Magnetism and Matter — Practice

Question 1

A bar magnet of moment 0.40 J per tesla is placed at 30 degrees to a uniform field of 0.25 T. Find the torque on it.

Accepted Answer

Apply torque = m B sin(theta), with theta measured between the moment and the field. Answer: Torque = 0.05 N m.

Question 2

A solenoid of 500 turns and cross-sectional area 2.0 x 10^-4 m squared carries a current of 3.0 A. Find its magnetic moment.

Accepted Answer

Use m = N I A, which is what makes the solenoid equivalent to a bar magnet. Answer: m = 0.30 J per tesla.

Question 3

A short bar magnet of moment 0.25 J per tesla is placed with its axis along the x-direction. Find the field at 20 cm along the axis.

Accepted Answer

Apply B = (mu_0/4 pi)(2m/r cubed), converting the distance to metres first. Answer: B = 6.3 x 10^-6 T, along the moment.

Question 4

A magnet of moment 0.60 J per tesla sits in a field of 0.40 T. Find its potential energy when aligned with the field and when antiparallel to it.

Accepted Answer

Use U = -m B cos(theta) at theta = 0 and theta = 180 degrees, and note which is the minimum. Answer: -0.24 J when aligned, stable; +0.24 J when antiparallel, unstable.

Question 5

A magnet of moment 1.2 J per tesla lies aligned with a field of 0.30 T. Find the work needed to turn it through 90 degrees.

Accepted Answer

The work equals the change in potential energy, so use W = -mB(cos theta2 - cos theta1). Answer: W = 0.36 J.

Question 6

For a short bar magnet of moment 0.48 J per tesla, find the ratio of the axial field to the equatorial field at the same distance, and compare their directions.

Accepted Answer

Write both expressions and divide; then read the direction of each relative to the moment. Answer: Ratio 2 to 1; axial along the moment, equatorial opposite to it.

Question 7

A magnetic material of susceptibility 4.0 x 10^-4 is placed in a magnetic intensity of 500 A per metre. Find its magnetisation, relative permeability, and class.

Accepted Answer

Use M = chi H and mu_r = 1 + chi, then read the class from the sign and size of chi. Answer: M = 0.2 A/m; mu_r = 1.0004; paramagnetic.

Question 8

A solenoid of 800 turns and area 2.5 x 10^-4 m squared carries 3.0 A and is free to turn in a horizontal field of 0.25 T, its axis at 30 degrees to the field. Find its moment and the torque on it.

Accepted Answer

First find m = N I A, then apply torque = m B sin(theta) with the angle taken from the field. Answer: m = 0.60 J per tesla; torque = 7.5 x 10^-2 N m.

Question 9

State Gauss's law for magnetism and use it to explain what happens when a bar magnet is cut into two pieces.

Accepted Answer

Start from the zero net flux statement and connect it to the non-existence of isolated poles. Answer: Net flux is zero through any closed surface, so no monopole exists and each piece is a complete magnet.

Practice — Magnetism and Matter

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