Electromagnetic Waves, Antennas, and RF: Faraday’s Law and Electromagnetic Induction - Exercises and Complete Worked Solutions

This companion article extends EM13 with a second self-study problem set on magnetic flux, Faraday’s law, Lenz’s law, motional emf, induced Electric Fields, and the Maxwell–Faraday equation. All exercises are stated first. Complete worked solutions follow in Part II.

The principal relations are

(1)

(2)

(3)

and, for a fixed contour and surface,

(4)

For a uniform field through a flat loop,

(5)

where 𝜃 is measured from the chosen surface normal. These conventions follow EM13 and standard treatments of electromagnetic induction [1, 2, 3, 5].

How to use this problem set

Attempt every exercise in Part I before reading Part II. For each induction problem:

1. choose and state the surface-normal orientation;
2. determine the signed magnetic flux;
3. differentiate the flux before applying the minus sign;
4. use Lenz’s law or the Lorentz force to interpret direction;
5. distinguish transformer emf from motional emf when the circuit itself moves.

Part I: Exercises

Exercise 1: signed magnetic flux and reversed orientation

A flat loop has area

(6)

A uniform magnetic field has magnitude

(7)

and the angle between B and the chosen surface normal is

(8)

Find:

1. the signed magnetic flux;
2. the signed flux after reversing the surface-normal convention.

Exercise 2: induced emf in a multiturn coil

A 120-turn coil has area per turn

(9)

The magnetic field is perpendicular to the coil and points along the chosen positive normal. It decreases at the constant rate

(10)

Find the signed emf for the positive loop traversal associated with the chosen normal.

Exercise 3: Lenz-law direction

A circular conducting loop is viewed from the front. An external magnetic field points into the page and is increasing.

Determine:

1. the direction of the induced magnetic field;
2. whether the induced current is clockwise or counterclockwise.

Exercise 4: rotating-coil generator

A 60-turn coil of area

(11)

rotates at frequency

(12)

in a uniform field

(13)

Take the magnetic flux to be maximum and positive at t = 0.

Find:

1. the angular frequency ω;
2. the peak emf ℰ₀;
3. the instantaneous emf at t = 1∕(8f).

Exercise 5: motional emf and rod polarity

A conducting rod of length

(14)

is oriented along the y axis and moves with

(15)

through

(16)

Find:

1. the motional-emf magnitude;
2. the direction of the magnetic force on positive charge;
3. which end of the rod becomes positive.

Exercise 6: energy balance in a sliding-rod circuit

A rod of length

(17)

slides at constant speed

(18)

through a uniform magnetic field

(19)

The rod completes a circuit with total resistance

(20)

Assume the rod, velocity, and field are mutually perpendicular.

Find:

1. the motional emf;
2. the current magnitude;
3. the electrical power dissipated;
4. the magnetic drag-force magnitude on the rod;
5. the mechanical power required to maintain constant speed.

Exercise 7: loop entering a magnetic-field region

A rectangular loop of height

(21)

and width

(22)

moves to the right at constant speed

(23)

into a region containing a uniform magnetic field

(24)

directed into the page. At t = 0, the leading edge of the loop just enters the field region.

Find:

1. the overlap area A(t) while 0 < t < w∕v;
2. the magnetic-flux magnitude during entry;
3. the emf magnitude during entry;
4. the emf after the loop is completely inside a uniform field region.

Exercise 8: changing loop area in a static field

A loop remains perpendicular to a uniform static magnetic field

(25)

Its area changes according to

(26)

in square metres, with t in seconds.

Find the emf magnitude and identify the physical origin of the flux change.

Exercise 9: induced electric field inside a changing-field region

A spatially uniform magnetic field fills a circular region and changes at the constant rate

(27)

Find the induced electric-field magnitude on a circular contour of radius

(28)

that lies entirely inside the changing-field region.

Exercise 10: induced electric field outside a changing-field region

A changing magnetic field occupies a circular region of radius

(29)

and changes at the rate

(30)

Find the induced electric-field magnitude on a circular contour of radius

(31)

Exercise 11: derive the Maxwell–Faraday differential equation

Starting from the stationary-contour integral law

(32)

use Stokes’ theorem to derive

(33)

State the assumption that permits the time derivative to be moved inside the surface integral.

Exercise 12: local curl from a specified time-varying field

At a particular spatial region, the magnetic field is

(34)

with t in seconds.

Find:

1. ∂B∕∂t;
2. ∇× E at t = 2.0 s;
3. the local circulation sense viewed from the +z side.

Exercise 13: simultaneous field change and geometry change

A rectangular loop enters a magnetic-field region. During the interval of interest, the overlap area is

(35)

and the field magnitude is also increasing according to

(36)

Use

(37)

Find at

(38)

the magnitudes of:

1. the contribution AdB∕dt;
2. the contribution B dA∕dt;
3. the total emf.

Exercise 14: RF loop receiving-field preview

A 10-turn loop has area per turn

(39)

A locally uniform sinusoidal magnetic field has magnitude

(40)

where

(41)

The field makes an angle

(42)

with the loop normal.

Ignoring loading, self-inductance, radiation, and spatial variation over the loop, find:

1. the flux per turn;
2. an expression for the induced emf;
3. the peak emf magnitude.

Part II: Complete Worked Solutions

Solution 1: signed magnetic flux and reversed orientation

For a uniform field through a flat loop,

(43)

Therefore,

ΦB	= (0.40)(1.5 × 10−2) cos 120∘	(44)
	= (0.40)(1.5 × 10−2)(−0.5)	(45)
	= −3.0 × 10−3 Wb.	(46)

Thus,

(47)

Reversing the surface normal changes the angle from 120^∘ to 60^∘, so the sign reverses:

(48)

The physical field and loop are unchanged. Only the orientation convention changes.

Solution 2: induced emf in a multiturn coil

Because the field is normal to the coil and points along the chosen positive normal,

(49)

For N turns,

(50)

Therefore,

ℰ	= −(120)(2.5 × 10−3)(−0.60)	(51)
	= +0.18 V.	(52)

Hence,

(53)

for the positive loop traversal associated with the chosen normal.

Solution 3: Lenz-law direction

The external magnetic field points into the page and is increasing. The induced response must oppose that increase, so the induced field points out of the page.

Therefore,

(54)

A counterclockwise current produces a magnetic field out of the page by the right-hand rule. Thus,

(55)

Solution 4: rotating-coil generator

The angular frequency is

(56)

Therefore,

ω	= 2π(30)	(57)
	= 188.5 rad/s.	(58)

So,

(59)

With maximum positive flux at t = 0,

(60)

so

(61)

The peak emf is

ℰ0	= (60)(4.0 × 10−3)(0.25)(188.5)	(62)
	= 11.31 V.	(63)

Thus,

(64)

At

(65)

the phase is

(66)

Therefore,

ℰ	= 11.31 sin	(67)
	= 8.00 V.	(68)

Hence,

(69)

Solution 5: motional emf and rod polarity

For mutually perpendicular rod, velocity, and field,

(70)

Thus,

|ℰ|	= (0.30)(0.50)(4.0)	(71)
	= 0.60 V.	(72)

Therefore,

(73)

For positive charge,

v × B	∝x ×z	(74)
	= −y.	(75)

So positive charge is driven toward the −y end of the rod:

(76)

Therefore, the −y end becomes positive relative to the +y end.

Solution 6: energy balance in a sliding-rod circuit

The motional emf is

ℰ	= BLv	(77)
	= (0.50)(0.40)(3.0)	(78)
	= 0.60 V.	(79)

Thus,

(80)

The current magnitude is

I	=	(81)
	=	(82)
	= 0.30 A.	(83)

Hence,

(84)

The electrical power dissipated is

Pelec	= I2R	(85)
	= (0.30)2(2.0)	(86)
	= 0.18 W.	(87)

So,

(88)

The magnetic force magnitude on the current-carrying rod is

FB	= ILB	(89)
	= (0.30)(0.40)(0.50)	(90)
	= 0.060 N.	(91)

Therefore,

(92)

The external force required to maintain constant speed has equal magnitude and opposite direction. Its mechanical power is

Pmech	= Fv	(93)
	= (0.060)(3.0)	(94)
	= 0.18 W.	(95)

Thus,

(96)

This equality demonstrates the energy-conservation content of Lenz’s law for the idealized circuit.

Solution 7: loop entering a magnetic-field region

During entry, the overlap width is

(97)

Therefore the overlap area is

(98)

The flux magnitude is

(99)

Thus,

(100)

The emf magnitude is

(101)

Numerically,

|ℰ|	= (0.60)(0.25)(2.0)	(102)
	= 0.30 V.	(103)

Therefore,

(104)

while the loop is entering.

The entry interval lasts

(105)

After the loop is fully inside a spatially uniform static field, the flux is constant, so

(106)

Solution 8: changing loop area in a static field

The field is static, but the area changes:

(107)

Therefore,

(108)

Since

(109)

we obtain

|ℰ|	= (0.80)(0.002)	(110)
	= 1.6 × 10−3 V.	(111)

Thus,

(112)

The flux change is caused by changing circuit geometry rather than by a time-varying magnetic field.

Solution 9: induced electric field inside a changing-field region

For a circular contour of radius r entirely inside the changing-field region,

(113)

Therefore,

(114)

Substitute:

|E|	= (4.0)	(115)
	= 6.0 × 10−2 V/m.	(116)

Hence,

(117)

Solution 10: induced electric field outside a changing-field region

For r > R, only the magnetic-field region contributes to the flux:

(118)

Therefore,

(119)

Substitute:

|E|	= (6.0)	(120)
	= 3.75 × 10−2 V/m.	(121)

Thus,

(122)

Solution 11: derive the Maxwell–Faraday differential equation

Start from

(123)

By Stokes’ theorem,

(124)

For a fixed surface, the time derivative may be moved inside the integral:

(125)

Therefore,

(126)

Because the surface is arbitrary,

(127)

The required assumption is that the chosen contour and spanning surface are fixed in space while taking the time derivative.

Solution 12: local curl from a specified time-varying field

The field is

(128)

Differentiate:

(129)

Thus,

(130)

At t = 2.0 s,

(131)

Therefore,

(132)

A curl in the −z direction corresponds to clockwise local circulation when viewed from the +z side.

Solution 13: simultaneous field change and geometry change

The magnetic flux magnitude is

(133)

Therefore,

(134)

At t = 0.50 s,

(135)

and

(136)

Also,

(137)

and

(138)

The field-change contribution is

A	= (0.15)(0.40)	(139)
	= 0.060 V.	(140)

The geometry-change contribution is

B	= (0.20)(0.30)	(141)
	= 0.060 V.	(142)

Therefore,

(143)

(144)

and the total emf magnitude is

(145)

Solution 14: RF loop receiving-field preview

The magnetic-field component normal to the loop is

(146)

The flux per turn is therefore

(147)

For N turns,

ℰ(t)	= −N	(148)
	= NAB0(2πf) cos 𝜃 sin(2πft).	(149)

Thus,

(150)

The peak magnitude is

(151)

Substitute

(152)

Then

ℰ0	= (10)(1.0 × 10−4)(1.0 × 10−6)(2π)(1.0 × 107)(0.5)	(153)
	= 3.14 × 10−2 V.	(154)

Hence,

(155)

This is a lumped induction preview only. A practical RF loop antenna additionally requires impedance, Resonance, loading, radiation, polarization, and full-wave analysis.

Common mistakes

• Using the angle from the plane rather than the surface normal in BA cos 𝜃.
• Ignoring the sign change when the surface orientation is reversed.
• Using Lenz’s law as “opposes the field” instead of “opposes the change in flux.”
• Applying BLv without checking the cross-product geometry.
• Forgetting that a static magnetic field can still produce emf if the circuit geometry changes.
• Treating transformer emf and motional emf as the same local mechanism.
• Using the full observation-contour area for the outside induced-E problem when the magnetic field occupies only radius R.
• Forgetting the factor N for a multiturn coil.

Reinforcement summary

Faraday’s law is fundamentally a flux-change law:

(156)

Flux can change because the field changes, the circuit geometry changes, the orientation changes, or several of these occur simultaneously. The local field form,

(157)

shows that a time-varying magnetic field produces circulating electric-field structure even when no wire is present.

These results form one half of the dynamical curl coupling required for electromagnetic waves. The complementary time-dependent magnetic-curl equation will arise from Maxwell’s correction to Ampère’s law.

References