Physics 2BLecture 29C

"Conscience is or magnetic compass; reason our chart"--Joseph Cook

From Last TimeIn the last example we explored the resulting motion of an electron in a magnetic field.The motion followed a helical path around the magnetic field lines.

We broke up the velocity of the electron into parallel and perpendicular components.

Even though v and B are not perpendicular to one another, the resulting force on the particle will be.

Helical PathThe angle between v and B did not change as time moved on.

This occurs because F is perpendicular to both v and B the entire time (RHR).

Also note that although the velocity vector changed constantly during the helical path, the speed of the particle never changed (the magnitudes v|| and v⊥ remained constant).

The magnetic force does no work on a charged particle.

Kinetic energy must then remain constant.

Force on a WireConsider a wire of length l with current I immersed in an external magnetic field, B.

A magnetic force is exerted on each moving charge in the wire due to the magnetic field.

The total force is the sum of all magnetic forces on all the individual charges producing the current.

where θ is the angle between B and the direction of the current I along the length l.

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F = IlBsinθ

Force on a WireIf the current in a wire is moving downward and the magnetic field is into the board apply Right Hand Rule:

Your positive current velocity points down (thumb)

Your magnetic field points into the board (forefinger).

Middle finger then points to the right.

Force is to the right since current is defined as positive charge moving.

Force on a WireIf the wire is not straight or the magnetic field is not uniform, then you must break up the wire into small straight segments, dl.

The force on one of these segments is:

The force on the wire as a whole is the vector sum of all differential forces on the different wire segments.

Torque on a Current LoopLet’s say we have a rectangular loop of wire (of length b and width a) with a clockwise current immersed in a magnetic field.

Use RHR on each section of the wire.

For the top section, the force would point upward.

X X X X XX X X X XX X X X XX X X X X

B fieldI Ftop

Fbot

Fleft Fright

What would be the direction of the magnetic force on this wire loop?

For the bottom, the force would point downward.For the left, the force would point to the left.For the right, the force would point to the right.

Torque on a Current LoopThe forces cancel out. So nothing will happen to the loop.

But, what if the wire loop was slightly tilted (θ) with respect to the magnetic field.

Now, we can create a torque due to these forces about a central axis.

The torque on the top section will be given by:

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τ top =r r

r F top sinθ

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τ top =a2Ftop sinθ

Torque on a Current LoopThe torque on the bottom section will be given by:

The force on the top and bottom wire will have the same magnitude given by:

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τ bot =r r

r F bot sinθ

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τ bot =a2Fbot sinθ

Also, by right hand rule, the torques will point in the same direction, such that the torque from the top and bottom add.

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Ftop = Fbot = IbBsin90° = IbB

Torque on a Current LoopTo find the total torque on the current loop, merely add the torques:

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τ tot = τ top + τ bot

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τ tot =a2Ftop sinθ +

a2Fbot sinθ

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τ tot = aFtop sinθ = a IbB( )sinθ

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τ tot = IABsinθwhere A is the area of the loop.

If we were to make N loops of this wire then our equation would become:

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τ tot = NIABsinθ

Magnetic MomentThe first part of right side of the last equation (NIA) is dependent on properties of the loop.

Because of this we define the magnetic dipole moment vector, μ, of the wire:

The magnetic dipole moment vector points perpendicular to the plane of the loop(s), a normal so to speak.

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µ = NIA

Magnetic MomentThe angle θ will be between μ and B such that:

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τ =r µ ×

r B = r

µ r B sinθ

A magnetic dipole moment located in an external magnetic field will have a magnetic potential energy, U.

U will depend on the dipole’s orientation in the field, such that:

U will be least value when μ and B are parallel.

U will be greatest when μ and B are anti-parallel.

Electric MotorAn application of this concept is the electric motor.

An electric motor converts electrical energy to mechanical energy (specifically rotational kinetic energy).A simple electric motor consists of a rigid current-carrying loop that rotates when placed in a magnetic field.

To provide continuous rotation in one direction, the current loop must periodically reverse, (AC current).

Torque A Current LoopExampleA circular coil of wire with average radius 5.00cm and exactly 30 turns lies in a vertical plane. It carries 5.00A of current, counterclockwise when viewed from above. The coil is in a uniform magnetic field directed to the right, with a magnitude of 1.20T. Find the magnetic moment and the torque on the coil. Which way does the coil tend to rotate?AnswerFirst, you must define a coordinate system.Here we must define the normal for the coil, we say that it is perpendicular to the coil (out of the board).

B field

Torque A Current LoopAnswerUse the magnetic moment equation:

The area is for the circular loop giving us:

Where the direction of this magnetic moment is the same as the direction of the normal of the coil.

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µ = NIA

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τ =r µ ×

r B = r

µ r B sinθ

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µ = NI πr2( )

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µ = 30( ) 5A( )π 0.05m( )2=1.18 A ⋅m2

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τ = 1.18 A ⋅m2( ) 1.20T( )sin90°

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τ =1.42 N ⋅m

For torque:

Torque A Current LoopAnswerFor direction, look at four separate points on the circular loop (top, left, bottom, and right parts):Using RHR on the top and bottom gives us:Zero magnetic force (I and B are parallel in both instances).

B field

Using RHR on the left gives us:A magnetic force that is out of board.Using RHR on the right gives us:A magnetic force that is into the board.So the coil will flip such that the right side goes into the board and the left side comes out of the board.

For Next Time (FNT)

Start reading Chapter 30

Finish working on the homework for Chapter 28