My lecture slides are posted at http://www.physics.ohio-state.edu/~humanic/

Information for Physics 1201 Midterm I Wednesday, February 20

1) Format: 10 multiple choice questions (each worth 5 points) and two show-work problems (each worth 25 points), giving 100 points total. 2) Closed book and closed notes. 3) Equations and constants will be provided on the midterm 4) Covers the material in Chapters 16, 17, 18, 19, and 20

Chapter 20

Magnetic Forces and Magnetic Fields

Magnetic Fields Produced by Currents

A solenoid is made up of many current loops which extend a finite distance along the axis of the loops characterized by the number of turns (number of loops) per unit length, n. The B-field in the interior of a long solenoid depends only on n and the current in the loops, I, as

Interior of a long solenoid nIB oµ=

B-field lines similar to those of a bar magnet with N and S poles shown. Use RHR

for direction of B inside

(B ~ 0 outside)

Example: Calculation of the B-field in a solenoid A 0.25 m long solenoid with 5000 turns has a current of 3.5 A applied to it. Find the B-field in the solenoid.

B = µonI

n = 5000 turns0.25m

= 2.0×104 turns m

B = 4π ×10−7( ) 2.0×104( ) 3.5( ) = 0.088T = 880G

Magnetic Materials

The intrinsic “spin” and orbital motion of electrons gives rise to the magnetic properties of materials è electron spin and orbits act as tiny current loops.

In ferromagnetic materials groups of 1016 - 1019 neighboring atoms form magnetic domains where the spins of electrons are naturally aligned with each other; magnetic domain sizes are ~ 0.01 - 0.1 mm.

An external magnetic field can induce magnetism in ferromagnetic materials by merging and aligning domains. Depending on the material, the induced magnetism may or may not become permanent. Putting iron in the center of a solenoid can create a strong electromagnet with fields 100x - 1000x the applied fields (also, can turn fields on and off).

Ampere’s Law

AMPERE’S LAW FOR STATIC MAGNETIC FIELDS For any current geometry that produces a magnetic field that does not change in time, sum up all of the segment lengths Δl multiplied by the B|| è this equals a constant multiplied by the current enclosed by the path. In mathematical notation,

IB oµ=Δ∑ ||

net current passing through surface bounded by path

Note: this is a messy thing to use for anything except in situations where the geometry of the problem has a lot of symmetry, e.g. the long straight wire

Ampere’s Law Example: An Infinitely Long, Straight, Current-Carrying Wire Use Ampere’s law to obtain the magnetic field è lots of symmetry!

IB oµ=Δ∑ ||

( ) IB oµ=Δ∑

IrB oµπ =2

rI

B o

πµ2

= è We get our familiar equation!

Can choose a path where B|| is constant

Since path is a circle of radius r

On RHS, I is the only current enclosed by path

Example: Ampere’s law applied to a solenoid

a b

c d

L

B||Δ∑ = µoI ⇒

Use closed path abcda

B||Δ( )ab + B||Δ( )bc + B||Δ( )cd + B||Δ( )da = µ0Ienclosed

B ≈ 0

B⊥ Δ

Since, outside solenoid along bc and da then,

B||Δ( )cd = µ0Ienclosed

I

n turns/length

B||ΔSince along cd

and Ienclosed = nLI

∴BL = µ0nLI ⇒ B = µ0nI

B

as seen earlier

The Force on a Current in a Magnetic Field

Example: The Force and Acceleration in a Loudspeaker The voice coil of a speaker has a diameter of 0.025 m, contains 55 turns of wire, and is placed in a 0.10-T magnetic field. The current in the voice coil is 2.0 A. (a) Determine the magnetic force that acts on the coil and the cone. (b) The voice coil and cone have a combined mass of 0.020 kg. Find their acceleration.

The Force on a Current in a Magnetic Field

(a)

(b) 2sm43kg 020.0N 86.0

===mFa

F = ILB sin θ = (2.0 A)[55π(0.025 m)](0.10 T)sin 90o = 0.86 N

The Torque on a Current-Carrying Coil

Consider the forces on a current-carrying loop in a magnetic field: The two forces on the loop have equal magnitude but an application of RHR shows that they are opposite in direction.

The Torque on a Current-Carrying Coil

The loop tends to rotate such that its normal becomes aligned with the magnetic field.

The Torque on a Current-Carrying Coil

( ) ( ) φφφτ sinsinsinNet torque 21

21 IABwILBwILB =+==

Derivation for the net torque on a current-carrying loop in a B field.

A = Lw

For a coil of N loops (turns) è

torque = (force) x (moment arm)

Magnetic moment of the coil

τ = NIABsinφ =MBsinφM = NIA

L

The Torque on a Current-Carrying Coil

Example: The Torque Exerted on a Current-Carrying Coil A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil.

(a) M = NIA = 100( ) 0.045 A( ) 2.0×10−4 m2( ) = 9.0×10−4 A ⋅m2

τ =MBsinϕ = 9.0×10−4 A ⋅m2( ) 0.15 T( )sin90 =1.4×10−4 N ⋅m(b)

The galvanometer revisited

How a galvanometer works. The coil of wire and pointer rotate when there is a current in the wire. The pointer stops when the magnetic torque on the coil, τ, is canceled by the opposing torque of the spring, τs .

τM = NIABsinφτ S = kΔφ

τ = τM −τ S = 0∑

NIABsinφ − kΔφ = 0 ⇒ Δφ =NIABsinφ

k⇒ Δφ∝ I

Δφ is the angular displacement of the needle k is the spring constant (Hooke’s Law)

The Torque on a Current-Carrying Coil

The basic components of a dc motor.

} Split-ring commutator

The Torque on a Current-Carrying Coil

How a dc motor works.

When a current exists in the coil, the coil experiences a torque.

Because of its inertia, the coil continues to rotate when there is no current.