Electromagnetic Waves, Antennas, and RF: Magnetic Fields - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM09, Magnetic fields. All exercises are stated first. Complete worked solutions follow in Part II.

The central point-charge magnetic-force law is

(1)

Its magnitude is

(2)

where 𝜃 is the angle between v and B. For perpendicular motion in a uniform field,

(3)

and

(4)

For a straight current-carrying segment,

(5)

The complete Lorentz force is

(6)

These are the same definitions and conventions developed in EM09 [1, 2, 3, 5].

How to use this problem set

Attempt every problem in Part I before consulting Part II. For magnetic-force problems, determine the vector direction before substituting numbers. A useful workflow is

1. identify the directions of v and B;
2. evaluate v × B with the right-hand rule;
3. reverse that direction if the charge is negative;
4. then calculate the magnitude.

For circular and helical motion, separate the velocity into components perpendicular and parallel to the field before choosing a formula.

Part I: Exercises

Exercise 1: vector Lorentz force on a proton

A proton moves with velocity

(7)

through a uniform magnetic field

(8)

Using

(9)

find the complete magnetic-force vector on the proton.

Exercise 2: same motion for an electron

An electron has the same velocity and moves through the same field as in Exercise 1.

Find the complete magnetic-force vector on the electron. Explain why its direction differs from the proton result even though v × B is unchanged.

Exercise 3: magnetic force at an oblique angle

A particle has charge magnitude

(10)

Speed

(11)

and moves through a magnetic field

(12)

The angle between v and B is 40^∘. Find the magnetic-force magnitude.

Exercise 4: parallel and perpendicular velocity components

A proton moves at speed

(13)

at an angle of 35^∘ to a uniform field

(14)

Find:

1. v_∥;
2. v_⊥;
3. the magnetic-force magnitude.

Exercise 5: special directions and zero force

For each situation below, state the magnetic-force magnitude relative to the maximum value |q|vB:

1. v ∥ B;
2. v ⊥ B;
3. v antiparallel to B;
4. the charge is stationary.

Explain each result using the cross product.

Exercise 6: show that a magnetic field does no work

Starting with

(15)

show that the instantaneous magnetic power

(16)

is zero. Explain what this implies for the speed and kinetic energy of a point charge acted on only by a magnetic field.

Exercise 7: circular orbit radius

A proton moves perpendicular to a uniform magnetic field with

(17)

Using

(18)

find the radius of the circular orbit.

Exercise 8: electron cyclotron frequency and period

An electron moves perpendicular to a uniform field

(19)

Using

(20)

find:

1. the cyclotron angular frequency ω_c;
2. the ordinary frequency f_c;
3. the period T_c.

Exercise 9: helical motion

A proton moves in a uniform magnetic field

(21)

with velocity components

(22)

Find:

1. the helix radius;
2. the cyclotron period;
3. the helix pitch.

Exercise 10: infer the magnetic field from an orbit

A proton with speed

(23)

moves perpendicular to an unknown uniform magnetic field and follows a circular orbit of radius

(24)

Find the magnetic-field magnitude.

Exercise 11: vector force on a current segment

A straight Conductor segment carries current

(25)

and has vector length

(26)

It lies in a uniform field

(27)

Find the complete force vector on the conductor segment.

Exercise 12: current segment at an oblique angle

A straight conductor of length

(28)

carries

(29)

through a uniform field

(30)

The angle between the conventional-current direction and the magnetic field is 55^∘. Find the force magnitude.

Exercise 13: relate B and H in vacuum

In vacuum the magnetic flux density and magnetic field intensity are related by

(31)

where

(32)

If

(33)

find H in A/m. State why B and H should not be treated as the same physical quantity even though they are proportional in vacuum.

Exercise 14: full Lorentz-force synthesis

A particle has positive charge

(34)

and moves with

(35)

The fields are

(36)

Find:

1. the electric-force vector;
2. the magnetic-force vector;
3. the total Lorentz-force vector;
4. the total force if the charge sign is reversed while the same velocity and fields are retained.

Part II: Complete Worked Solutions

Solution 1: vector Lorentz force on a proton

The force law is

(37)

First determine the direction:

(38)

Because the proton charge is positive, the force points in the same direction as v × B.

The magnitude is

FB	= evB	(39)
	= (1.602 × 10−19)(2.5 × 106)(0.18)	(40)
	= 7.21 × 10−14 N.	(41)

Therefore,

(42)

Solution 2: same motion for an electron

The cross product is unchanged:

(43)

But an electron has negative charge,

(44)

Multiplying by a negative scalar reverses the force direction. The magnitude remains the same as in Exercise 1:

(45)

Thus

(46)

The right-hand rule gives the direction of v × B; the charge sign must still be applied afterward.

Solution 3: magnetic force at an oblique angle

Use

(47)

Substituting,

FB	= (3.0 × 10−6)(250)(0.40) sin 40∘	(48)
	≈ 1.93 × 10−4 N.	(49)

Therefore,

(50)

The result is smaller than the maximum value |q|vB because only the velocity component perpendicular to B contributes.

Solution 4: parallel and perpendicular velocity components

When 𝜃 is measured from the field direction,

(51)

and

(52)

Thus

v∥	= (5.0 × 105) cos 35∘	(53)
	≈ 4.10 × 105 m/s,	(54)

so

(55)

Similarly,

v⊥	= (5.0 × 105) sin 35∘	(56)
	≈ 2.87 × 105 m/s.	(57)

Therefore,

(58)

Only v_⊥ enters the magnetic-force magnitude:

FB	= ev⊥B	(59)
	= (1.602 × 10−19)(2.87 × 105)(0.12)	(60)
	≈ 5.51 × 10−15 N.	(61)

Thus

(62)

Solution 5: special directions and zero force

The magnitude law is

(63)

For parallel motion,

(64)

so

(65)

For perpendicular motion,

(66)

so

(67)

For antiparallel motion,

(68)

so again

(69)

For a stationary charge,

(70)

and therefore

(71)

These are direct consequences of the cross product.

Solution 6: show that a magnetic field does no work

Instantaneous mechanical power is

(72)

Insert the magnetic force:

PB

= q(v × B) ⋅ v.

(73)

By definition, v × B is perpendicular to v. The dot product of perpendicular vectors is zero, so

(74)

Therefore

(75)

In the nonrelativistic case,

(76)

so the speed remains constant. A magnetic field acting alone can bend the trajectory by changing the direction of velocity, but it cannot change the particle’s kinetic energy.

Solution 7: circular orbit radius

For perpendicular motion,

(77)

Substitute the proton values:

r	=	(78)
	≈ 5.37 × 10−2 m.	(79)

Thus

(80)

Solution 8: electron cyclotron frequency and period

The cyclotron angular frequency is

(81)

For the electron,

ωc	=	(82)
	≈ 1.41 × 1010 rad/s.	(83)

Therefore,

(84)

The ordinary frequency is

fc	=	(85)
	≈ 2.24 × 109 Hz.	(86)

Thus

(87)

The period is

Tc	=	(88)
	≈ 4.47 × 10−10 s.	(89)

Hence

(90)

Solution 9: helical motion

The radius is determined by the perpendicular velocity:

(91)

Therefore,

r	=	(92)
	≈ 5.01 × 10−2 m.	(93)

Thus

(94)

The cyclotron period is

Tc	=	(95)
	=	(96)
	≈ 2.62 × 10−7 s.	(97)

Hence

(98)

The pitch is the parallel distance traveled during one period:

p	= v∥Tc	(99)
	= (8.0 × 105)(2.62 × 10−7)	(100)
	≈ 2.10 × 10−1 m.	(101)

Therefore,

(102)

Solution 10: infer the magnetic field from an orbit

Start with

(103)

Solve for B:

(104)

For the proton,

B	=	(105)
	≈ 0.131 T.	(106)

Thus

(107)

Solution 11: vector force on a current segment

The force is

(108)

The direction follows

(109)

The magnitude is

F	= ILB	(110)
	= (4.0)(0.30)(0.20)	(111)
	= 0.24 N.	(112)

Therefore,

(113)

Solution 12: current segment at an oblique angle

Use

(114)

Then

F	= (2.5)(0.50)(0.30) sin 55∘	(115)
	≈ 0.307 N.	(116)

Thus

(117)

Solution 13: relate B and H in vacuum

In vacuum,

(118)

Therefore,

(119)

Convert the field magnitude:

(120)

Then

H	=	(121)
	≈ 39.8 A/m.	(122)

Thus

(123)

The quantities are related in vacuum but are not identical: B is magnetic flux density and is measured in tesla, while H is magnetic field intensity and is measured in A/m. Their distinction becomes especially important in material media.

Solution 14: full Lorentz-force synthesis

The total Lorentz force is

(124)

The electric force is

FE	= qE	(125)
	= (2.0 × 10−6)(300)x N	(126)
	= 6.0 × 10−4x N.	(127)

Hence

(128)

For the magnetic term,

(129)

Its magnitude is

FB	= qvB	(130)
	= (2.0 × 10−6)(200)(0.50)	(131)
	= 2.0 × 10−4 N.	(132)

Thus

(133)

Both contributions point in +x, so

F	= FE + FB	(134)
	= 8.0 × 10−4x N.	(135)

Therefore,

(136)

If the charge sign is reversed while E, v, and B remain unchanged, both the electric and magnetic force terms reverse together. Hence

(137)

Common mistakes

• Applying the right-hand rule directly to a negative charge. First find v × B; then reverse the direction if q < 0.
• Using the full speed in helical-motion radius calculations. The orbit radius depends on v_⊥ only.
• Using v_∥ in the magnetic-force magnitude. Parallel velocity contributes no magnetic force.
• Assuming a magnetic field changes kinetic energy. Magnetic force is perpendicular to velocity, so it does no work on a point charge.
• Forgetting that L points with conventional current.
• Treating B and H as interchangeable symbols. They have different units and different roles in material media.

What EM09E reinforces

The central magnetic-force relation is

(138)

Its cross-product structure determines both magnitude and direction:

(139)

For perpendicular motion,

(140)

and

(141)

For a straight current segment,

(142)

These results prepare the next main lesson, EM10, which asks the complementary question: how do moving charges and currents generate magnetic fields?

References