule gives the pattern of magnetic field lines due to current carrying wire.
(i) Straight current (ii) Circular current
Thumb In the direction of current Curling fingers In the direction of current,
Curling fingers Gives field line pattern Thumb Gives field line pattern
Case I : wire in the plane of the paper Case I : wire in the plane of the paper
I
Magnetic ACW CWfield lines
B B
Towards observer or Away from the observer
perpendicular or perpendicularI out-wards inwards
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Case II : Wire is to the plane of the paper. Case II : Wire is to the plane of the paper
ACW concentric & CW concentric &circular field lines circular field lines
ACW CW
S N N S B I
Towards observer Away from the observer
GOLDEN KEY POINTS
• When current is straight, field is circular
• When current is circular, field is straight (along axis)
• When wire is in the plane of paper, the field is perpendicular to the plane of the paper.
• When wire is perpendicular to the plane of paper, the field is in the plane of the paper.
APPLICATION OF BIOT-SAVART LAW :• Magnetic field surrounding a thin straight current carry g conductor
AB is a straight conductor carrying current i from B to A. At a P, whosepoint
perpendicular distance from AB is OP =a, the direction of field is perpendicularto the plane of paper, inwards (represented by a cross)=a tan dl=a sec2 d...(i)
=90°– & r=asec
• By Biot-Savart’s law
0 id sin (due rrent element id
dB to c at point P)4 r 2
B= dB
0
id sin (due to wire AB) B= 0 i
cos d 4 r 2 4
EdujournalTaking limits of integration as –2 2 to 1
Aid r
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B
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0 i 1 0 i sin 10 i sin 1 sin 2 (inwards)
B cos d 4 a 4 a
2
4 a
2
Ex am p l e
Magnetic field due to infinite length wire at point 'P'
Solu ti on
I 2
= 90°
0 I M 90°P 0 I
BP = [sin90° + sin90°] 1
= 90° BP =4d 2d
d
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Magnetic field due to semi infinite length wire at point 'P'
I
0 I [sin + sin90°] 90° 2=90° 0 I [sin + 1]
Sol . B = B =4 d M P P 4dP 1=
L d
Ex am p l e
Magnetic field due to special semi infinite length wire at point 'P'
Solu ti on
B = 0 I [sin0° + sin90°] B = 0 I4d 4dP P
I
90°
2
=90°
M
1
=d°
P
Ex am p l e
N
Magnetic field due to special finite length wire at point 'P' 90°2=°
MP
d1=0°
Solu ti onB = 0 I [sin0° + sin] ; B = 0 I sin
4 d 4dP P
Ex am p l e
If point ‘P’ lies out side the line of wire then magnetic field at point ‘P’ :
d P2
I 1
Solu ti onB P
0 I sin 90 1 sin 90 2 0 I (cos 1 cos 2 )
4 d 4 d
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• Magnetic field due to a loop of current
Magnetic field lines due to a loop of wire are shown in the figure
iB
i
i
The direction of magnetic field on the axis of current loop can be determined by right hand thumb rule. Iffingers of right hand are curled in the direction of current, the stretched thumb is in the direction ofmagnetic field.
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• Calculation of magnetic field in z-axis
0 id sin 900 O
Consider a current loop placed in y-z plane carrying current in id
anticlockwise sense as seen from positive x-axis. Due to a small R i y-axis
current element id shown in the figure, the magnetic field at P 22
Edujournal r= R+xis given by dB
.
4 r2 dB
xP
i x-axisThe angle between id
andr is 900 because id is along y-
axis, while lies in x-z plane. The directionof dB
is perpendicularr
to as shown. The vector dB can be res lved into two components,r
dB cos along z-axis and dB sin al ng x-axis.For any two diametrically opposite current elements, the components along x-axis add up, while the other two
components cancel out. Therefore, the field at P is due to x-component of field only. Hence, we haveB= dB sin
0 id sin = 0 id R B= 0 iR d 2
R
4 r 2 4 r 2 r 4 r 3 z-axisdBcos
0 i 2R 2
0
i 2 R 2 dB
B = = r R2
x2
4 R 2 x2 3 / 2P4 r 3 x-axis
dBsin
(a) At the centre, x=0, B =0 i
centre 2R i x2
3 / 2 0 i 3 x2 (b) At points very close to centre, x<<R B= 0
1 = 1 2
2 R2
2 2R
(c) At points far off from the centre, x>>R B= 0 2R2
4 x3
(d) The result in point (c) is also expressed as B = 0
2M
4 x3
where M= R 2 , is called magnetic dipole
moment.
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E x a m p l eFind the magnetic field at the centre of a current carrying conductor bent in the form of an arc subtending angle at its centre. Radius of the arc is R.
Solu ti onLet the arc lie in x-y plane with its centre at the origin.Consider a small current element id as shown.
The field due to this element at the centre is
0 id sin 900
dB= id and R are perpendicular4 R 2
Now d Rd
dB
0 iRd dB = 0 i d
4 R2
4 RThe direction of field is outward perpendicular to plane of paper
0 i
0 i 0
0 iTotal magnetic field B= dB B d B=
4R4 R 0 4R
y-axis
idi
R
d
R
x-axisO
Ex am p l eFind the magnetic field at the centre of a current carrying conductor be t in the form of an arc subtendingangle 1 and 2 at the centre. (N, I, R)
EdujournalACW
Solu ti onMagnetic field at the centre of arc abc and adc wire of circuit loop b
B abc
I I B I a gle = arc length 1 10 1 1
and B adc 0 2 2
abc 1 1
=2 2
14 4 B adc I 2 2 radiusr a I1
I1 R2 I1 2 2
V=I1R1=I2R2 I 2 R 1 I2
1 R A R ) d
I2 cB
abc
2
1
B1
1
Badc
=
B 1 1 2 2
HELMHOLT'Z COILS ARR ANG MENT (N, I, R) R
This arrangement is used to produce uniform magnetic field O1
O2 ACW
of short range. It consists :-M
• Two identical co–axial coils (N, I, R same) (uniform magnetic field
• Placed at distance (center to center) equal to radius ('R') of coils of short range BHC)P S Q
• Planes of both coils are parallel to each other.
• Current direction is same in both coils (observed from same side) O1
R/2 M O2
otherwise this arrangement is not called "Helmholtz coil R/2
arrangement".Ex am p l e
A pair of stationary and infinitely long bent wires are placed in the x-y plane I Q
as shown in fig. The wires carry currents of 10 ampere each as shown. The R O S
segments L and M are along the x-axis. The segments P and Q are parallel toL M
the y-axis such that OS = OR = 0.02 m. Find the magnitude and direction ofP
the magnetic induction at the origin O. I
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S o l u t i o n
As point O is along the length of segments L and M so the field at O due to these segments will be zero. Further,
0 I ˆ
0 I ˆas the point O is near one end of a long wire, B R B P B Q = (k) + (k) [as RO = SO = d]
4 d 4 d0 2I ˆ –7 2
10ˆ –4 Wb ˆ
so, B R 10 × = 10(k )
Substituting the given data, BR
0.02(k)
m2 (k)
4 d
B = 10–4 T and in (+z) direction.
Ex am p l e
Calculate the field at the centre of a semi-circular wire of radius R in situations depicted in figure (i), (ii) and (iii)if the straight wire is of infinite length.
b I I a I a
RR R
a b I Oc O I
I O I I bI
c R c
(i) R(ii) (iii)
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Solu ti onThe magnetic field due to straight current carring wire of infinite ength, for a point at a distance R from one of itsends is zero if the point is along its length and 0 I if the poi t is on a line perpendicular to its length while at the
4 R0 I
centre of a semicircular coil is so net magnetic field at the centre of semicircular wire is B R B a B b B c4R
0 I 0 I(i) BR = 0 + + 0 = ( into the page)
4 R 4R(ii) B
R
= 0 I + 0 I + 0 I = 0 I [ + 2] (out of the page)
4 R 4 R 4 R 4 R(iii) BR = 0 I +0 I
+0 I = 0 I [ – 2]
(into the page)
4 R 4 R 4 R 4 R
Edujournal
Ex am p l eCalculate the magnetic induction at the point O, if the current carrying wire is b in the shape shown in figure. The radius of the curved part of the wire is a andlinear parts are assumed to be very long and parallel. c I d
Solu ti onMagnetic induction at the point O due to circular portion of the wire a e
µ0 I 0 i 3 3
B1 = = (out of the page) ( = )4R 4 a 22
Magnetic induction at O due to wire cd will be zero since O lies on the line cd itself when extended backward. Magnetic induction at O due to infinitely long straight wire ae is
µ i µ0 i µ i B =
0
[sin 1 sin 2 ] where r = a, = 0, = B = sin 0 sin =0
4r 2 4 2 4 a2 1 2 2 a
Because both the fields are in same direction i.e. perpendicular to plane of paper and directed upwards, hence
the resultant magnetic induction at O is B = B1 + B2 = 0 i 3 1
24a
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E x a m p l eIn the frame work of wires shown in figure, a current i is allowed to flow. B i
ACalculate the magnetic induction at the centre O. If angle is equal to 90°, R
2 C
then what will be the value of magnetic induction at O ? O D
Solu ti on i R1
0 iMagnetic induction at O due to the segment BC is B1 = E
4R2
Similarly, the magnetic induction at O due to circular segment AED is B2 = 0 i (2)
4R1
Magnetic field due to segments AB and CD is zero, because point 'O' lies on axis of these parts.
Hence resultant magnetic induction at O is B=B1+B2 = 0 i 2 ,
4R2 R 1
If = 90° = i 3 i 1 3 , then B = 0
= 0
2 4 2R22R 1 8 R 2 R 1
Ex am p l e
Two concentric circular coils X and Y of radii 16 cm and 10 cm respectively lie in the same verticalplane containing the north-south direction. Coil X has 20 tur s a d carries a current of 16 A; coilY has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, andin Y clockwise, for an observer looking at the coils facing the west.What is the magnitude and direction of the magnetic fie d at their common centre
(i) Due to coil X alone ? (ii) Due to coil Y lone ? (iii) Due to both the coils ?
Solu ti on
According to the figure the magnitude of the magnetic field at the centre of coil X is
Bx
= 0 x N x 2 107 × 16 20 =4×10–4T N
2 rx
2 0.16 Ix Coil X
Coil YSince the current in coil X is anticlockwise, the direction of Bx
is towards the east as shown in fig re. W E
The magnitude of magnetic field at the centre of the coil Y Iy By B
x
0 N 4
10
7 18 25
is given by BY =
Y Y
= × =9 × 10–4 T S2 Y 2 2Edujournal
since the current in coil Y is clockwise, the direction of field BY is towards the west (see fig.). Since the
two fields are collinear and oppositely directed. The magnitude of the resultant field =difference betweenthe two fields and its direction is that of the bigger field. Hence the net magnetic field at the common centreis 5 × 10–4 T and is directed towards the west.
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Ex am p l e
A long wire bent as shown in the figure carries current I. If the radius of the
semi-circular portion is "a" then find the magnetic induction at the centre C.
Solu ti on X
0 I ˆ
Due to semi circular part B
1 4a i
Z I
O a Y
CI I
Ad
van
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0 I 0 I ˆ 0 I ˆdue to parallel parts of currents B 2 2
4a(k
ˆ)
, Bnet =BC= B1B2 = 4a (i) + (k)
2a0 I
2 4magnitude of resultant field B = B12 B2
2 = 4 a
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E x a m p l eA piece of wire carrying a current of 6 A is bent in the form of a circular arc of radius 10.0 cm, and it subtends an angle of 120° at the centre. Find the magnetic field due to this piece of wire at the centre.
Solu ti onµ0 I ,=120°= 2
Magnetic field at centre of arc B = rad4R 3 I
0 I 0 I 7 120°B = 2 = = 4 10 6 100 T = 12.57 µT
4R 6R3 6 10Ex am p l e
An infinitely long conductor as shown in fig. carrying a current I with a semicircular loop on X-Y plane and two straight parts, one parallel to x-axis and another coinciding with Z-axis. What is the magnetic field induction at the centre C of the semi-circular loop.
Solu ti onThe magnetic field induction at C due to current through straight part of the conductor parallel to X-axis is
Z
Y
I
I C r
O X
I
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0
I I I ˆ
sin 0
0 0
B = 4 r / 2 sin = 2 r acting along + Z direction. i.e. B1= 2 r k
1 2
inThe magnetic field induction at C due to current through the semi-circular loop in X-Y plane is0 I 0I 0I ˆB
2
= 4 r / 2 = Edujournal k r
acting along + Z-direction i.e. B2 =
2 r .2rThe magnetic field induction at C due to current through the str ight part of the conductor coinciding with Z-axis is
0
I 0 I 0 I ˆ
B3 =
4 r / 2 sin sin 0 = acting along (–X)-axis i.e.
B 3 i
2 2r
2 r
BB1 B2 B3= 0 I ˆ 0 I ˆ 0 I ˆ 0 I ˆ ˆ
Total magnetic field induction at C is 2r k + 2 k –
2r i = 2r
1
k
i
Ex am p l eA conductor carrying a current i is bented as shown in figure. Find the magnitude of magnetic field at the origin
Solu ti on i ˆ 1 0 i ˆ
Field at O due to part 1 B 1 0
B 2 y
4i Field at O due to part 2
k
4 2r 1
i2 iWire 3 passes through origin when it is extended backwards B 3 0
3xO
0 i ˆ ˆ z i
B0 B1 B2 B3 = i
k
4r 2
Ex am p l e 0.03m
A conductor of length 0.04 m is tangentially connected to a circular 4A
loop of radius 0.03 m perpendicular to its plane. Find the magnetic
4m mfield induction at the centre of the loop if 4 ampere current is passed
00.
50
through the conductor as shown in fig..
0Solu ti on
Magnetic field induction at the centre of the loop due to the straight current-carrying conductor, I 4107
4
0.04
B = 0sin sin = sin 0 = 1.07 × 10–5 T
4 r 1 2
4 0.03 0.05
Magnetic fields due to the two halves of the loop are equal in magnitude and opposite in direction. So, themagnetic field induction due to the loop at the centre of the loop is zero. So, the magnetic field induction atthe centre of the loop is 1.07 × 10–5T.
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E x a m p l eFigure shows a right-angled isosceles triangle PQR having its base equal to a. A
current of I ampere is passing downwards along a thin straight wire cutting the
plane of the paper normally as shown at Q. Likewise a similar wire carries an
equal current passing normally upwards at R. Find the magnitude and direction of
the magnetic induction at P. Assume the wires to be infinitely long.
Solu ti ona
Let r = PQ = PR and a2 = r2 + r2 = 2r2 or r =
2
0 I 0 I2 IMagnetic induction at P due to conductor at Q is B = r =
0
=
1 2 2 a 2 a
Magnetic induction at P due to conductor at R is B2 = 0 I (along PQ)
2 a
Pr 90° r
Q45°
a R
(along PR)
0I 2 0I 2 0 I 0 I2Now, resultant of these two is B = B 1
2 B22 = = =
a 2 a
2 a 2a
The direction of B is towards the mid-point of the line QR.
AMPERE'S CIRCUITAL LAWAmpere's circuital law state that line integral of the magnetic field round I4 n
I3o
i
any closed path in free space or vacuum is equal to times of net W t
I5a
l
0 C ucurrent or total current which crossing through the a ea bou ded by the A c
r
I1
i
I2
c
closed path. Mathematically B . 0I I=(I1I2+I3) Positive
Negative
This law independent of size and shape f the closed path.Any current outside the closed path is n t included in writing the right hand side of lawNote :• This law suitable for infinite long and symmetrical current distribution.• Radius of cross section of thick cylinderical conductor and current density must be given to apply this law.
MAGNETOMOTIVE FORCE (M.M.F.)
Edujournal
H . d IB . d 0 I , where
B 0 H , 0 H .
d
0I
The line integral of magnetising field around any closed path is equal to net current crossing through the area
bounded by the closed path, also called 'magnetomotive force'. Magnetomotive force (M.M.F.) = H. d
APPLICATION OF AMPERE'S CIRCUITAL LAW• Magnetic field due to infinite long thin current carrying straight conductor
curr
ent
& M
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Consider a circle of radius 'r'. Let XY be the small element of length d .
because direction of along the tangent of the circle. By A.C.L.
B . d 0 I , B d cos
0
I
(where = 0°)
B d
0
(where d 2 r
)
B d cos 0
0
I I
(2r) = 0I B
0
I
2r
B and d are in the same direction
Ir O B
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• Magnetic field due to infinite long solid cylinderical conductor
• For a point inside the cylinder r < R, Current from area R2 is = I
so current from area r2 is = I (r2 ) = I r2
R 2 R2
By Ampere circuital law for circular path 1 of radius r
I r2 0 Ir
Bin (2r) = 0I' = 0 Bin = Bin rR2 2R2
• For a point on the axis of the cylinder (r = 0); B axis 0
• For a point on the surface of cylinder (r = R)
By Ampere circuital law for circular path 2 of radius R
0 I
Bs (2 R) = 0I Bs = 2 R (it is maximum)B
I
R
1 2
3
axis
Cross-sectionalview
I1
2 3
R
• For a point outside the cylinder (r > R) :-
inB
max
By Ampere circuital law for circular path 3 of radius r
(2 r) = I B
out
0 I 1
B B r = 02r outout 0 r .
Magnetic field outside the cylinderical conductor does not depend upon nature(thick/thin or solid/hollow) of the conductor as well as its radius of cross section.
• Magnetic field due to infinite long hollow cylinderical conductor
• For a point at a distance such that r < a < b B1 0
• For a point at a distance such that < r < b
r2 a2
B2(2r) =0 I' B2(2r) = 0 I b2
a 2
22Edujournal
B2 0 I r a
2 r 2
a 2
b
r
B
r < R
1B
out
r
r = R r > R
b a
1 2 3
AxisSide View
I
a 1 2 3
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ory.
p65 b
0 I Cross sectional
• For a point at a distance r such that r > b > a, B (2r) = I B
3 view
3 2r0
• For a point at the axis of cylinder r = 0 Baxis
= 0
Magnetic field at specific positions for thin hollow cylinderical conductor I
At point 1 B1 = 0 O 1 2 3
0 IR
At point 2 B2= (maximum) [outer surface] and
2R
B2= 0 (minimum) [inner surface] B
1B
0 I out rAt point 3 B3 = (for the point on axis Baxis =0)2r r
r = 0 r = R
E11
JEE-Physics
Magnetic field due to an infinite plane sheet of current
dBsindB
-z P P
dB dBcos
x dBcos
y rx x x-axis
dBsin
An infinite sheet of current lies in x-z plane, carrying current along-z axis. The field at any point Pon y is along a line parallel to x-z plane. We can take a rectangular amperian loop as shown. If youtraverse the loop in clockwise direction, inward current will be positive.By Ampere circuital law, = 0
enclosed ....(i)B.d
ina
PQRS BP Q
Let represents current per unit length. bThe current enclosed is given by
enclosed=a
x-axis.
Now, B.d B.d B.d B.d B.d b
PQRS PQ QR RS SR S RB
Now, B.d B.d 0 as B dQR SP
0
Also, B.d B.d = 2 B a as B d 2B a 0 a B 2
PQ RS
Ex am p l e Edujournal
A long straight solid conductor of radi s 5 cm carries a current of 3A, which is uniformly distributed over itscircular cross-section. Find the magnetic field induction at a distance 4 cm from the axis of the conductor.Relative permeability of the con uctor = 1000.
Solu ti onImagine a circular path of radius whose centre lies on the axis of I=3A
solid conductor such that the point P lies on it. Ir2
the current threading this closed path I'= I
r2 =
R 2 R 2 r
Magnetic field B acts tangential to the amperian circular path at P and is same inmagnitude at every point on circular path. P B
Ir 2 0 r I r R=
0
r 5cm
Using Ampere circuital law B.d
= µ0µrI' B (2r) = R 2 B = 2R2
I
B = 4 107 1000 3 0.04
= 9.6 × 10–3 T.
2 0.05 2
Ex am p l eA current I flows along a thin walled tube of radius R with a long longitudinal slit of
width b (<<R). What is the magnetic field induction at a distance r (< R) ?
Solu ti onUsing principle of superposition,field due to strip in place of slit + field due to tube with slit B = 0
so B = field due to strip in place of slit= 0 I b
(2R b)2r
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12E
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MAGNETIC FIELD DUE TO SOLENOID
It is a coil which has length and used to produce uniform magnetic field of long range. It consists a conducting
wire which is tightly wound over a cylinderical frame in the form of helix. All the adjacent turns are electricallyinsulated to each other. The magnetic field at a point on the axis of a solenoid can be obtained by superposition
of field due to large number of identical circular turns having their centres on the axis of solenoid.
Magnetic field due to a long solenoid
A solenoid is a tightly wound helical coil of wire. If length of solenoid is large, as compared to its radius,then in the central region of the solenoid, a reasonably uniform magnetic field is present. Figure shows apart of long solenoid with number of turns/length n.We can find the field by using Ampere circuital law.
Consider a rectangular loop ABCD. For this loop
B.d
0 ienc
ABCD
Now
B.d B.d B.d B.d B.d B aABCD AB BC CD DA
This is because B.d B.d 0, B d .AB CD
EdujournalAnd,
B.d 0 ( B outside the solenoid is negligibleDA
Now, ienc = n a i s B
0 n a i B 0
i
Finite length solenoid :
Its length and diameter are comparable.
2 r B
1 (uniform)E P M E'
End-I End-II
aA D
b
B C
(Magnetic field lines)
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ory.
p65 By the concept of BSL magnetic field at the axial point 'P' obtained as : B P
0 nI (cos 1 cos 2 )
2
Angle 1 and 2 both measured in same sense from the axis of the solenoid to end vectors.
Infinite length solenoid :
Its length very large as compared to its diameter i.e. ends of solenoid tends to infinity.
( a ) Magnetic field at axial point which is well inside the solenoid
1 0° and 2 180° B 0 nI [cos 0° – cos 180°]0 nI [(1) – (–1)] 0 nI
22( b ) Magnetic field at both axial end points of solenoid
1 = 90° and 2 180° B 0 nI [cos 90° – cos 180°] 0 nI [(0) – (–1)] 0 nI
2 2 2
E13
JEE-Physics
E x a m p l eThe length of solenoid is 0.1m. and its diameter is very small. A wire is wound over it in two layers. Thenumber of turns in inner layer is 50 and that of outer layer is 40. The strength of current flowing in twolayers in opposite direction is 3A. Then find magnetic induction at the middle of the solenoid.
Solu ti on
Direction of magnetic field due to both layers is opposite, as direction of current is opposite so
Bnet = B1 – B2 = 0n1I1 – 0n2I2 = 0 N1 I – 0 N2 I ( I1=I2=I)
0 I 4 10–
7
3
= (N 1 – N 2 ) = 0 1
(50 – 40) = 12 × 10–5 T
Ex am p l e
A closely wound, solenoid 80 cm. long has 5 layers of winding of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0A. Estimate the magnetic field
(a) Inside the solenoid (b) Axial end points of the solenoidSolu ti on Edujournal
(a) Magneticfield inside the solenoid
N 4 107
2000 8
B = nI = I, (N=400× 5= 2000) = =8×10–3Tin 0 0 (80
10
2)
0 nI 810 3
(b) Magnetic field at axial end points of solenoid B ends = = =4×10–3T2 2
Ex am p l e
A straight long solenoid is produced magnetic field 'B' at its centre. If it cut into two equal parts and same number
of turns wound on one part in ouble layer. Find magnetic field produced by new solenoid at its centre.Solu ti on
Magnetic field produced by long solenoid is B = 0nI, where n = N/
Same number of turns wound over half length N 0NI
Magnetic field produced by new solenoid is B' = I=2 =2B
/ 20
Ex am p l e
Find out magnetic field at axial point ‘P’ of solenoid shown in figure
(where turn density ‘n’ and current through it is I)
Solu ti on
Magnetic field at point ‘P’ due to finite length solenoid 30° 60°P
0 nIBP = [cos 1 – cos 2], where 1 = 30° (CW),2
2= (180°–60°) = 120° (CW) = 0 nI [cos 30°–cos 120°] end end
2 30° 60° 3 1
0 nI 0 nI P
= = ( 3 + 1)2 2 2 4
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ory.
p65
14E
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E x a m p l e
A uniform magnetic fieldˆ
exists in a region. A current carrying wire is placed in x-y plane as shown. B B 0 k
yB=B0k a
a a
i
a BA a
x
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Solu ti onThe conductor consists of 5 straight sections viz AC, CD, DE, EF and FB as shown.As the field is uniform, force on the sections is given by
ˆ ˆ ˆ
Bˆ
F i i B . Thus, FAC i ai B0 k 0 iajˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
F CD i a j
B0 k B 0 ia i , F DE i a i B 0 k B 0 ia j , F EF a j B 0 k B 0 ia i
ˆ Edujournalxˆ ˆ Net force, F FAC FCD FDE FEF FFB
ˆ
F FB i a j B
0
k B 0 ia j 3B 0 iaj
CURRENT CARRYING CONDUCTOR IN MAGNETIC FIELDWhen a current carrying conductor placed in mag etic field, magnetic force exerts on each free electronwhich are present inside the conductor. The resultant of these forces on all the free electrons is called magneticforce on conductor.
• Magnetic force on current elementThrough experiments Ampere established that when current element I d is placed in magnetic field B , it
experiences a magnetic force dFm I (d B )
Id (Cw)
dF B
m
(external)
• Current element in a magnetic field does not experience any force if the current in it is parallel oranti–parallel with the field = 0° or 180° dFm = 0 (min.)
• Current element in a magnetic field experiences maximum force if the current in it is perpendicular withthe field = 90° dFm = BId (max.)
• Magnetic force on current element is always perpendicular to the current element vector and magnetic
field vector. dFm I d and dFm B (always)
• Total magnetic force on straight current carrying conductor in uniform magnetic field given asf f
Fm = dFm d = I × B , Fm I(L B )i i
i I f
L L N
Where L = d , vector sum of all length elements from initial to final point, which is in accordance with thei
law of vector addition and | L | = length of the condutor.E
15
JEE-Physics
• Total magnetic force on arbitrary shape current carrying conductor in uniform magnetic field B isf
f
dF
m
I
d × B , Fm I(L B ) (L = ab)
i i I
Initial Finalpoint
point a L bf
Where L = i
d , vector sum of all length elements from initial to final point or displacement between
free ends of an arbitrary conducter from initial to final point.
GOLDEN KEY POINT• A current carrying closed loop (or coil) of any shape placed in uniform magnetic
field then no net magnetic force act on it (Torque may or may not be zero)f I
L =
i
d = 0 or
d = 0i=f
So net magnetic force acting on a current carrying closed loop Fm 0 (always)• When a current carrying closed loop (or coil) of any shape p aced non uniform magnetic field then net
.inmagnetic force is always acts on it (Torque may or m y not be zero)
Edujournal
Ex am p l eA wire bent as shown in fig carries current i and is placed in a uniform field of magnetic induction B thatemerges from the plane of the figure. Calc late the force acting on the wire.
R R R
i i
Solu ti on
The total force on the whole wire is Fm = I| L |B = I(R + 2R + R)B = 4RIB
Ex am p l e
A square of side 2.0 m is placed in a uniform magnetic field B 2.0 T in a direction perpendicular to the plane
of the square inwards. Equal current i = 3.0 A is flowing in the directions shown in figure. Find the magnitude of magnetic force on the loop.
x x x B x
C Dx x x xx x x x
x Ax x
E x
Solu ti on
Net force on the loop = 3 ( FAD ) Force on wire ACD = Force on AD = Force on AED
Fnet = 3(i) (AD) (B) = (3) (3.0) (2 2 )(2.0) N = 36 2 N. Direction of this force is towards EC.
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16E
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E x a m p l eA metal rod of mass 10 gm and length 25 cm is suspended on two springs as shown in figure. The springsare extended by 4 cm. When a 20 ampere current passes through the rod it rises by 1 cm. Determine the
magnetic field assuming acceleration due to gravity to be 10 m/s2.
x kx x
kF=bil x
T Tx x
Mg I x
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Sol . Let tension in each spring is = T0
Initially the rod will be in equilibrium if 2T0 = Mg then T0 = kx0 ...(i)Now when the current I is passed through the rod it will experience a forceF = BIL vertically up; so in this situation for its equilibrium,
2T + BIL = Mg with T = kx...(ii) (x = 4 – 1 = 3cm)So from eq. (i) and eq.(ii) T Mg BIL x 1 BIL in
2 2
T Mg x 0 Mg0
.Mg x 0 x 10 103 10 310
2
s
B =Edujournal
ILx0 = 20 25 10 4 10
= 1.5 × 10–2T
Ex am p l e
Two conducting rails are connected to source of e.m.f. a d form an incline as shown in fig. A bar of mass 50g slides without friction down the incline through a ve tical magnetic field B. If the length of the bar is 50 cm anda current of 2.5 A is provided by the battery, for what value of B will the bar slide at a constant velocity ? [g =
10 m/s2]I v I
= Bco
ns
tF
mc RSolu ti on o m
gForce on current carrying wire F = BIL Fm
sin
The rod will move down the plane with constant velocity only if mg
F cos = mg sin BIL cos = mg sin
mg 50 103 10 3
or, B = tan = =0.3TIL 2.5 50 102 4
Ex am p l eA wire PQ of mass 10g is at rest on two parallel metal rails. The separation between the rails is 4.9 cm. Amagnetic field of 0.80 tesla is applied perpendicular to the plane of the rails, directed downwards. Theresistance of the circuit is slowly decreased. When the resistance decreases to below 20 ohm, the wirePQ begins to slide on the rails. Calculate the coefficient of friction between the wire and the rails.
P
6V
4.9cm
QE
17
JEE-Physics
S o l u t i o nWire PQ begins to slide when magnetic force is just equal to the force of friction, i.e.,µ mg = i B sin ( = 90°) so i = E 6 i B 0.3 4.9 102 0.8
= =0.3A so µ = = = 0.12
R 20 mg 10 103 9.8 Ex am p l e
A wire abcdef with each side of length ' ' bent as shown in figure and carrying a current I is placed in a uniform magnetic field B parallel to +y direction.What is the force experienced by the wire.
Zc
b a
dY
e f B
X
Solu ti on
Magnetic force on wire abcdef in uniform magnetic field is Fm = I (LB) ,
L is displacement between free ends of the conductor from n t al to
. ˆ
ˆ Edujournal 10
2I 1
ˆ
I
L B = BIL
ˆ ˆ (k )
= BI , along +z directionfinal point. L = ( ) i and B =(B) j ; Fm = ( i j)in=
BIMAGNETIC FORCE BETWEEN TWO PAR ALLEL CURRENT CARRYING CONDUCTORS
Like currents u like currents
1 2 1 2
I1
I1 Repulsion
B2 B1 dF12 B2 B1 dF21
dF12
dF21
dThe net magnetic force acts on current carrying conductor due to its own field is zero. So consider two infinite
long parallel conductors separated by distence 'd' carrying currents I and I .
Magnetic field at each point on conductor (ii) due to current I1 is B1
= [uniform field for conductor (2)]2 d
Magnetic field at each point on conductor (i) due to curent I2 is B2 = 0 I2 [Uniform field for conductor (1)]
2 d
consider a small element of length 'd ' on each conductor. These elements are right angle to the external magnetic field, so magnetic force experienced by elements of each conductor given as
0 I 2 B2)dF12 = B2 I1 d = I1 d ...(i) (Where I1 d
2 d I 1
dF21 = B1 I2 d = 0
I2 d ... (ii) (Where I2 d B1) 2 d
Where dF12 is magnetic force on element of conductor (i), due field of conductor (i) and dF21 is magnetic force on element of conductor (ii), due to field of conductor (i).Magnetic force per unit length of each conductor is dF12 = dF21 =
0 I1
I2
d d 2 df =
0 I1 I2 N/m (in S.I.) f = 2 I1I2 dyne/cm (In C.G.S.)
2 d d
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p65
18E
JEE-PhysicsDefinition of ampere : 1 2
Magnetic force/unit length for both infinite length conductor gives as
0
I1
I2
(4 107 ) (1) (1) 1A 1A
f = 2 d
= 2 (1) = 2 × 10–7 N/m
'Ampere' is the current which, when passed through each of two parallelinfinite long straight conductors placed in free space at a distance of 1 mfrom each other, produces between them a force of 2 × 10–7 N/m
• Force scale f= 0 I1 I2 is applicable when at least one conductor must be of infinite
2 dlength so it behaves like source of uniform magnetic field for other conductor.
I I 2
Magnetic force on conductor 'LN' is FLN = f × FLN 0 1
2 d
1m
L
I1
I2
d
(source) N
• Equillibrium of free wireCase I : Upper wire is free : Consider a long horizontal wire which is rigidly fixed another wire is placeddirectly above and parallel to fixed wire.
(free) I2 fm Finite length (m,)(Stable gEquilibrium)
h
Edujournal Infinite length
– (fixed) I1 +(Source)
Magnetic force per unit length of free wire f = 0 I1 I2 , and it is repulsive in nature because currents are unlike.
2hm
Free wire may remains suspended if the magnetic force per unit length is equal to weight of its unit lengthAt balanced condition f = W'. Weight per unit length of free wire =
0 I1 I2m g (stable equillibrium condition)
2hm
If free wire is slightly pluked and released then it will executes S.H.M. in vertical plane.
The time period of motion is T 2 hg
Case II : Lower wire is free : Consider long horizontal wire which is rigidly fixed. Another wire is placed
directly below and parallel to the fixed wire.
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– I1
(Source) Infinite length
+(fixed)
d (depht)
(Unstable (free) I2
fm Finite length (m,)
Equilibrium) g
Magnetic force per unit length of free wire is f = 0I1I2 , and it is attractive in nature because currents are like.
m 2d
Free wire may remains suspended if the magnetic force per unit length is equal to weight of its unit lengthAt balanced condition fm = W'Weight per unit length of free wire 0 I1 I2 m g (unstable equillibrium condition)
2d
E19
JEE-Physics
E x a m p l eTwo horizontal parallel straight conductors, each 20 cm long, are arranged one vertically above the otherand carry equal currents in opposite directions. The lower conductor is fixed while the other is free tomove in guides remaining parallel to the lower. If the upper conductor weights 1.20 g, what is theapproximate current that will maintain the conductors at a distance 0.75 cm apart.
Solu ti onIn equilibrium magnetic force Fm will balance weight mg. P Fm Q
i
So mg = Fm mg = 0 i2 mg
2 d d
2mgd
2 1.2 103 9.8 0.75 i = = R i S=
2205 = 47 A0 4 107 20 102
MAGNETIC DIPOLE MOMENTA magnetic dipole consists of a pair of magnetic poles of equal and opposite strength separated by small distance. Ex. Magnetic needle, bar magnet, solenoid, coil or loop.
• Magnetic moment of Bar magnet
Neutral point AMagnetic -m +m
axis
Edujournal
S N Cross sectional
0 view
The magnetic moment of a bar magnet is defined as a vector quantity having magnitude equal to the product of pole
strength (m) with effective length ( ) and directed along the axis of the magnet from south pole to north pole.
It is an axial vectorM = m
S.I. unit : - A.m2
GOLDEN KEY POINTS
• Attractive property : A bar magnet attracts certain magnetic substances (eg. Iron dust). The attractingpower of the bar magnet is maximum at two points near the ends called poles. So the attracting powerof a bar magnet at its poles called 'pole strength'
• The 'pole strength' of north and south pole of bar magnet is conventionally represented by +m and –mrespectively.
• The 'pole strength' is a scalar quantity with S.I. unit A–m.
• The 'pole strength' of bar magnet is directly proportional to its area of cross section. m A
• The attracting power of a bar magnet at its centre point is zero, so it is called 'neutral point'.
• Magnetic poles are always exists in pairs i.e. mono pole does not exist in magnetism. So Gauss law in
magnetism given as B ds 0
• Effective length or magnetic length :– It is distance between two poles along the axis of a bar magnet.
As pole are not exactly at the ends, the effective length ( ) is less than the geometrical length ( 0) of the
bar magnet. ~091 0
• Inverse square law (Coulomb law) : The magnetic force between two isolated magnetic poles of strength
m1 and m2 lying at a distance 'r' is directly proportional to the product of pole strength and inverselyproportional to the square of distance between their centres. The magnetic force between the poles canbe attractive or repulsive according to the nature of the poles.
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20E
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Fm mm12 mm µ0 (S.I.)Fm=k 4
F 1 1 2 where k2
r 1 (C.G.S.)m
r2
Inverse square law of Coulomb in magnetism is applicable only for two long bar magents becasue isolated poles cannot exist.
• If a magnet is cut into two equal parts along the length then pole strength is reduced to half and
m M
length remains unchanged. New magnetic dipole moment M'=m'( )= 2 2 .
The new magnetic dipole moment of each part becomes half of original value.m' = (m/2), × '= m' = m '= /2
-m +m S NS N S NS M N
S N MM'= m×
M = m × M' = (m/2) × '= M/22=
2
• If a magnet is cut into two equal parts transverse to the length then pole strength remains unchanged
• The magnetic dipole moment of a magnet is equal to product of pole strength and distance betweenand length is reduced to half. New magnetic dipole moment M '
m M .
2 2
The new magnetic dipole moment of each part becomes half of original value.
Edujournal 1 2
poles. M=m
N S
N SN S
• As magnetic moment is vector, in case of two magnets having magnetic moments M and M with
angle between them, the resulting magnetic moment.
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ory.
p65
Exam p le
The forcebetween two magnetic polesin air is 9.604 mN. If one pole is 10 timesstronger than the other, calculate the pole strength of each if distance between two polesis 0.1m?
Solu ti on
Ex am p l e
E21
JEE-Physics
S o l u t i o n MIf m is the pole strength then M=m.L m
L r LIf it is bent into a semicircular arc then L= r
So new magnetic moment M ' m 2r
M 2
L 2M
L
Ex am p l eTwo identical bar magnets each of lengthother with the north pole of one touchingthe system.
L and pole strength m are placed at right angles to each the south pole of other. Evaluate the magnetic moment of
Solu ti onS M2
2 2
M1 =M2 = mL M R M 1 M 2 2M1M2 cos 2 m L M1
2 MR
N S NM sin 90 M1
and tan
1i.e. = tan–11 = 45° M2
M M cos 90MAGNETIC MOMENT OF CURRENT CARRYING COIL ( LOOP)
Current carrying coil (or loop) behaves like magnetic dipole. The face of coil in which current appears to flowEdujournal
anti clock wise acts as north pole while face of coil in which current appears to flow clock wise acts as southpole.• A loop of geometrical area 'A', carries a current 'I' then magentic moment of coil M= IA
• A coil of turns 'N', geometrical area 'A', carries current 'I' then magentic moment M = N I A
Magnetic moment of current carrying coil is an axial vector M NIA where
A is a area vector perpendicu-
lar to the plane of the coil and along its axis. SI UNIT : A-m2 or J/TACW CW
M M
M
Direction of M find out by right hand thumb rule
• Curling fingers In the direction of current • Thumb Gives the direction of MFor a current carrying coil, its magnetic moment and magnetic field vectors both are parallel axial vectors.
Ex am p l eFind the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i = 2.0 A
C D
B E
A F
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S o l u t i o nBy assuming two equal and opposite currents in BE, two current carrying loops (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other.
C D
B E
A F
Hence, Mnet = M 2 M2 = 2M where M = iA = (2.0)(0.1)(0.1) = 0.02 A-m2
Mnet = ( 2 )(0.02) A-m2 = 0.028 A-m2
Ex am p l e
The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current I as shown in fig. What is the magnetic induction at the centre O and magnetic moment of the loop in cases (A) and(B) ?
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(A) (B) I
R2 R2 ni.
R O
S R R1I Q P1
E duj o ur n alS R O Q P
Solu ti on
As the point O is along the length of the straight wires, so the field at O due to them will be zero and hence.0 I I 0
1 1 1 2 1 2 1 2 2
B I R 2 R 1 I R R (A) i.e., B & M NIS =1× I = 2
4 R1 4 2 2 2 1
R
2
R
1
R 2
0
1 1 1 2
2 (B) Following as in case (A), in this situati n, B I and, M I 2 R
4 R
1R
2 2
MAGNETIC DIPOLE IN MAGNETIC FIELDTorque on magnetic dipole
N
Fm=mB CW+m
B=(uniform)
CWS
Fm=mB-m
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= force × perpendicular distance between force couple = (mB) ( sin) , where M = m
= MBsin =90° = MB (maximum)
Vector form MB
= 0° or 180° = 0 (minimum)
( b ) Coil or Loop
MB NIA B = BINAsin =90° = BINA (maximum)
= 0° or 180° = 0 (minimum)
IM CW
O
B(uniform)
Plane of coil
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GOLDEN KEY POINTS
• Torque on dipole is an axial vector and it is directed along axis of rotation of dipole.
• Tendency of torque on dipole is try to align the M in the direction of B or tries to makes the axis of dipole
parallel to B or makes the plane of coil (or loop) perpendicular to B .
Fnet=0 (no translatory motion)• Dipole in uniform magnetic field
may or may not be zero (decides by )
Fnet 0 (translatory motion)• Dipole in non uniform magnetic field
may or may not be zero (decides by )
• When a current carrying coil (or loop) is placed in longitudenal magnetic field then maxmium torque acts on it.
=90° M B = MB = BINAmax
• When a current carrying coil (or loop) is placed in transverse magnetic field the no torque acts on it.
= 0° M
B or = 180° Manti B min = 0
• Work done to rotate a dipoleEdujournalinmagneticfieldisstoredintheform of potential energy of magnetic dipole.WORK DONE IN ROTATING A MAGNETIC DIPOLE
Work done in rotating a dipole in a uniform magnetic field through small angle 'd'
dW = .d = MBsind
So work done in rotating a dipole from angular position 1 to 2 with respect to the Magnetic field direction
W = 2 dW = 2 MB sin d = MB(c s 1 c s 2 )1 1
• If magnetic dipole is rotated from field direction i.e. 1 = 0° to position 2=
then work done is W = MB (1 – cos) = 2MB sin2 /2
in one rotation = 0° or 360° W = 0 in 1/4 rotation = 90° W=MB
in half rotation = 180° W = 2MB in 3/4 rotation = 270° W=MB
POTENTIAL ENERGY OF MAGNETIC DIPOLEThe potential energy of dipole defined as work done in rotating the dipole from a direction perpendicular to
the given direction. U = W – W90° U = MB (1 – cos) – MB = MB cos In vector form U M . B
GOLDEN KEY POINTS
• When M and B are parallel ( = 0°), the dipole has minimum potential energy and it is in stable equillibrium.
U MB (minimum)
• When M and B are anti parallel ( = 180°), the dipole has maximum potential energy and it is in unstable
equillibrium. (maximum)
• When M and B are prpendicular to each other ( = 90°), the dipole has potential energy U=0 and in this situation maximum torque acts on it hence no equillibrium.
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E x a m p l eA circular coil of 25 turns and radius 6.0 cm, carrying a current of 10 A, is suspended vertically ina uniform magnetic field of magnitude 1.2 T. The field lines run horizontally in the plane of the coil.Calculate the force and torque on coil due to the magnetic field. In which direction should a balancingtorque be applied to prevent the coil from turning ?
Solu ti on
Magnetic force Fm= I d B
For coil or close loop d 0 so Fm 0
IThe torque on a coil of any shape having N turns and B
I
current in a magnetic field B is given by = NIABsin = 25 ×10××6 × 6 × 10–4 × 1.2 × sin90° = 3.39 NThe direction of is vertically upwards. To prevent the coil from turning,an equal and opposite torque must be applied.
Ex am p l eA uniform magnetic field of 5000 gauss is established along the positive z-direction. A rectangular loopof side 20 cm and 5 cm carries a current of 10 A is suspended in this magnetic field. What is thetorque on the loop in the different cases shown in the followi g figures ? What is the force in eachcase ? Which case corresponds to stable equilibrium ?
Z Z Z
I
B B BY I Y Y
O O I O
X (a) X (b) X (c)
Z Z Z
B B BO O Y O Y
Y30°I
X (d) (e) (f)
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(a) Torque on loop, =BIA sin
Here, =90°; B=5000 gauss=5000× 10–4 tesla = 0.5 tesla = 10 ampere, A=20 × 5cm2=100 × 10–4=10–2 m2
Now, = 0.5 × 10× 10–2=5 × 10–2 Nm It is directed along –y-axis(b) Same as (a).(c) =5 × 10–2 Nm along –x-direction(d) =5×10–2 N m at an angle of 240° with +x direction.(e) is zero. [ Angle between plane of loop and direction of magnetic field is 90°](f) is zero.Resultant force is zero in each case. Case (e) corresponds to stable equilibrium.
Ex am p l eA circular coil of 100 turns and having a radius of 0.05 m carries a current of 0.1 A. Calculate thework required to turn the coil in an external field of 1.5 T through 180° about an axis perpendicularto the magnetic field ? The plane of coil is initially at right angles to magnetic field.
Solu ti onWork done W = MB (cos1–cos2) =NAB (cos1–cos2) W = Nr2B (cos1– cos2) = 100 × 0.1 × 3.14 × (0.05)2 × 1.5 (cos 0° –cos) = 0.2355J
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E x a m p l eA bar magnet of magnetic moment 1.5 JT–1 lies aligned with the direction of a uniform magnetic field of 0.22 T.(a) What is the amount of work required to turn the magnet so as to align its magnetic moment.
(i) Normal to the field direction? (ii) Opposite to the field direction?(b) What is the torque on the magnet in case (i) and (ii)?
Solu ti onHere, M = 1.5 JT–1, B = 0.22 T.(a) P.E. with magnetic moment aligned to field = – MB
P.E. with magnetic moment normal to field = 0P.E. with magnetic moment antiparallel to field = + MB
(i) Work done = increase in P.E. = 0 – (–MB) = MB = 1.5 × 0.22 = 0.33 J.(ii) Work done = increase in P.E. = MB – (–MB) = 2MB = 2 × 1.5 × 0.22 = 0.66 J.b) We have = MBsin(i) = MB sin = 1.5 × 0.22 × 1 = 0.33 J. (=90° sin=1) This torque
will tend to align M with B.(ii) = MB sin = 1.5 × 0.22 × 0=0 ( =180° sin=0)
Ex am p l eA short bar magnet of magnetic moment 0.32 J/T is placed in uniform fieldof 0.15 T. If the baris free to rotate in plane of field then which orientation would correspo d to its (i) stable and (ii) unstableequilibrium? What is potential energy of magnet in each case?
Solu ti on Orbital current :-Edujournal I(current)(a)
Umin = – MB(i) If M is parallel to B then =0°. So potential energy U =U
min = –MB = –0.32 × 0.15 J= –4.8 × 10–2 J (stable equilibrium)(ii) If M is antiparallel to B then = ° So potenti l energy
An atomic orbital electron, which doing bo nded niform circular motion around nucleus. A current constitueswith this orbital motion and hence orbit behaves like current carrying loop. Due to this magnetism produces atnucleus position. This phenomenon called as 'atomic magnetism.Bohr's postulates : v
mv2 kze2 h ACW(i) = (ii) L = mvr = , where n = 1, 2,3 .......
r r2 +Ze 2
Basic elements of atomic magnetism : r e–
I ef e ev e Fe
T 22r(b) Magnetic induction at nucleus position :- As circular orbit behaves like current
carrying loop, so magnetic induction at nucleus position BN =
0 I 2r
B
0
ef
0
e
0 ev 0 e
N 2r 2Tr 4 r2 4r(c) Magnetic moment of circular orbit :- Magnetic dipole moment of circular orbit
M = IA where A is area of circular orbit. M ef r 2
er2 evr er2
T2 2
• Relation between magnetic moment and angular momentum of orbital electron
Magnetic moment M = evr m eL ( angular momentum L = mvr) I(current)
2m2 mr
Vector form M eL
2m eFor orbital electron its M and L both are antiparallel axial vectors. M
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BOHR MAGNETON (B)
According of Bohr's theory, angular momentum of orbital electron is given byL = nh , where n = 1, 2, 3 ........ and h is plank's constant.
2 eheL
Magnetic moment of orbital electron is given by M = = n2m 4 m
eh
• If n = 1 then M = 4m , which is Bohr magneton denoted by B
• Definition of B :
Bohr magneton can be defined as the magnetic moment of orbital electron which revolves in first orbit of an atom.
= eh 16. 1019
6.6 1034
• = = 0.923 × 10–23 A.m.2 3.14 91. 10 314m 4B
• Basic elements of atomic magnetism for first orbit of H-atom (n=1, z = 1)
(a) Accurate form :- (v = 2.18 × 106 m/sec, f = 6.6 × 1015 cy/sec. r = 0.529Å)• Orbital current I = 0.96 mA• Magnetic induction at nucleus position BN = 12.8 T
• Magnetic moment of orbital electron M = 0.923 × 10–23 A.m2
Edujournal
(b)Simple form :- (v 2 × 106 m/sec, f 6 × 1015cy/sec, 0 5Å)• Orbital current I 1mA• Magnetic induction at nucleus position BN 4T
• Magnetic moment of orbital electron M = µB
A.m2
A NONCONDUCTING CHARGED BODY IS ROTATED WITH SOME ANGULAR SPEED.
In this case the ratio of magnetic moment and angular momentum is constant which is equal to q2m
here q = charge and m = the mass of the body.
Ex am p l e
In case of a ring, of mass m, ra ius R and charge q distributed on it circumference.
Angular momentum L = I = (mR2)() ... (i)
Magnetic moment M = iA = (qf) (R2)F I R 2 q
M = (q) G J (R2) = q ...(ii)++
+++ + ++
++ ++H 2 K 2 R
+ ++ +
M q ++++ ++ +++
++
f = 2From Eqs. (i) and (ii) L 2m
Although this expression is derived for simple case of a ring, it holds good for other bodies also. For example, fora disc or a sphere. M = qL M
qbIg , where L = I
2m
2m
Rigid body R i n g D i s c Solid sphere Spherical shell
Moment of inertia (I) mR 2 mR2 2 mR2 2 mR2
5 32Magnetic moment= q I
q R2
q R 2
qR 2
qR 2
2 m 2 4 5 3E
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FORCE ON A CHARGED PARTICLE IN A MAGNETIC FIELDForce experienced by
a current element din magnetic field B is given by dF = Id × B ....(i)
Now if the current element d is due to the motion of charge particles, each particle having a chargeq moving with velocity v through a cross-section A, [with volume dV=A d ]
d = nqAv d nqdV v
From eqn (i) we can writedF =ndV q (v B)
ndV=the total number of charged particles in volume dV
1 d F
(n=number of charged particles per unit volume), force on a charged particle F = =q ( v B )n dV
GOLDEN KEY POINT• The force
F isalways perpendicular to both the velocity v and the field B
• A charged particle at rest in a steady magnetic field does not experience any force.If the charged particle is at rest then so
v 0 , v B
0• A moving charged particle does not experience any force in a magnetic field if its motion is parallel
or antiparallel to the field.
q q=0° F Fv B
B B
v q O O 90° vv
=180° q qF=0 F= vBsin Fmax=qvB(A) (B) (C)
• If the particle ismoving perpendicular to the field. In this situation all the threevectors F ,
v and
B are mutuallyperpendicular to each ther. Then sin = max = 1, i.e., = 90°,
The force will be maximum Fmax = q v B
• Work done by force due to magnetic field in motion of a charged particle is always zero.When a charged particle move in magnetic field, then force acts on it is always perpendicular todisplacement, so the work done,W = cos 90 0 (as = 90°),
F.ds Fds
Edujournal 1 2
And as by work-energy theorem W = KE, the kinetic energy mv remains unchanged and2
hence speed of charged particle v remains constant. v ofHowever, in this situation the force changes the direction of motion, so the direction of velocity the
charged particle changes continuously.
MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD
Motion of a charged particle when it is moving collinear with the field magnetic field is not affected by the field (i.e. if motion is just along or opposite to magnetic field) ( F = 0) Only the following two cases are possible :• Case :
When the charged particle is moving perpendicular to the field.
The angle between B and v is =90°. So the force will be maximum (= qvB) and always perpendicular tomotion (and also field); Hence the charged particle will move along a circular path (with its plane
perpendicular to the field). Centripetal force is provided by the force qvB, So mv2
qvB r mv
qBrv qB
Angular frequency of circular motion, called cyclotron or gyro-frequency. = r m
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and the time period, T= 2= 2
m i.e., time period (or frequency) is independent of speed of particle
qBand radius of the orbit. Time period depends only on the field B and the nature of the particle,i.e., specific charge (q/m) of the particle.This principle has been used in a large number of devices such as cyclotron (a particle accelerator),
bubble-chamber (a particle detector) or mass-spectrometer etc.
• Case :
The charged particle is moving at an angle to the field : ( 0°, 90° or 180°). Resolving the velocity ofthe particle along and perpendicular to the field.The particle moves with constant velocity v cos alongthe field ( no force acts on a charged particle when it moves parallel to the field).And at the same time it is also moving with velocity v sin perpendicular to the field due to whichit will describe a circle (in a plane perpendicular to the field)
m (v sin ) 2 r 2 m
Radius of the circular path r = and Time period T= =v sin qBqBSo the resultant path will be a helix with its axis parallel to the field B as shown in fig.
v sin Helical Path
B r
v cos p (vcos)= 2 m (v cos )
The pitch p of the helix = linear distance t avelled in one rotation p=T
qB
Ex am p l e
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV enters a
region with uniform magnetic field of 0.15 T. Determine the radius of the trajectory of the electron if the field is –
(a) Transverse to its initial velocity(b) Makes an angle of 30° with the initial velocity [Given : me = 9 × 10–31 kg]
Solu ti on
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1 mv2 = eV v = 2eV 2 1.6 1019 2 103 8 7
= = × 10 m/s2 9 10
31 3mb g
10(a) Radius r = mv = 91031 8 3 7= 10–3 m = 1mm
1 qB 1.6 1019 0.15
(b) Radius r2 = mv sin = r1sin =1× sin30° =1 × 1 = 0.5 mm
qB 2
MOTION OF CHARGED PARTICLE IN COMBINED ELECTRIC AND MAGNETIC FIELDS
Let a moving charged particle is subjected simultaneously to both electric field E and magnetic field B .
force Fe qE
The moving charged particle will experience electric and magnetic force Fm
=q ( v B )
.
Net force on the charge particle "Lorentz- force"F q(E v
B)Depending on the direction of v,E and B various situation are possible and the motion in general
is quite complex.Case :
v ,E and B all the three are collinear :
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JEE-Physics
Ev v'
q q
BAs the particle is moving parallel or antiparallel to the field. The magnetic force on it will be zero
F qE
and only electric force will act So, acceleration of the particle a m m
Hence, the particle will pass through the field following a straight line path (parallel to the field) withchange in its speed.
In this situation speed, velocity, momentum and kinetic energy all will change without change in directionof motion as shown in figure above.Case : v , E and B are mutually perpendicular :
E
q v Fe q v
Fm
B
If in this situation Edujournal re such that
direction and magnitude of E nd BResultant force F F
e
Fm 0 F 0
mThen as shown in fig., the particle will pass throughthe field with same velocity Fe=Fm i.e. qE=qvB v E
BEx am p l e
A beam of protons is deflected sideways. Could this deflection be caused by
(i) a magnetic field (ii) an electric field? If either possible, what would be the difference?
Solu ti onYes, the moving charged particle (e.g. proton, -particles etc.) may be deflected sideway either by anelectric or by a magnetic field.(i) The force exerted by a magnetic field on the moving charged particle is always perpendicular
to direction of motion, so that no work is done on the particle by this magnetic force. Thatis the magnetic field simply deflects the particle and does not increase its kinetic energy.
(ii) The force exerted by electric field on the charged particle at rest or in motion is always along
the direction of field and the kinetic energy of the particle changes.
Ex am p l eA neutron, a proton, an electron an -particle enter a region of constant magnetic field with equalvelocities. The magnetic field is along the inwards normal to the plane of the paper. The tracks of theparticles are shown in fig. Relate the tracks to the particles.
C× × × ×
B×× × ×
A × × × ×
× × × × D
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JEE-PhysicsS o l . Force on a charged particle in magnetic field
F =q(v B)
For neutron q=0, F=0 hence it will pass undeflected i.e., tracks C corresponds to neutron.
If the particle is negatively charged, i.e. electron. F e(v B)
It will experience a force to the right; so track D corresponds to electron.
If the charge on particle is positive. It will experience a force to the left; so both tracks A and Bcorresponds to positively charged particles (i.e., protons and -particles). When motion of chargedparticle perpendicular to the magnetic field the path is a circle with radius
mv r m m r= i.e. and as
qB q q
4m while
2e
m m
e q p
m m
q q p
So r>rp track B to -particle and A corresponds to proton.
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Ex am p l eAn electron does not suffer any deflection while passing through a region. Are you surethat there
is no magnetic field? Is the reverse definite?Solu ti on
If electron passing through a certain region does not suffer any deflection, thenwe are not sure that
there is no magnetic field in that region. This is due to that electron suffers no force when it movesparallel or antiparallel to magnetic field. Thus the m gnetic field may exist parallel or antiparallel tothe direction of motion of electron.The reverse is not true since an electron can also be deflected by the electric field.
Ex am p l e
In a chamber, a uniform magnetic field of 8 × 10–4 T is maintained. An electron with a speed of
4.0 × 106m/s enters the chamber in direction normal to the field.(a) Describe the path of the electr n.
(b) What is the frequency of revol tion of the electron?
(c) What happens to the path of the electron if it progressively loses its energy due to collisions
with the atoms or molecules of theenvironment ?Solu ti on Edujournal
9.1 10
31 4
10
6
mv
(a) The path of the electron is a circle of radius r= = = 2.8 × 10–2mBe 1.6 1019
8 104
The sense of rotation of the electron in its orbit can be determined from the direction of the centripetal
force. F = – e ( v
× B ) . So, if we look along the direction of B , the electron revolves clockwise.
(b) the frequency of revolution of the electron in its circular orbit
eb1.6 10
19 8.0 10
4
f = = Hz = 22.4 MHz2 m 2 9.1 10
31
(c) Due to collision with the atomic consistent of the environment, the electron progressively loses
its speed. If the velocity vector of the electron remains in the same plane of the initial circularorbit after collisions, the radius of the circular orbit will decrease in proportion to the decreasingspeed. However, in general, the velocity of the electron will not remain in the plane of the initialorbit after collision. In that case, the component of velocity normal to B will determinethe radiusof the orbit, while the component of velocity parallel to B remains constant. Thus, the path
of the electron, between two collisions is, in general, helical. But an important fact must be noted:the frequency of orbital revolution remains the same, whatever be the speed of the electron.
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E x a m p l e
A beam of protons with velocity 4 × 105 m/s enters a uniform magnetic field of 0.3 tesla at an angleof 60° to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of helix. Mass of proton=1.67× 10–27kg.
Solu ti on
v sin Radius of helix r= (component of velocity to field isvsin)qB
vsin
(1.67 1027
)(4 105 )
3 2= 2 = × 10–2m = 1.2cm r
3(1.6 10
19 )0.3 B
2r vcos
p=vcos ×Again, pitch T (where T= ) p
v sin
p= v cos 2r cos 60 2 (1.2
102 )
= = 4.35× 10–2m=4.35cmv sin sin 60
Ex am p l e
The region betwen x = 0 and x = L is filled with uniform, steady magnetic field ˆ . A particleB 0 k
of mass m, positive charge q and velocity ˆ travels along X-axis and enters the region of magneticv 0 i
field. Neglect the gravity throughout the question.(a) Find the value of L if the particle emerges from the region of magnetic field with its final velocity
at an angle 30° to its initial vel city.
(b) Find the final velocity of the particle and the time spent by it in the magnetic field, if the magnetic field now extends upto 2.1 L.
Solu ti on
(a) The particle is moving with velocity v 0ˆi , perpendicular to
magnetic field B 0 kˆ . Hence the particle will move along x=0
x=L
mv0a circular arc OA of radius r =
qB0
D
A30°
Let the particle leave the magnetic field at A.
AD L rm v0
From CDA, sin60°= CA r L = rsin30°= 2 L = 2qB0
r
30°
C
(b) As the magnetic field extends upto 2.1 L i.e., L > 2r,
so the particle completes half cycle before leaving the magnetic
field, as shown in figure.
The magnetic field is always perpendicular to velocity vector,
therefore the magnitude of velocity will remain the same.
r
Final velocity = v0 ( ˆi) =– v 0
ˆiTime spent in magnetic field= v0
yv0i
mv
0
i
= qB0
N
OD
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32E
JEE-Physics
E x a m p l eA uniform magnetic field with a slit system as shown in fig. is to be used as a momentum filter for highenergy charged particles. With a field of B tesla it is found that the filter transmits -particle each ofenergy 5.3 MeV. The magnetic field is increased to 2.3 B tesla and deuterons are passed into the filter.What is the energy of each deuteron transmitted by the filter ?
Source Detector
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Solu ti onIn case of circular motion of a charged particle in a magnetic fieldEk =
r2 q 2 B 2
2mb g b g b b2.3g
2 B 2
g (E k )D 4
So according to the given problem (Ek) = r2 2e and (Ek)D = r 2 e 2 2.3B 2 i.e. 2
2 b 4m g 2 2m g2
(E k ) 2 2b
Edujournal (Ek)D = (5.3) × 5.29 = 14.02 MeV
2
Ex am p l eA charged sphere of mass m and charge q starts slidi g from rest on a vertical fixed circular track of radius R
from the position as shown in figure . There existsniform and constant horizontal magnetic field of inductionB. Find the maximum force exerted by the track on the sphere.
q m
B
R O
Sol . Magnetic force on sphere Fm = qvB (directed radially outward)
mv 2
mv 2 N R
Fm
N – mg sin – qvB = N = + mg sin + qvB mgsin
R mg
RHence, at = /2 we get Nmax 2mgR
= + mg + qB 2gR = 3mg + qB 2gRR
Ex am p l eAn electron gun G emits electrons of energy 2keV travelling in the positive x-direction. The electrons arerequired to hit the spot S, where GS = 0.1m, and the line GS makes an angle of 60° with the x-axis, asshown in the figure. A uniform magnetic field B parallel to GS exists in the region outside the electron gun.Find the minimum value of B needed to make the electron hit S.
B S
60° x
GE
33
JEE-Physics
S o l . The velocity of the electrons emitted by electron gun along x-axis, is v = 2E K E = 1 mv2
m 2
The velocity of the electron can be resolved into two components v cos and v sin , parallel and
perpendicular to the magnetic field respectively. Due to component v cos electron will move in the direction
of magnetic field with constant speed v cos but due to component v sin , it will move on a circular path in
the plane to magnetic field. Hence electron will move on a spiral path. As electrons are required to hit the spot
S, hence distance travelled by electron in one time period along the direction of magnetic field must be just equal
to GS. The electrons may also hit the spot S after two or more time periods but minimum value of B is required
As per above discussion,G S = Distance travelled along the direction of magnetic field in one time period
2m 2m 1
2E K 2E
q
bGSg= (v cos ) × T = v cos × qB = × cos × qB B = × cos × 2m ×m m
cos 60 2 3.14 91. 1031
= 2 2 1.6
1016
× = 4.68 × 10–3 T
91. 10
31 1.6 1019 0.1
MOVING COIL GALVANOMETER
A galvanometer is used to detect the current and h s moderate resistance.EdujournalNiS
Principle. When a current carrying coil is placed in magnetic field, it experiences a torque given
by = NiAB sin where is the angle between normal to plane of coil and direction of magnetic
field. In actual arrangement the coil is suspended between the cylindrical pole pieces of a strong magnet.
The cylindrical pole pieces give the field radial such that sin =1 (always). So torque =NiAB If C is torsional rigidity
(i.e., restoring couple per unit twist of the suspension wire), then for deflection of coil =C. In equilibrium we
have external couple = Restoring couple i.e.C=NiAB or
NAB i i.e., iC
In words the deflection produced is directly proportional to current in the coil.
NABThe quantity = is called the current sensitivity of the galvanometer. Obviously for greater
i C
sensitivity of galvanometer the number of turns N, area of coil A and magnetic field B produced by pole
pieces should be larger and torsional rigidity C should be smaller. That is why the suspension wire is used
of
phos
phor
bron
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for
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torsi
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rigidi
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is
smal
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CON VERSION OF GALVANOMETER INTO A MMETER
An ammeter is a low resistance galvanometer; used to measure current directly in amperes and is always
connected in series with the circuit. To convert a galvanometer into ammeter, a low resistance, called
shunt, is connected in parallel to the galvanometer as shown in figure.
S
i a ig G b i
Let ig be the current in galvanometer for its full scale deflection and G the resistance of galvanometer.Let i is the range of ammeter and is the current in shunt S. Then potential difference across a and
b is
Vab = ig G = iSS. ...(i)
At junction a, i=iS + ig i.e., iS=i–igTherefore from (i) igG = (i–ig)S or ig(S+G) = iS i.e., ig = S ...(ii)
S G
This is the working equation for conversion of g lv nometer into ammeter. Here ig < i.
i g G i g
From (ii) shunt required S= . If ig << i, S= Gi ig i 1 1 1 RA= SG
The resistance of ammeter RA so f rmed is given by = + ....(iii)
R A
G S GS
Note: Equation (ii) may alsobe used to increase the range of given ammeter. Here G will be resistance
of given ammeter, S shunt applied, ig its initial range and i the new range desired.
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CON VERSION OF GALVANOMETER INTO VOLTMETER
A voltmeter is a high resistance galvanometer and is connected between two points across which
potential difference, is to be measured i.e., voltmeter is connected in parallel with the circuit. To convert a
galvanometer into voltmeter, a high resistance R in series is connected to the galvanometer.
ig G igR
V V
If V is range of voltmeter, then i = or resistance in series R = G ...(i)R Gg
ig
This is working equation for conversion of galvanometer into voltmeter.
The resistance of voltmeter so formed is RV = R+G ...(ii)
Note: Equation (i) may also be used to increase the range of voltmeter. If V0 is initial range and V
is new range of voltmeter, then ig = V 0 V
R GGE
35
JEE-Physics
SOME WORKED OUT EXAMPLES
E x a m p l e # 1Current i = 2.5 A flows along the circle x2 + y2 = 9 cm2 (here x & y in cm) as shown. Magnetic field at point (0, 0, 4 cm) is
(A) 36 10
7
T kˆ (B) 36
10
7 T k
ˆ
9 10
7
ˆ 9
107 ˆ
(C) 5
T k (D) T k
5 Solu ti on Ans. (A)
Magnetic field on the axis of a circular loop0
Edujournal ˆ2 iR 2 10 7 2
2.53 2 104 ˆ 9
105
107 T ˆ
B
R
2
z2 3 / 2
k T k 36 k
4 125 10
6 25
Ex amp le # 2There are constant electric field ˆ ˆ present between plates P and P'. A particle of mass
0 j & magnetic field Bkm is projected from plate P' along y axis with velocity v1. After moving on the curved path, it passes through
point A just grazing the plate P with vel city v2. The magnitude of impulse (i.e. F t p ) provided bymagnetic
force during the motion of particle from rigin to point A is :–y
0 B0
A m v2 P
v1
m x
P'
(A) m|v –v | (B) m v 2 v2 (C) mv 1 (D) mv 2
2 1 1 2
Solu ti on Ans. (D)Electric force is only responsible for the change in momentum along y–axis. Therefore impulse provide bymagnetic force is JB = mv2.
Ex amp le # 3Three identical charge particles A, B and C are projected perpendicular to the uniform magnetic field withvelocities v1, v2 and v3 (v1 < v2 < v3) respectively such that T1, T2 and T3 are their respective time period ofrevolution and r1, r2 and r3 are respective radii of circular path described. Then :-(A) r1
r2 r3 (B)T1<T2<T3 (C) r1
r2 r3 (D) r1 = r2 = r3
T1 T2 T3 T1 T2T3
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36E
JEE-PhysicsS o l u t i o n Ans. (C)
2 m mv r
T = & r = vqB qB T
Ex amp le # 4An infinitely long straight wire is bent as shown in figure. The circular portion has a radius of 10 cm with itscenter O at a distance r from the straight part. The value of r such that the magnetic field at the center Oof the circular portion is zero will be :-
mc
0
1O
r
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10 20 1 5(A) cm (B) cm (C) cm (D) cm
5
Sol u ti on Ans. (A)B circular loop
Bwire 0 I
0 I r
10 cm in2 10 2 r
.Ex amp le # 5
Two cylindrical straight and very long non magnetic co ductors A and B, insulated from each other, carry acurrent I in the positive and the negative z–direction espectively. The direction of magnetic field at origin is
y
A B x
(A) ˆ (B) ˆ (C) ˆ ˆ
(D) – ji i
Solu ti on Edujournal Ans. (C)
y
A BA BB B x
Ex amp le # 6The magnetic force between wires as shown in figure is :-
i Lx
I iI2 x L iI2 2x L 0 iI x L
(A) 0
n (B) 0
n (C) n (D) None of these
2 2x 2 2x 2 x
E37
JEE-Physics
S o l u t i o n0 I
Magnetic field at dr, B =
Force on small element at a distance r of wire of length L is
F 0 i I x L dr 0 i I x L n
2 xr
2 x
Ans. (C)
i Ldr
0IdF = i(dr)
r 2r
x
I
Ex amp le # 7A wire carrying a current of 4A is bent in the form of a parabola x2 + y = 16 as shown in figure, where x and
ˆy are in meter. The wire is placed in a uniform magnetic field B 5 k tesla. The force acting on the wire is
y
Edujournalx
(A) ˆ (B) – 80 ˆ N ˆ ˆ
80 j N (C) 160 j N (D) 160 j NSolu ti on Ans. (C)
ˆ ˆ ˆ
F I B 4 8 i
5k 160 j N
Ex amp le # 8A conducting coil is bent in the form of equilateral triangle of side 5 cm. Current flowing through it is 0.2 A. Themagnetic moment of the triangle is :-(A) 3 × 10–2 A–m2 (B) 2.2 × 10–4 A–m2 (C) 2.2 × 10–2 A–m2 (D) 3 × 10–4 A–m2
Solu ti on Ans. (B)Magnetic moment of current carrying triangular loop M = IA
1 5 102 2 3
M 0.2 510 = 2.2 × 10–4 A-m2
2 2
Ex amp le # 9A disc of radius r and carrying positive charge q is rotating with angular speed in a uniform magneticfield B about a fixed axis as shown in figure, such that angle made by axis of disc with magnetic field is .Torque applied by axis on the disc is
Disc
B
Fixed axis
q r 2 B sin q r 2 B sin
(A) , clockwise (B) , anticlockwise2 4q r 2 B sin q r 2 B sin
(C) , anticlockwise (D) ,clockwise42
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38E
JEE-PhysicsS o l u t i o n Ans. (D)
2 2
M q M q mr M B q r B sin clockwise L 2m 2m 2 4
E x a m p l e # 1 0
A particle of mass m and charge q is thrown from origin at t = 0 with velocity 2 i 3 j 4k units in a region with
m
uniform magnetic field 2 iunits. After time t = qB , an electric field E is switched on, such that particle moves
on a straight line with constant speed. E may be(A) ˆ ˆunits ˆ ˆ units
–8 j 6k (B) 6 i 9k
Solu t i onn ˆ ˆ ˆ
At t = qB , v 2i 3 j 4k
; For net force to be zero
E x a m p l e # 1 1
(C) ˆ ˆ units (D) ˆ ˆ units 12 j 9k 8 j 6k
Ans. (A)ˆ ˆ
qv B qE
0 E v
B 8j
6k
A particle of specific charge q 1010 C kg–1
is projected from
mEdujournal 3 ˆ
5 ms –1 in a uniform magnetic field B210 kx–axis with a velocity of 10(A) The centre of the circle lies on the y–axis(B) The time period of revolution is 10–7 s.
(C) The radius of the circular path is 5 mm
(D) The velocity of the particle at t = 1 107 s is105
ˆ m / s
4Solu ti on
B,5 ˆ ,
B2103 ˆ Force will be in y–direction
F q v v 10 i k Motion of particle will be in xy plane
Time period T
2m 2 107 s
10
10 3
qB 2 10
mv 105 5 3 5Radius of path = 10 m mm
qB
10
10 2
10 3
the origin along the positive
tesla. Choose correctalternative(s)
Ans. (A,B,C,D)
y
t= T4
T 107 xAt t = s. Velocity of particle will be in +y direction. t=0
4 4
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E x a m p l e # 1 2A circular current carrying loop of radius R is bent about its diameter by 90° and
placed in a magnetic field B B0 (iˆ
ˆj) as shown in figure.
(A) The torque acting on the loop is zero
IR2 ˆ ˆ(B) The magnetic moment of the loop is i j
2(C) The angular acceleration of the loop is non zero.
IR2 ˆ ˆ(D) The magnetic moment of the loop is i j
2
y
I
xI
z
E39
JEE-Physics
S o l u t i o n Ans. (A,B)IR2 ˆ ˆ
M 2 i j; MB0
E x a m p l e # 1 3A current–carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the planeof the ring–(A) There is no net force on the ring.(B) The ring will tend to expand.(C) The ring will tend to contract.(D) Either (B) or (C) depending on the directions of the current in the ring and the magnetic field.
Solu ti on Ans. (A,D)Net force = 0 and ring will tend to expand/contracts depending on I & B.
E x a m p l e # 1 4In the given figure, B is magnetic field at P due to shown segment AB of an infinitecurrent carrying wire. A loop is taken as shown in figure. Which of the followingstatement(s) is/are correct. in
(A) B d 0 i(B) B=
0
2 r
(C) Magnetic field at P will be tangential (D) None of theseSolu ti on Ans. (C)
Magnetic field at P is tangential.
Example#15 to 17Two charge particles each of mass ‘m’, carrying cha ge +q and connected with each other by a massless inextensible string of length 2L are describing circular path in the plane of paper, each with speed
v qB 0 L (where B0 is constant) about their centre of mass in the region in which an uniform magnetic field B
mexists into the plane of paper as shown in figure. Neglect any effect of electrical & gravitational forces.
1 5 .
1 6 .
1 7 .
The magnitude of the magnetic field such that no tension is developed in the string will be
(A) B 0 (B) B0 (C) 2B0 (D) 0
2If the actual magnitude of magnetic field is half to that of calculated in part (i) then tension in the string will be
3 q 2 B 20 L q 2 B 20 L 2q 2 B 20 L
(A) (B) zero (C) (D)
2m m4 m3 q 2 B 2 L
Given that the string breaks when the tension is T =0
. Now if the magnetic field is reduced to such a4 m
value that the string just breaks then find the maximum separation between the two particles during theirmotion(A) 16 L (B) 4L (C) 14L (D) 2L
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S o l u t i o n1 5 . Ans. (B)
mv2 q + T+ 2q
T + qvB =L qvB
qB 0 L B mv2
T = 0; q = B = Bm L 0
1 6 . Ans. (C)
+ T qvB mv2 m (qB 0L )2 q(qB 0 L ) B q 2 B 20 L
0 T 0T + ;
m 2 R m 2mqv(B0/2) 2 R 2
1 7 . Ans. (C)Maximum reperation = 2R + (2R – 2L)
2L v
VR
T + qvB = mv2 B = B0/4 R = 4L maximum seperat on =16L – 2L= 14LR
EdujournalExample#18 to 20
Curves in the graph shown give, as functions of radial dist nce r, the magnitude B of the magnetic field insideand outside four long wires a, b, c and d, carryi g curre ts that are uniformly distributed across the crosssections of the wires. The wires are far from one a other.
B
a
b c
r
1 8 . Which wire has the greatest radius?
(A) a (B) b (C) c (D) d1 9 . Which wire has the greatest magnitude of the magnetic field on the surface?
(A) a (B) b (C) c (D) d
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2 0 . The current density in wire a is(A) greater than in wire c(C)equal to that in wire c
Solu ti on1 8 . Ans. (C)
Inside the cylinder : B 0 I r ...(i)
R2
2Outside the cylinder B 0 I ...(ii)
2 rInside cylinder B r and outside B 1
r
(B) less than in wire c(D) not comparable to that in wire c due to lack of information
R
I
r
r
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So from surface of cylinder nature of magnetic field changes.Hence it is clear from the graph that wire ‘c’ has greatest radius.
E41
JEE-Physics
1 9 . Ans. (A)Magnitude of magnetic field is maximum at the surface of wire ‘a’.
2 0 . Ans. (A)
Inside the wire B(r) 0 I r ; dB 0 Ii.e. slope
I current density.
R 22 R 2 dr 2 R 2
It can be seen that slope of curve for wire a is greater than wire c.E x a m p l e # 2 1
Column-I gives some current distributions and a point P in the space around these current distributions. Column-
II gives some expressions of magnetic field strength. Match column-I to corresponding field strength at point P given in column-II
Column – I Column – II
(A) A conducting loop shaped as regular hexagon of sidex, carrying current i. P is the centroid of hexagon
(B) A cylinder of inner radius x and outer radius 3x, carryingcurrent i. Point P is at a distance 2x from the axis ofthe cylinder
(C) Two coaxial hollow cylinders of radii x and 2x, each carry g
current i, but in opposite direction. P is a point at distanceEdujournal
1.5x from the axis of the cylinders .in(D) Magnetic field at the centre of an -sided regul r
polygon, of circum circle of radius x, carryi g currenti, n , P is centroid of the polygon. Ans. (A) (Q))
Solu ti on
0 i 0 i
For A : BP 6 sin 30 sin 30 3
4 x sin 60 x
(P) 3
0 i
32 x
(Q) 3 0 i
x
(R) 0
i
2 x
(S) 0
i
3 x
(T) Zero
; (B) (P) ; (C) (S) ; (D) (R)
P
x
60°x
0 3 i 3 0 i8For B: B P
2 2x 32 x
For C : B P 0 i 0 i 3x2 1.5x
For D : If n , n sided polygon circle so B =
0 i 2 x
2x P
x
3x
1.
5
x P
x
3x
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E x a m p l e # 2 2A very small current carrying square loop (current I) of side 'L' is placed in y-z plane with centre at origin ofthe coordinate system (shown in figure). In column–I the coordinate of the points are given & in column–IImagnitude of strength of magnetic field is given. Then
