MAGNETISMMAGNETISM

MADE BY:MADE BY:NIKHIL AND AMITNIKHIL AND AMITFAITH ACADEMYFAITH ACADEMY

2008-092008-09

MagnetismMagnetism

We are probably all familiar with the facts that We are probably all familiar with the facts that magnets repel and attract each other. This is magnets repel and attract each other. This is similar to electric charges. This suggests that similar to electric charges. This suggests that we can propose the following law we can propose the following law (like (like Newton’s Law of Gravity and Coulomb’s Law for Newton’s Law of Gravity and Coulomb’s Law for Electricity):Electricity):

FFmagneticmagnetic = X p = X p11 p p22 / r / r121222

where where XX would be the magnetic constant, would be the magnetic constant, analogous to analogous to GG and and kk, that describes the , that describes the strength of the magnetic force.strength of the magnetic force.

Magnetic PolesMagnetic PolesThe “The “pp” in the previous equation stands for ” in the previous equation stands for

polespoles. In magnetism, we have two kinds . In magnetism, we have two kinds of poles, and we call them of poles, and we call them NorthNorth and and SouthSouth. . Like poles repelLike poles repel and and unlike poles unlike poles attractattract, just as electric charges do., just as electric charges do.

However, However, unlike chargesunlike charges, we always have , we always have two polestwo poles! If we break a magnet (which ! If we break a magnet (which has two poles) in half, we have not has two poles) in half, we have not separated the two poles, rather we have separated the two poles, rather we have two (smaller) magnets that both have two two (smaller) magnets that both have two poles!poles!

MagneticMagnetic PolesPoles

NN S S

N S N SN S N S

Break one bar magnet in half, and you have two Break one bar magnet in half, and you have two smaller bar magnets!smaller bar magnets!

Magnetic Moment Of ElectronMagnetic Moment Of Electron

I=q/t (I=current,q=charge)I=q/t (I=current,q=charge)

T=2T=2r/vr/v

I=ve/2I=ve/2rr

mvr=nh/2mvr=nh/2v=nh/2v=nh/2mrmr

I=nhe/4I=nhe/4mrmr22

M=IAM=IA

Magnetic Force LawMagnetic Force LawMagnitude:Magnitude: FFmagneticmagnetic = q v B sin(q = q v B sin(qvBvB))

Direction:Direction: right hand rule:right hand rule:

thumbthumb = = handhand fingersfingers

Point your Point your right handright hand in the direction of in the direction of vv, , curl you curl you fingersfingers in the direction of in the direction of BB, and , and the the forceforce will be in the direction of your will be in the direction of your thumbthumb; ; if the charge is negative, the if the charge is negative, the force direction is oppositeforce direction is opposite that of your that of your thumb (or use you left handthumb (or use you left hand).).

UnitsUnitsFFmagneticmagnetic = q v B sin(q = q v B sin(qvBvB))

This law effectively defines the magneticThis law effectively defines the magneticfield, B field, B (just like F(just like Felectricelectric=qE defined E)=qE defined E)..

The MKS The MKS unitsunits of B are: of B are:

Tesla = Nt-sec / Coul-mTesla = Nt-sec / Coul-m . .

This unit turns out to be a very large one. We This unit turns out to be a very large one. We have a smaller unit:have a smaller unit:

10,000 Gauss = 1 Tesla10,000 Gauss = 1 Tesla . .

ExampleExampleA proton is moving at a speed of 3 x 10A proton is moving at a speed of 3 x 1044 m/s m/s

towards the West through a magnetic field towards the West through a magnetic field of strength 500 Gauss directed South. What of strength 500 Gauss directed South. What is the strength and direction of the magnetic is the strength and direction of the magnetic force on the proton at this instant?force on the proton at this instant?

qqprotonproton = +e = 1.6 x 10 = +e = 1.6 x 10-19 -19 Coul.Coul.

v = 3 x 10v = 3 x 1044 m/s, West m/s, West

B = 500 Gauss * 1 Tesla/10,000 GaussB = 500 Gauss * 1 Tesla/10,000 Gauss

= .05 T, South= .05 T, South

Magnitude:Magnitude: FFmagneticmagnetic = q v B sin( = q v B sin(vBvB))

Direction:Direction: right hand ruleright hand rule

Magnitude: Magnitude: FF = = (1.6 x 10(1.6 x 10-19-19 Coul) * (3 x 10 Coul) * (3 x 1044 m/s) * m/s) *

(.05 T) * sin(90(.05 T) * sin(90oo)) = = 2.4 x 102.4 x 10-16-16 Nt Nt..

DirectionDirection: thumb = hand x fingers: thumb = hand x fingers

= West x South = = West x South = UPUP..

Note: although the force looks small, Note: although the force looks small, consider the acceleration: consider the acceleration: aa = F/m = = F/m =

2.4 x 102.4 x 10-16-16 Nt / 1.67 x 10 Nt / 1.67 x 10-27-27 kg kg = = 1.44 x 101.44 x 101111 m/s m/s22..

Magnetic Force and MotionMagnetic Force and Motion

Since the magnetic field is perpendicular to Since the magnetic field is perpendicular to the velocity, and if the magnetic force is the velocity, and if the magnetic force is the only force acting on a moving charge, the only force acting on a moving charge, the force will cause the charge to go in a the force will cause the charge to go in a circle:circle:

F = maF = ma, , FFmagmag = q v B = q v B, and , and a = a = 22r = vr = v22/r/r

gives: q v B = mvgives: q v B = mv22/r, or /r, or r = mv/qBr = mv/qB . .

Mass SpectrographMass SpectrographWe can design an instrument in which we can We can design an instrument in which we can

control the magnetic field, B. If we ionize control the magnetic field, B. If we ionize (almost always singly) the material, we know (almost always singly) the material, we know the charge, q. We can use known voltages to the charge, q. We can use known voltages to get a known speed for the ions, v. We can get a known speed for the ions, v. We can then have the beam circle in the field and hit a then have the beam circle in the field and hit a target, and from that we can measure the target, and from that we can measure the radius, r. Hence, we can then calculate the radius, r. Hence, we can then calculate the mass using mass using r = mv/qBr = mv/qB . .

ExampleExampleTo see if this is really feasible, let’s try using To see if this is really feasible, let’s try using

realistic numbers to see what the radius for a realistic numbers to see what the radius for a proton should be:proton should be:

q = 1.6 x 10q = 1.6 x 10-19-19 Coul; Coul; B = .05 Teslas (500 G)B = .05 Teslas (500 G)

m = 1.67 x 10m = 1.67 x 10-27-27 kg; kg; VVaccacc = 500 volts gives: = 500 volts gives:

(1/2)mv(1/2)mv22 = qV, or v = [2qV/m] = qV, or v = [2qV/m]1/21/2 = 3.1 x 10 = 3.1 x 1055 m/s m/s

r = mv/qB r = mv/qB = (1.67 x 10= (1.67 x 10-27-27kg)*(3.1 x 10kg)*(3.1 x 1055 m/s) / m/s) / (1.6 x 10(1.6 x 10-19-19 Coul)*(0.05 T) = .065 m = Coul)*(0.05 T) = .065 m = 6.5 cm6.5 cm..

Mass SpectrometerMass Spectrometer

Heavier masses will give bigger radii, but we Heavier masses will give bigger radii, but we can shrink the radii if they become too big can shrink the radii if they become too big by using bigger magnetic fields. Note that by using bigger magnetic fields. Note that by measuring quantities that we can easily by measuring quantities that we can easily measure (charge, radius, Voltage, magnetic measure (charge, radius, Voltage, magnetic field), we can determine very tiny masses! field), we can determine very tiny masses! In one of our experiments in lab (Charge to In one of our experiments in lab (Charge to Mass of the Electron), we will essentially Mass of the Electron), we will essentially determine the mass of an electron!determine the mass of an electron!

Other usesOther uses

A cyclotron is an instrument used to A cyclotron is an instrument used to accelerate charged particles to very high accelerate charged particles to very high speeds. It uses magnetic fields to bend speeds. It uses magnetic fields to bend the charges around in circles to keep them the charges around in circles to keep them in one place while they’re being in one place while they’re being accelerated.accelerated.

Magnetic bottles can be used to contain high Magnetic bottles can be used to contain high energy plasmas in fusion research.energy plasmas in fusion research.

BIOT-SAVART LAWBIOT-SAVART LAW db->I(l=Length of the wire)db->I(l=Length of the wire)

db->dldb->dl

db->1/rdb->1/r22 (r =Radius of the wire) (r =Radius of the wire)

db->sindb->sin

db= db= ooidl*ridl*r->->/4/4rr33

(where -> denotes directly proportional sign)(where -> denotes directly proportional sign)

Magnetic FieldsMagnetic FieldsB = (B = (oo/4/4) q v sin() q v sin(vrvr) / r) / r22

Direction: right hand rule Direction: right hand rule

thumb = hand x fingers thumb = hand x fingers

Point your Point your handhand in the direction of in the direction of vv, curl , curl you you fingersfingers in the direction of in the direction of rr, and the , and the fieldfield will be in the direction of your will be in the direction of your thumbthumb; ; if the charge is negative, the field if the charge is negative, the field direction is oppositedirection is opposite that of your thumb. that of your thumb.

Magnetic ConstantMagnetic ConstantThe constant The constant ((oo/4/4)) is a seemingly strange way of is a seemingly strange way of

writing a constant that serves the same purpose writing a constant that serves the same purpose as G and k, but that is exactly what it does.as G and k, but that is exactly what it does.

The value: The value: ((oo/4/4) = 1 x 10) = 1 x 10-7-7 T*m*sec/Coul T*m*sec/Coul (or 1 (or 1

x 10x 10-7-7 T*m/Amp). T*m/Amp).In fact, the constant k is sometimes written as:In fact, the constant k is sometimes written as:

k = 1/(4k = 1/(4oo)). .

[In both cases, the sub zero on [In both cases, the sub zero on and and indicates the field indicates the field is in vacuum.]is in vacuum.]

Loop of CurrentLoop of CurrentFor the field at the For the field at the center center of a current loop of a current loop

we have the special equation:we have the special equation:BBat center of loopat center of loop = = oo N I / 2R N I / 2R

where N is the number of turns in the wire,where N is the number of turns in the wire,

II is the current in each loop, and is the current in each loop, and

RR is the radius of the loop. is the radius of the loop.

Forces on Forces on rectangular current looprectangular current loop

Consider the situation in the figure below:Consider the situation in the figure below:

A A current loopcurrent loop (with current direction going (with current direction going

counter-clockwise) is situated in a counter-clockwise) is situated in a MagneticMagnetic

FieldField going from North to South poles going from North to South poles..

NN B B S S

I I

We need to consider the forces on each of the four sides We need to consider the forces on each of the four sides of the current loop.of the current loop.

The The forceforce on the on the top top andand bottom bottom of the loop are of the loop are zerozero, , since the field and current are either parallel or anti-since the field and current are either parallel or anti-parallelparallel ( (IBIB = 0 = 0oo or 180 or 180oo).).

I I FFtoptop=0=0

NN B B S S

I I FFbottombottom=0=0

The current on the The current on the leftleft side is going down while side is going down while the field is directed to the right, so that means the field is directed to the right, so that means thethe forceforce is directed out of the screen, and the is directed out of the screen, and the magnitude is: magnitude is: FFleftleft = I L B sin(90 = I L B sin(90oo) = ) = II LL BB..

NN B B S S LL

The current on the The current on the rightright side is going up while side is going up while the field is directed to the right, so that means the field is directed to the right, so that means the the forceforce is directed into the screen, and the is directed into the screen, and the magnitude is: magnitude is: FFrightright = I L B sin(90 = I L B sin(90oo) = ) = II LL BB..

FF

NN B B S S LL

Net ForceNet ForceFFtoptop = 0 = 0 FFbottombottom = 0 = 0

FFleftleft = ILB out = ILB out FFrightright = ILB in = ILB in

As we can see, the As we can see, the NETNET FORCE FORCE ((F) is F) is zerozero..

However, since the force is pushing out on However, since the force is pushing out on the left and in on the right, there is a the left and in on the right, there is a TorqueTorque! The loop will tend to rotate about ! The loop will tend to rotate about an axis through the center.an axis through the center.

Torque on Torque on rectangular current looprectangular current loopRecall that torque is: Recall that torque is: = r F sin( = r F sin(rFrF)). In the . In the

figure below we can see that figure below we can see that r = w/2r = w/2. Thus . Thus the the FFleftleft gives a torque of gives a torque of (w/2)ILB(w/2)ILB, and the , and the

FFrightright also gives a torque of also gives a torque of (w/2)ILB(w/2)ILB..

rr

FF F F

NN B B S S LL

ww

Since both torques are trying to rotate the loop Since both torques are trying to rotate the loop in the same direction, the total torque is:in the same direction, the total torque is:

= wILB= wILB. We note that . We note that wL = AwL = A (width times (width times

length = Area).length = Area). Also, we can have several Also, we can have several loops (loops (NN) that will each give a torque.) that will each give a torque.

FF F F

NN B B S S LL

w w rr

Torque on a loopTorque on a loop

The final result for this loop is: The final result for this loop is: = N A I B sin(= N A I B sin(IBIB) sin() sin(rFrF)) . .

In this orientation, qIn this orientation, qIB IB = 90= 90oo and q and qrF rF = 90= 90oo . .

If the loop does rotate, we see that If the loop does rotate, we see that qqIB IB remains remains

at 90at 90oo (the current still goes up and down, the (the current still goes up and down, the field still goes to the right), but field still goes to the right), but qqrF rF changes changes as as

the loop rotates!the loop rotates!

t = N A I B sin(qt = N A I B sin(qrFrF)) . .

THE ENDTHE END