8/9/2019 Electromagnetism and magnetic circuit 6
1/30
8/9/2019 Electromagnetism and magnetic circuit 6
2/30
From the equation for the magnetic field of a moving chargedparticle, it is easy to show that a current I in a little length dl
of wire gives rise to a little bit of magnetic field.
dB
r
r
dl
The Biot-Savart Law
4
vr
r l0
2
I d rdB =
r
I
You may see the equation written using rr =r r .
16-Apr-10 2Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
3/30
Applying the Biot-Savart Law
I
ds
r
r
U
dB
4
vr rr
0
2
I s r rr r
r r
4
0
2I s si
r
r rB = dB
Homework Hint: if you have a tiny piece of a wire, just calculate dB; no need to integrate.16-Apr-10 3Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
4/30
Example: calculate the magnetic field at point P due to a thinstraight wire of length L carrying a current I. (P is on the
perpendicular bisector of the wire at distance a.)
4
vr
r0
2
I ds rdB =
r
vrds r ds si k
4
0
2
I ds sidB
r
ds is a i fi itesimal qua tity i the directio ofdx, so
4
0
2
I dx sidB
r
I
y
r
x
dBP
ds
Ur
x
z
a
L
16-Apr-10 4Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
5/30
2 2r = x +aa
sin =
r
I
y
r
x
dBP
ds
Ur
x
z
4
0
2
I dx sindB =
r
a
4 4
0 0
3/232 2
I dx I dxdB = =
r x
4L/2
0
3/2-L/2 2 2
I dx aB =
x +a
4 L/2
0
3/2-L/2 2 2
I dxB =
x
L
16-Apr-10 5Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
6/30
I
y
r
x
dBP
ds
Ur
x
z
a
4 L/2
0
3/2-L/2 2 2
I a dxB =
x +a
look integral up in tables, use theweb,or use trig substitutions
3/2 1/2
2 2 2 2 2
dx x=
x +a a x +a
4
L/2
0
1/22 2 2
-L/2
I xB =
ax
a
4
-
0
1/2 1/22 22 2 2 2
I a L/2 -L/2=
a L /2 +a a -L /2 +a
L
16-Apr-10 6Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
7/30
I
y
r
x
dBP
ds
Ur
x
z
a
4
-
0
1/22 2 2
I a 2L/2B =
a L /4 + a
4
0
1/22 2
I L 1B =
a L /4 + a
0
2 2
I L 1B =
2 a L + 4a
0
2
2
I 1B =
2 a 4a1+
L
16-Apr-10 7Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
8/30
I
y
r
x
dBP
ds
Ur
x
z
a
0
2
2
I 1B =
2 a 4a1+
L
When Lpg,
.
0I
B =2 a
0I
B =2 r
or The r in this equation has a differentmeaning than the r in the diagram!
16-Apr-10 8Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
9/30
Example: Long StraightConductor
What is the
magnetic field 4mm
from a long straight
conducting wire
carrying a 5 amp
current?
T
rIB
4
3
7
0
105.21042
5104
2
v! vv
vv!
!
T
TT
Q
8/9/2019 Electromagnetism and magnetic circuit 6
10/30
I
B
r
Magnetic Field of a Long Straight Wire
Weve just derived the equation for the magneticfield around a long, straight wire*
0IB =
2 r
with a direction given by a new right-handrule.
*Dont use this equation unless you have a long, straight wire!
16-Apr-10 10Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
11/30
Looking down along the wire:I
B
The magnetic field is notconstant.
At a fixed distance r from the wire, the magnitude of themagnetic field is constant.
The magnetic field direction is always tangent to the
imaginary circles drawn around the wire, and perpendicularto the radius connecting the wire and the point where thefield is being calculated.
16-Apr-10 11Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
12/30
Magnetic Field of a Current-Carrying Wire
It is experimentally observed that parallel wires exert forces oneach other when current flows.
I1 I2
F12 F21
I1 I2
F12 F21
16-Apr-10 12Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
13/30
The magnitude of the force depends onthe two currents, the length of the wires,and the distance between them.
0 1 2I I L
F = 2 d
I1 I2
F12 F21
d
L
The wires are electrically neutral,so this is not a Coulomb force.
We showed that a long straight wire carrying a currentI gives rise to a magnetic field B at a distance r from
the wire given by
0IB =2 r
I
B
rThe magnetic field of one wire exerts a force on anearby current-carrying wire.
This is NOT a
starting equation
16-Apr-10 13Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
14/30
Example: use the expression for B due to a current-carryingwire to calculate the force between two current-carrying wires.
I1 I2
F12
d
L
0 22
IB = k
2 d
vr r r
12 1 1 2F =I L B
L2L1
B2
vr
0 212 1
I= I Lj k
2 d
x
y
r0 1 2
12
I I L
= i2 d
The force per unit length of wire is
r
0 1 212 I I= i.L 2 d
16-Apr-10 14Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
15/30
I1 I2
F12 F21
d
L
r0 1
1
IB = - k
2 d
vr r r
21 2 2 1F =I L B
L2L1
B18
v
r0 1
21 2
IF = I Lj k
2 d
x
y
r0 1 2
21
I I L
F =- i2 d
The force per unit length of wire is
r
0 1 221 I IF = - i.L 2 d
16-Apr-10 15Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
16/30
If the currents in the wires are in the opposite direction, theforce is repulsive.
I1 I2
F12 F21
d
L
L2L1y
16-Apr-10 16Chetan Upadhyay
8/9/2019 Electromagnetism and magnetic circuit 6
17/30
8/9/2019 Electromagnetism and magnetic circuit 6
18/30
Force between two parallel
current-carrying straight wires (1)
1. Parallel wires with current flowing
in the same direction, attracteach other.
2. Parallel wires with current flowing
in the opposite direction, repel
8/9/2019 Electromagnetism and magnetic circuit 6
19/30
Force between two parallel
current-carrying straight wires (2)
Note that the force exerted on I2 by I1is equal but opposite to the force
exerted on I b I .
a
II
F
o
T
Q
2
21 N!
8/9/2019 Electromagnetism and magnetic circuit 6
20/30
Problem 1
Two parallel conductors of 20 mm
diameter each carrying 120 A in opposite
directions are separated by an air space of
60 mm. The conductors are 10 m long.
Find the force on each conductor.
16-Apr-10 Chetan Upadhyay 20
F = 0.36 N
vv
--1 2
12 21 1 2
4 10 I I L LF =F = =2 10 I I
2 d d
8/9/2019 Electromagnetism and magnetic circuit 6
21/30
Problem 2
Force between two wires carrying
currents in opposite direction is 20.4 kg/m.
When they are placed parallel with their
axis 5 cm apart. Calculate the current in
one conductor when the current flowing
through the other conductor is 5 KA.
16-Apr-10 Chetan Upadhyay 21
F = 20.4 x 9.81= 200.124 N
d=0.05 mI1 = 5000 A
vv
--1 2
12 21 1 2
4 10 I I L LF =F = =2 10 I I
2 d d
8/9/2019 Electromagnetism and magnetic circuit 6
22/30
16-Apr-10 Chetan Upadhyay 22
Chapter 3
LIFTING POWER OF MAGNET
8/9/2019 Electromagnetism and magnetic circuit 6
23/30
16-Apr-10 Chetan Upadhyay 23
N
S
A = area of cross section ofeach pole
P = force in Newtonbetween two poles
N
S
P
8/9/2019 Electromagnetism and magnetic circuit 6
24/30
16-Apr-10 Chetan Upadhyay 24
N
S
A = area of cross section ofeach pole
P = force in Newtonbetween two poles
N
S
P
8/9/2019 Electromagnetism and magnetic circuit 6
25/30
16-Apr-10 Chetan Upadhyay 25
N
S
A = area of cross section of each poleP = force in Newton between two
poles
N
S
dxWork done= P . dx
Volume increased= A . dx
P
8/9/2019 Electromagnetism and magnetic circuit 6
26/30
Energy in a
Magnetic Field Let the energy stored in the inductor at
any time be E.
E = U = L I2
8/9/2019 Electromagnetism and magnetic circuit 6
27/30
Energy stored in
a Magnetic Field Given E = U = L I2,
So, the energy stored per volume
This applies to any region in which a magneticfield exists not just the solenoid
22
21
2 2o
o o
B BU n A A
nQ
Q Q
! !
l l
2
2B
o
U Bu
V Q! !
8/9/2019 Electromagnetism and magnetic circuit 6
28/30
16-Apr-10 Chetan Upadhyay 28
N
S
dx
Energy stored into air magneticfield = B2/ 20 joule / m
3
So energy stored into airmagnetic field
= B2 {A . dx} / 20
Work done= B2 {A .dx} / 20 = P . dx
So, the lifting power of magnet= B2A/ 20
P
8/9/2019 Electromagnetism and magnetic circuit 6
29/30
16-Apr-10 Chetan Upadhyay 29
8/9/2019 Electromagnetism and magnetic circuit 6
30/30
16-Apr-10 Chetan Upadhyay 30
THANK YOU
PLEASE e-mail :
sir, my name is ..andmy roll no. is..
Please refer :-www.reachatkalol.blog.com