Electromagnetic Waves, Antennas, and RF: Magnetic Forces on Charges and Currents - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM10, Magnetic forces on charges and Currents. All exercises are stated first. Complete worked solutions follow in Part II.

The central microscopic magnetic-force law is

(1)

For a thin current element,

(2)

and for a continuous current distribution,

(3)

where f_B is force per unit volume. For a planar N-turn current loop,

(4)

(5)

and

(6)

These are the same definitions and conventions developed in EM10 [1, 2, 3, 5].

How to use this problem set

Attempt every problem in Part I before consulting Part II. For vector-force problems, determine the cross-product direction before inserting numerical values. For current-loop problems, distinguish carefully among

(7)

A loop may have zero net force while still having nonzero torque.

Part I: Exercises

Exercise 1: magnetic force on a negative moving charge

A charge

(8)

moves with

(9)

through

(10)

Find the complete magnetic-force vector.

Exercise 2: from electron drift to force on a wire

A metal wire has mobile-electron number density

(11)

The electron charge is

(12)

and the drift velocity is

(13)

The wire cross-sectional area is

(14)

A length L = 0.20 m of the wire lies in

(15)

Find:

1. the current-density vector J = nqv_d;
2. the conventional current I;
3. the magnetic force on the wire segment.

Exercise 3: vector force on a straight conductor

A straight wire carries

(16)

along a vector length

(17)

The magnetic field is

(18)

Find the complete force vector.

Exercise 4: force magnitude for an oblique wire

A straight Conductor has

(19)

The angle between the current direction and the magnetic field is 60^∘. Find the magnetic-force magnitude.

Exercise 5: magnetic force density

A conductor carries uniform current density

(20)

in a magnetic field

(21)

Find the magnetic force density

(22)

Exercise 6: total force from a uniform force density

Inside a small conductor volume,

(23)

and

(24)

are uniform. The conductor volume is

(25)

Find the total magnetic force.

Exercise 7: curved wire in a uniform field

A wire carries current

(26)

from

(27)

to

(28)

along an arbitrary curved path. The magnetic field is uniform:

(29)

Use

(30)

to find the net force.

Exercise 8: closed loop in a uniform magnetic field

Starting from

(31)

show that a closed current loop in a uniform magnetic field has

(32)

Explain why this result does not imply zero torque.

Exercise 9: torque on a multiturn rectangular loop

A 20-turn loop carries

(33)

and encloses area

(34)

Its normal makes an angle 40^∘ with a uniform field

(35)

Find the torque magnitude.

Exercise 10: magnetic dipole moment

A circular 50-turn coil carries

(36)

and each turn encloses area

(37)

When viewed from the +z side, the current is counterclockwise.

Find:

1. the magnitude of μ;
2. its direction.

Exercise 11: vector torque from a dipole moment

A magnetic dipole has

(38)

in a field

(39)

Find the complete torque vector.

Exercise 12: magnetic potential-energy change

A dipole with

(40)

is in a uniform field

(41)

It rotates from

(42)

to the aligned state

(43)

Find U_i, U_f, and ΔU = U_f − U_i.

Exercise 13: when zero net force no longer follows

A closed current loop is placed first in a uniform magnetic field and then in a strongly nonuniform magnetic field.

For each case, discuss whether the following must be zero:

1. net magnetic force;
2. magnetic torque.

Explain which step in the uniform-field proof fails when B varies along the loop.

Exercise 14: synthesis - coil force, torque, and energy

A 100-turn planar coil carries

(44)

and each turn has area

(45)

The coil is placed in a uniform field

(46)

with the coil normal at

(47)

to the field.

Find:

1. the magnetic dipole moment magnitude;
2. the net magnetic force on the complete closed loop;
3. the torque magnitude;
4. the magnetic potential energy at 30^∘;
5. the energy change if the coil rotates quasistatically to alignment.

Part II: Complete Worked Solutions

Solution 1: magnetic force on a negative moving charge

First evaluate the cross product for a positive charge:

(48)

The magnitude is

FB	= |q|vB	(49)
	= (2.0 × 10−6)(4.0 × 104)(0.15)	(50)
	= 1.2 × 10−2 N.	(51)

Because the charge is negative, the force direction is opposite to v × B. Therefore,

(52)

Solution 2: from electron drift to force on a wire

The current density is

(53)

Both q and the x component of v_d are negative, so the current density points in +x:

J	= (8.5 × 1028)(−1.602 × 10−19)[−(2.0 × 10−4)x]	(54)
	= 2.7234 × 106x A/m2.	(55)

Thus

(56)

The conventional current is

I	= JA	(57)
	= (2.7234 × 106)(1.0 × 10−6)	(58)
	= 2.7234 A.	(59)

Therefore,

(60)

For the wire segment,

(61)

Hence

F	= IL × B	(62)
	= (2.7234)(0.20)(0.30)(x ×z)	(63)
	= −0.1634y N.	(64)

Thus

(65)

Solution 3: vector force on a straight conductor

Use

(66)

The direction is

(67)

The magnitude is

F	= ILB	(68)
	= (5.0)(0.40)(0.20)	(69)
	= 0.40 N.	(70)

Therefore,

(71)

Solution 4: force magnitude for an oblique wire

The magnitude is

(72)

Thus

F	= (3.0)(0.50)(0.40) sin 60∘	(73)
	≈ 0.520 N.	(74)

Hence

(75)

Solution 5: magnetic force density

The force density is

(76)

Since

(77)

we obtain

fB	= (1.5 × 106)(0.080)x	(78)
	= 1.2 × 105x N/m3.	(79)

Therefore,

(80)

Solution 6: total force from a uniform force density

First,

(81)

Because

(82)

we obtain

(83)

The force is uniform, so

F	= fBV	(84)
	= (−1.0 × 105)(3.0 × 10−6)y	(85)
	= −0.30y N.	(86)

Thus

(87)

Solution 7: curved wire in a uniform field

The endpoint displacement is

(88)

Therefore,

F	= I(r2 − r1) × B	(89)
	= 2.5(0.30x + 0.40y) × (0.30z).	(90)

Use

(91)

Then

F	= 0.75[−0.30y + 0.40x]	(92)
	= 0.30x − 0.225y N.	(93)

Hence

(94)

Solution 8: closed loop in a uniform magnetic field

Start from

(95)

Because B is uniform, it can be moved outside the path integral:

(96)

For any closed path,

(97)

Therefore,

(98)

This does not imply zero torque. Equal and opposite forces can act at different points on the loop and form a couple, producing rotation even though their vector sum is zero.

Solution 9: torque on a multiturn rectangular loop

The magnetic moment magnitude is

μ	= NIA	(99)
	= 20(1.5)(0.030)	(100)
	= 0.90 A m2.	(101)

The torque magnitude is

τ	= μB sin 𝜃	(102)
	= (0.90)(0.20) sin 40∘	(103)
	≈ 0.1157 N m.	(104)

Thus

(105)

Solution 10: magnetic dipole moment

The magnitude is

μ	= NIA	(106)
	= 50(0.25)(6.0 × 10−4)	(107)
	= 7.5 × 10−3 A m2.	(108)

Hence

(109)

Viewed from +z, the current is counterclockwise. Curling the right-hand fingers with the current makes the thumb point in +z. Therefore,

(110)

Solution 11: vector torque from a dipole moment

Use

(111)

Then

τ	= (0.040)(0.30)(y ×z)	(112)
	= 0.012x N m.	(113)

Thus

(114)

Solution 12: magnetic potential-energy change

The potential energy is

(115)

Initially,

Ui	= −(0.050)(0.40) cos 120∘	(116)
	= +1.0 × 10−2 J.	(117)

So

(118)

At alignment,

Uf	= −(0.050)(0.40) cos 0	(119)
	= −2.0 × 10−2 J.	(120)

Thus

(121)

The change is

ΔU	= Uf − Ui	(122)
	= −0.020 − 0.010	(123)
	= −3.0 × 10−2 J.	(124)

Therefore,

(125)

The aligned state is lower in potential energy.

Solution 13: when zero net force no longer follows

For a closed loop in a uniform magnetic field,

(126)

The torque need not be zero. Unless μ is parallel or antiparallel to B, the loop can have

(127)

For a strongly nonuniform field, neither quantity must vanish. Different portions of the loop can sample different field magnitudes or directions, so the local forces need not cancel completely. The uniform-field proof fails at the step

(128)

because a spatially varying B cannot be taken outside the integral.

Solution 14: synthesis - coil force, torque, and energy

The magnetic moment magnitude is

μ	= NIA	(129)
	= 100(0.40)(5.0 × 10−4)	(130)
	= 2.0 × 10−2 A m2.	(131)

Hence

(132)

Because the field is uniform and the coil is closed,

(133)

The torque magnitude is

τ	= μB sin 30∘	(134)
	= (0.020)(0.25)(0.5)	(135)
	= 2.5 × 10−3 N m.	(136)

Thus

(137)

At 30^∘,

U30	= −μB cos 30∘	(138)
	= −(0.020)(0.25)(0.8660)	(139)
	≈−4.33 × 10−3 J.	(140)

Therefore,

(141)

At alignment,

(142)

The potential-energy change is

ΔU	= U0 − U30	(143)
	= −5.00 × 10−3 − (−4.33 × 10−3)	(144)
	≈−6.70 × 10−4 J.	(145)

Thus

(146)

The coil lowers its magnetic potential energy as it moves toward alignment.

Common mistakes

• Using electron drift direction in I dℓ. The current-element vector follows conventional current.
• Dropping the cross product. Magnetic force is perpendicular to both the current direction and the magnetic field.
• Treating J × B as total force. It is force per unit volume and must be integrated over volume.
• Assuming zero net force means zero torque. A current loop in a uniform field can have zero net force and nonzero torque.
• Using A instead of NA for a multiturn coil. The magnetic moment is μ = NIA.
• Measuring 𝜃 from the plane of the loop. In τ = μB sin 𝜃 and U = −μB cos 𝜃, 𝜃 is the angle between the loop normal and B.
• Applying the closed-loop zero-force proof to a nonuniform field. The field may not be taken outside the path integral when it varies with position.

What EM10E reinforces

The sequence from microscopic to macroscopic magnetic force is

(147)

For a continuous current distribution,

(148)

For a closed planar coil,

(149)

(150)

and

(151)

These ideas complete the force side of the current-field interaction. EM11 reverses the direction of reasoning and asks how electric currents generate magnetic fields through the Biot–Savart law.

References