Faraday’s Law of Induction

Q: What is the difference between magnetic field and magnetic flux?

Magnetic field B is a vector defined at each point and measured in tesla. Magnetic flux φ is a scalar that counts how much of that field crosses a chosen surface, φ = BA cos θ, measured in weber. A strong field can still give zero flux if it runs parallel to the surface.

Q: How is Faraday's law tested in NEET?

NEET usually asks single-step numericals: emf from a given φ(t), average emf when a field collapses in a given time, induced charge, or the peak emf of a rotating coil. Statement questions on where the emf peaks and on the non-conservative induced electric field also appear.

PhysicsElectromagnetic InductionFor JEE aspirants

Faraday's law of induction says that an emf appears in a circuit whenever the magnetic flux through it changes, and that the size of the induced emf equals the rate at which the flux changes: $ε = - \frac{d ϕ}{d t}$ . Nothing needs to touch the circuit and no battery is needed: a moving magnet, a growing current nearby or a turning coil is enough. Faraday's law of induction is the starting point for generators, transformers and induction cookers, and it is a scoring chapter in both JEE Main and NEET.

On this page1Magnetic flux2Faraday's experiments3Faraday's law4Induced charge5Three ways to change flux6Rotating coil7Induced electric field

Key Formulas - Quick Reference

★ Must learn Magnetic flux: $ϕ = B \cdot A = B A cos θ$ , measured in weber ( $1 Wb = 1 T m^{2}$ )
★ Must learn Faraday's law: $ε = - \frac{d ϕ}{d t}$ ; for a coil of $N$ turns, $ε = - N \frac{d ϕ}{d t}$
Induced current: $i = \frac{∣ ε ∣}{R} = \frac{N}{R} \frac{d ϕ}{d t}$
★ Must learn Induced charge: $q = \frac{N ∣Δ ϕ ∣}{R}$ , which does not depend on how fast the flux changed
★ Must learn Coil rotating at angular speed $ω$ : $ϕ = N B A cos ω t$ and $ε = N B A ω sin ω t$
Peak values for that coil: $ε_{0} = N B A ω$ and $i_{0} = \frac{N B A ω}{R}$
Torque needed to keep it turning: $τ = \frac{N ^{2} B ^{2} A ^{2} ω}{R} sin^{2} ω t$
Induced electric field: $\oint E \cdot d l = - \frac{d ϕ}{d t}$
★ Must learn Inside a circular region of changing field ( $r < R$ ): $E = \frac{r}{2} \frac{d B}{d t}$
Outside that region ( $r > R$ ): $E = \frac{R ^{2}}{2 r} \frac{d B}{d t}$

1. Magnetic Flux: the Quantity That Matters

Induction is not about how strong the magnetic field is. It is about how much of the field passes through the circuit, and whether that amount is changing. The quantity that counts the field lines crossing a surface is called magnetic flux.

For a flat surface of area $A$ placed in a uniform field $B$ , the magnetic flux is

ϕ = B \cdot A = B A cos θ

where $θ$ is the angle between $B$ and the normal to the surface. The SI unit is the weber (Wb), and $1 Wb = 1 T m^{2}$ . Flux is a scalar, and it can be positive, negative or zero.

Figure 1: Magnetic flux

ϕ = B \cdot A = B A cos θ

, with

θ

measured between

B

and the normal

\overset{n}{^}

(the direction of

A

), not the plane of the loop.

Magnetic field

B

A vector defined at every point. Unit tesla (T). Tells how strong the field is at a point.

Magnetic flux

ϕ

A scalar for a whole surface, $ϕ = B A cos θ$ . Unit weber (Wb). Tells how much field passes through the loop; zero if $B$ lies in its plane.

1.1 Rules for the area vector

The area vector $A$ is always perpendicular to the surface, never along it. The angle $θ$ in $B A cos θ$ is measured from this normal.
For an open surface you may choose either direction for $A$ , but once chosen you must keep it for the whole problem, because it fixes the sign of $ϕ$ and therefore the sense of the induced current.
For a closed surface the outward normal is taken as positive.
If the field is not uniform, add up the contributions: $ϕ = \int B \cdot d A$ .
For any closed surface $\oint B \cdot d A = 0$ , because magnetic field lines always close on themselves. Whatever goes in must come out.

Figure 2: Flux counts the field lines that cross the loop. Tilting the normal by

6 0^{\circ}

halves the count (

cos 6 0^{\circ} = 0.5

); with

B

in the plane of the loop (

θ = 9 0^{\circ}

) no line crosses, so

ϕ = 0

however strong

B

is.

Exam Trick

If the field lies in the plane of the loop, the flux is zero, whatever the value of $B$ . Students lose marks by using $ϕ = B A cos 3 7^{\circ}$ when the diagram shows the lines lying flat in the plane; the correct angle from the normal is then $9 0^{\circ}$ , so $ϕ = 0$ .

Key idea

Flux counts field lines through the loop:

ϕ = B A cos θ

with

θ

from the normal. A field lying in the plane gives

ϕ = 0

2. What Faraday Actually Observed

In 1831 Michael Faraday (and independently Joseph Henry) ran three experiments with a coil connected only to a galvanometer, with no cell in the circuit:

Pushing a bar magnet into the coil deflected the needle. Pulling it out deflected the needle the other way. Holding the magnet still, however close, gave no deflection at all.
Replacing the magnet with a second coil carrying a steady current gave the same result: deflection only while the second coil was moving.
Keeping both coils fixed but switching the current in the second coil on and off also deflected the needle, at the moment of switching.

Figure 3: Faraday's observations. The galvanometer deflects only while the flux through the coil is changing: magnet moving, or current in the primary being switched on or off. A steady field, however strong, gives no reading.

The common thread is motion or change, not the field itself. A steady field, however strong, induces nothing. What produces an emf is a changing flux.

A coil sitting at rest in a field that is non-uniform in space but steady in time has no emf induced in it. The number of lines through it is odd-looking but constant, and only a change with time matters.

3. Faraday's Law of Induction

Whenever the magnetic flux through the area bounded by a closed conducting loop changes, an emf is induced in the loop, equal in magnitude to the rate of change of that flux:

ε = - \frac{d ϕ}{d t}

If the coil has $N$ turns wound so that the same flux $ϕ$ threads each of them, the emfs add up:

ε = - N \frac{d ϕ}{d t}

The minus sign is Lenz's law written into the formula: it fixes the direction of the induced emf, which always opposes the change that produced it. For numerical work you normally use magnitudes, $∣ ε ∣ = N \frac{d ϕ}{d t}$ , and settle the direction separately with Lenz's law.

Figure 4:

ε = - \frac{d ϕ}{d t}

is minus the slope of the

ϕ

t

graph, not its height. Rising flux gives

- 0.3 V

, flat flux (even at

0.6 Wb

) gives zero, and the steeper fall gives

+ 0.6 V

Exam Trick

Graph questions: the emf is minus the slope of the $ϕ$ - $t$ graph, never its height. A straight rising segment gives a constant negative emf, a flat segment gives zero (however large the flux), and the steepest segment gives the largest emf. Draw the emf graph as horizontal steps, one per straight piece.

Symbol	Meaning	SI unit
$ϕ$	Magnetic flux through the circuit	weber (Wb)
$B$	Magnetic field (magnetic induction)	tesla (T)
$A$	Area of the loop	$m^{2}$
$θ$	Angle between $B$ and the normal to the loop	radian or degree
$ε$	Induced emf	volt (V)
$N$	Number of turns in the coil	none
$R$	Total resistance of the circuit	ohm ( $Ω$ )

3.1 Induced current

The induced emf drives a current through the loop. If the total resistance of the circuit is $R$ , then

i = \frac{ε}{R} = - \frac{N}{R} \frac{d ϕ}{d t}

Notice that the emf does not depend on $R$ at all. Resistance only decides how much current that emf can push. An open loop still has an induced emf across its ends; it just carries no current.

3.2 Induced charge: the time drops out

Often a question asks how much charge flows while the flux changes, rather than the current at an instant. Because current is the rate of flow of charge, the time cancels:

At any instant $i = \frac{d q}{d t} = \frac{N}{R} \frac{d ϕ}{d t}$ .
So $d q = \frac{N}{R} ∣ d ϕ ∣$ .
Integrating from the initial flux $ϕ_{1}$ to the final flux $ϕ_{2}$ :
$q = \frac{N ∣ ϕ _{2} - ϕ _{1} ∣}{R} = \frac{N ∣Δ ϕ ∣}{R}$

Figure 5: Charge is the area under the

i

t

graph. Doubling the time halves the current, so the area stays

q = \frac{N ∣Δ ϕ ∣}{R} = \frac{8.0 \times 1 0 ^{- 3}}{5} = 1.6 mC

(numbers of Solved Example 2).

Exam Trick

Charge depends only on the total change in flux, not on how quickly it happened. Halve the time and the emf and current both double, but the charge is exactly the same. This is why a search coil with a ballistic galvanometer can measure flux without any timing at all.

Quick Recall: tap to check

A coil sits at rest in a very strong but steady field. What emf is induced?

Zero. Only a change of flux induces an emf.

The flux through a loop is

0.6 Wb

and constant. What is the emf?

Zero: the slope of the

ϕ

t

graph is zero.

The same flux change happens in half the time. What happens to

ε

i

and

q

ε

and

i

double;

q = N ∣Δ ϕ ∣/ R

is unchanged.

Does the induced emf depend on the resistance of the loop?

No.

R

decides only the current and the charge.

Key idea

ε = - N \frac{d ϕ}{d t}

gives the size from the rate of change; the charge

q = \frac{N ∣Δ ϕ ∣}{R}

needs only the total change.

4. Three Ways to Change the Flux

Since $ϕ = B A cos θ$ , there are exactly three things that can change, and every induction problem you will meet is one of them (or a combination):

Figure 6: Since

ϕ = B A cos θ

, an emf appears whenever

B

A

θ

changes with time. Every induction problem is one of these three, or a mix of them.

4.1 Changing the field: $ϕ = A B (t)$

With the loop fixed and flat-on to the field,

∣ ε ∣ = A \frac{d B}{d t}

This covers a loop near a wire whose current is growing, a coil inside a solenoid whose current is being switched, and any question that gives you $B$ as a function of time.

When the field is not uniform over the loop, as beside a long straight wire, split the loop into thin strips over which $B$ is constant and add up $d ϕ = B d A$ .

Figure 7: The field of the wire falls as

1/ x

, so split the frame into strips:

d ϕ = \frac{μ _{0} i}{2 π x} l d x

and

ϕ = \frac{μ _{0} i l}{2 π} ln \frac{r _{2}}{r _{1}}

. With

i = 10 A

l = 0.20 m

r_{2} / r_{1} = 2

ϕ = 2.77 \times 1 0^{- 7} Wb

(Solved Example 7).

4.2 Changing the area: $ϕ = B A (t)$

With the field steady,

∣ ε ∣ = B \frac{d A}{d t}

A rod sliding on rails, a loop being pulled out of a field region, or a circular loop whose radius is shrinking all belong here. When the change of area comes from something physically moving, the same emf can also be found from the motional-emf formula $ε = B v l$ , and the two routes must agree.

4.3 Changing the angle: the a.c. generator

Let a coil of $N$ turns and area $A$ spin at a constant angular speed $ω$ in a uniform field $B$ , about an axis in its own plane and perpendicular to $B$ . Then $θ = ω t$ , so

Flux at time $t$ : $ϕ = N B A cos ω t$ .
Differentiate: $ε = - \frac{d ϕ}{d t} = N B A ω sin ω t$ .
Peak emf: $ε_{0} = N B A ω$ , reached when the plane of the coil is parallel to $B$ (that is, when the flux is momentarily zero).

Figure 8: For a coil turning at constant

ω

ϕ = N B A cos ω t

and

ε = N B A ω sin ω t

. The emf peaks a quarter period after the flux (it lags by

T /4

): largest when the coil's plane is parallel to

B

and the flux is zero.

This sinusoidal emf is exactly what an a.c. generator delivers, and it is why mains supply is alternating. The current follows $i = \frac{N B A ω}{R} sin ω t = i_{0} sin ω t$ .

JEE Advanced

Keeping the coil turning costs work. The induced current $i$ in the field $B$ feels a torque opposing the rotation, so an external torque $τ$ must be supplied, and all of that mechanical power appears as heat:

τ ω = i^{2} R \Rightarrow τ = \frac{N ^{2} B ^{2} A ^{2} ω}{R} sin^{2} ω t

Averaged over a cycle, $⟨ sin^{2} ω t ⟩ = \frac{1}{2}$ , so the mean power needed is $\frac{N ^{2} B ^{2} A ^{2} ω ^{2}}{2 R}$ . This is the energy-conservation side of Lenz's law.

Figure 9: Faraday's-law problems in one chart. Differentiate a given

ϕ (t)

; otherwise identify what changes (

B

A

θ

). The charge needs only

Δ ϕ

, never the time.

Key idea

Every emf problem asks one question: what is changing,

B

A

θ

? For a rotating coil the emf is largest when the flux is zero.

5. The Induced Electric Field

Take a loop lying at rest and switch on a changing magnetic field. The free electrons in the wire start to drift, so a force must be acting on them. It cannot be a magnetic force, because a magnetic field exerts no force on a charge at rest. The only remaining possibility is an electric field created by the changing magnetic field.

A magnetic field changing with time sets up an induced electric field whose line integral around any closed path equals the rate of change of flux through that path:

\oint E \cdot d l = - \frac{d ϕ}{d t}

This field is very different from the electrostatic field of charges. Its lines are closed loops with no start and no finish, it is non-conservative, and no potential can be defined for it. It exists whether or not a wire is there: the wire only makes the effect visible as a current.

Figure 10: A changing

B

sets up closed loops of induced electric field, even outside the region where

B \neq = 0

. With

B

into the page and growing,

E

circulates anticlockwise.

Figure 11: Two kinds of electric field. Charges give lines that begin and end (

\oint E \cdot d l = 0

); a changing magnetic field gives closed loops (

\oint E \cdot d l = - \frac{d ϕ}{d t}

), so it can drive a current round a closed wire.

Electrostatic field

Produced by charges. Lines start on $+$ and end on $-$ . Conservative: $\oint E \cdot d l = 0$ , so a potential $V$ exists.

Induced electric field

Produced by a changing $B$ . Lines are closed loops. Non-conservative: $\oint E \cdot d l = - \frac{d ϕ}{d t}$ , so no potential can be defined.

5.1 The standard cylindrical case

A field $B$ confined to a cylinder of radius $R$ changes at a steady rate $\frac{d B}{d t}$ . By symmetry the induced $E$ is the same at every point of a circle of radius $r$ about the axis and points along it, so $\oint \vec{E}\cdot d\vec{l} = E(2\pi r)$.

Inside ( $r < R$ ), the flux enclosed is $B π r^{2}$ , so $E (2 π r) = π r^{2} \frac{d B}{d t}$ , giving
$E = \frac{r}{2} \frac{d B}{d t}$
Outside ( $r > R$ ), only the field inside the cylinder contributes, so $E (2 π r) = π R^{2} \frac{d B}{d t}$ , giving
$E = \frac{R ^{2}}{2 r} \frac{d B}{d t}$
The two expressions agree at $r = R$ , where $E$ has its largest value $\frac{R}{2} \frac{d B}{d t}$ .

Figure 12:

E = \frac{r}{2} \frac{d B}{d t}

for

r < R

(straight line) and

E = \frac{R ^{2}}{2 r} \frac{d B}{d t}

for

r > R

(falls as

1/ r

). The maximum

E_{ma x} = \frac{R}{2} \frac{d B}{d t}

is at

r = R

; at

r = 2 R

it is exactly half.

The direction follows from Lenz's law. If $B$ points into the page and is increasing, the induced $E$ circulates anticlockwise, which is the direction in which a positive charge placed there would be pushed.

Quick Recall: tap to check

Is there an induced electric field outside a cylinder of changing field, where

B = 0

Yes:

E = \frac{R ^{2}}{2 r} \frac{d B}{d t}

for

r > R

Where is the induced electric field strongest?

At the edge of the field region,

r = R

Can a potential be defined for the induced electric field?

No. Its line integral round a closed loop is not zero, so it is non-conservative.

Key idea

A changing magnetic field makes a closed-loop electric field:

E \propto r

inside the region,

E \propto 1/ r

outside, largest at

r = R

Figure 13: Mind map of the whole concept. Revise from it before the solved examples.

6. Solved Examples

Solved Example 1

A circular loop of area

A

is held in a uniform magnetic field

B

so that the field lines lie flat in the plane of the loop, making

3 7^{\circ}

with one of its diameters. Find the magnetic flux through the loop.

Solution:

Given: the field lies in the plane of the loop.

Key idea: $θ$ in $ϕ = B A cos θ$ is the angle between $B$ and the normal to the loop, not the angle drawn inside the plane.

Since $B$ lies in the plane, it is perpendicular to $A$ , so $θ = 9 0^{\circ}$ and

ϕ = B A cos 9 0^{\circ} = 0

Answer: $ϕ = 0$ . The $3 7^{\circ}$ in the question is a distractor: no field line crosses the surface, so no line is counted.

Solved Example 2

A loop of area

20 cm^{2}

and resistance

5 Ω

is placed in a uniform field of

2.0 T

with its plane perpendicular to the field. It is then turned through

18 0^{\circ}

. Find the induced emf, the induced current and the charge that flows if the turn takes (a)

0.01 s

, (b)

0.02 s

Solution:

Given: $B = 2.0 T$ , $A = 20\,\text{cm}^{2} = 20 \times 10^{-4}\,\text{m}^{2} $,$ R = 5\,\Omega $,$ N = 1$.

Flux before and after. Taking the normal along $B$ at the start,

ϕ_{i} = B A cos 0^{\circ} = (2.0) (20 \times 1 0^{- 4}) = 4.0 \times 1 0^{- 3} Wb

ϕ_{f} = B A cos 18 0^{\circ} = - 4.0 \times 1 0^{- 3} Wb

∣Δ ϕ ∣ = 8.0 \times 1 0^{- 3} Wb

(a) $Δ t = 0.01 s$ : $∣ ε ∣ = \frac{8.0 \times 1 0 ^{- 3}}{0.01} = 0.8 V$ , $i = \frac{0.8}{5} = 0.16 A$ , $q = i Δ t = 1.6 \times 1 0^{- 3} C$ .

(b) $Δ t = 0.02 s$ : $∣ ε ∣ = 0.4 V$ , $i = 0.08 A$ , $q = 1.6 \times 1 0^{- 3} C$ .

Answer: (a) $0.8 V$ , $0.16 A$ ; (b) $0.4 V$ , $0.08 A$ ; the charge is $1.6 mC$ in both cases. Doubling the time halves the emf and the current, but leaves the charge untouched, exactly as $q = ∣Δ ϕ ∣/ R$ predicts.

Solved Example 3

A conducting circular loop of variable radius

r

lies in a uniform field

B = 0.020 T

with its plane perpendicular to the field. The radius is shrinking at the steady rate

1.0 mm s^{- 1}

. Find the emf induced at the instant the radius is

2.0 mm

Solution:

Given: $B = 0.020 T$ , $\frac{d r}{d t} = 1.0 \times 1 0^{- 3} m s^{- 1}$ , $r = 2.0 \times 1 0^{- 3} m$ .

Only the area is changing, so $ϕ = B π r^{2}$ and

∣ ε ∣ = B \frac{d A}{d t} = B \cdot 2 π r \frac{d r}{d t}

Substituting:

∣ ε ∣ = (0.020) (2 π) (2.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 3}) = 2.5 \times 1 0^{- 7} V

Answer: $∣ ε ∣ \approx 2.5 \times 1 0^{- 7} V$ , that is about $0.25 μ V$ . Differentiate $π r^{2}$ properly: the factor is $2 π r$ , not $π r$ .

Solved Example 4

A rectangular coil of

N

turns, area

A

and resistance

R

rotates with constant angular velocity

ω

about an axis in its own plane, perpendicular to a uniform field

B

. The flux is maximum at

t = 0

. Find (a) the peak emf and peak current, (b) the torque that must be applied to keep the rotation steady.

Solution:

(a) Flux and emf. Since the flux is maximum at $t = 0$ ,

ϕ = N B A cos ω t

ε = - \frac{d ϕ}{d t} = N B A ω sin ω t

So the peak emf is $ε_{0} = N B A ω$ and the peak current is $i_{0} = \frac{N B A ω}{R}$ .

(b) Torque. All the mechanical power supplied is dissipated in the resistance:

τ ω = i^{2} R = \frac{N ^{2} B ^{2} A ^{2} ω ^{2}}{R} sin^{2} ω t

τ = \frac{N ^{2} B ^{2} A ^{2} ω}{R} sin^{2} ω t

Answer: $ε_{0} = N B A ω$ , $i_{0} = N B A ω / R$ , and $τ = \frac{N ^{2} B ^{2} A ^{2} ω}{R} sin^{2} ω t$ . The torque is largest exactly when the current is largest, which is when the coil's plane is parallel to $B$ .

Solved Example 5

The magnetic flux through a closed circuit of resistance

20 Ω

varies with time as

ϕ = 7 t^{2} - 4 t

, with

ϕ

in weber and

t

in second. The magnitude of the induced current at

t = 0.25 s

is
(A)

25 mA

(B)

0.025 mA

(C)

47 mA

(D)

175 mA

Solution:

Differentiate the flux:

∣ ε ∣ = \frac{d ϕ}{d t} = ∣14 t - 4∣

At $t = 0.25 s$ : $|\varepsilon| = |14(0.25) - 4| = |3.5 - 4| = 0.5\,\text{V}$.

i = \frac{∣ ε ∣}{R} = \frac{0.5}{20} = 0.025 A = 25 mA

Answer: (A). Option (D) comes from substituting $t$ into $ϕ$ instead of into $d ϕ / d t$ , which is the usual trap in this question.

Solved Example 6

A long straight wire carries a current that grows as

i = k t

with

k > 0

. A circular loop of radius

a

and resistance

R

lies in the same plane, with its centre at a distance

d

from the wire, where

a ≪ d

. Find the induced current in the loop.

Solution:

Field at the loop. Since $a ≪ d$ , the field over the loop is almost uniform and equal to its value at the centre:

B = \frac{μ _{0} i}{2 π d} = \frac{μ _{0} k t}{2 π d}

Flux:

ϕ = B (π a^{2}) = \frac{μ _{0} a ^{2} k t}{2 d}

emf and current:

∣ ε ∣ = \frac{d ϕ}{d t} = \frac{μ _{0} a ^{2} k}{2 d} \Rightarrow i_{ind} = \frac{μ _{0} a ^{2} k}{2 d R}

Answer: $i_{ind} = \frac{μ _{0} a ^{2} k}{2 d R}$ , constant in time. If the loop lies on the side of the wire where the field points into the page, that flux is growing, so by Lenz's law the induced current runs anticlockwise to push flux back out of the page.

Solved Example 7

A rectangular wire frame of length

0.20 m

and width

0.05 m

lies in the plane of a long straight wire carrying

10 A

, with its near side

5 \times 1 0^{- 2} m

from the wire and its long sides parallel to it. Find the flux through the frame. If the current then falls uniformly to zero in

0.20 s

, find the induced emf.

Solution:

Set up the integral. The field varies across the frame, so take a strip of width $d x$ at distance $x$ from the wire, where $B = \frac{μ _{0} i}{2 π x}$ and the strip area is $l d x$ with $l = 0.20 m$ :

ϕ = \int_{r_{1}}^{r_{2}} \frac{μ _{0} i l}{2 π x} d x = \frac{μ _{0} i l}{2 π} ln \frac{r _{2}}{r _{1}}

Substitute $r_{1} = 0.05 m$ , $r_{2} = 0.05 + 0.05 = 0.10 m$ , so $r_{2} / r_{1} = 2$ :

ϕ = (2 \times 1 0^{- 7}) (10) (0.20) ln 2 = 2.77 \times 1 0^{- 7} Wb

emf. The current falls uniformly, so the flux falls uniformly too:

∣ ε ∣ = \frac{∣Δ ϕ ∣}{Δ t} = \frac{2.77 \times 1 0 ^{- 7}}{0.20} = 1.39 \times 1 0^{- 6} V

Answer: $ϕ = 2.77 \times 1 0^{- 7} Wb$ and $∣ ε ∣ = 1.39 \times 1 0^{- 6} V$ (about $1.4 μ V$ ).

Solved Example 8

A generator coil has

N = 50

turns, each of area

200 cm^{2}

, and spins at

50

revolutions per second in a uniform field of

0.10 T

. Find the peak emf, and state the position of the coil when this peak occurs.

Solution:

Given: $N = 50$ , $A = 200\,\text{cm}^{2} = 2.0 \times 10^{-2}\, \text{m}^{2} $,$ B = 0.10\,\text{T} $,$ f = 50\,\text{Hz}$.

Angular speed: $ω = 2 π f = 2 π (50) = 314 rad s^{- 1}$ .

Peak emf:

ε_{0} = N B A ω = (50) (0.10) (2.0 \times 1 0^{- 2}) (314) = 31.4 V

Answer: $ε_{0} \approx 31 V$ , reached when the plane of the coil is parallel to $B$ , that is, when the flux through it is momentarily zero. This is the single most tested point in the whole topic: peak emf goes with zero flux.

Solved Example 9

A magnetic field is confined to a cylindrical region of radius

R = 10 cm

and is increasing at the steady rate

0.20 T s^{- 1}

along the axis. Find the magnitude of the induced electric field at

r = 4 cm

and at

r = 20 cm

from the axis.

Solution:

Inside ( $r = 0.04 m < R$ ):

E = \frac{r}{2} \frac{d B}{d t} = \frac{0.04}{2} (0.20) = 4.0 \times 1 0^{- 3} V m^{- 1}

Outside ( $r = 0.20 m > R$ ):

E = \frac{R ^{2}}{2 r} \frac{d B}{d t} = \frac{( 0.10 ) ^{2}}{2 ( 0.20 )} (0.20) = 5.0 \times 1 0^{- 3} V m^{- 1}

Answer: $4.0 mV m^{- 1}$ inside and $5.0 mV m^{- 1}$ outside. The field outside the region is not zero, even though $B$ there is zero: the induced electric field spreads beyond the region that contains the magnetic field.

Solved Example 10

A thin non-conducting ring of mass

m

and radius

r

carries a charge

q

spread uniformly around it and can turn freely about its own axis. It starts at rest with no magnetic field present. A field

B

is then switched on suddenly, perpendicular to the plane of the ring. Find the angular speed the ring acquires.

Solution:

Why it turns. The sudden change of flux creates an induced electric field along the ring, which pushes the charge and so exerts a torque.

Induced field at the ring: $E (2 π r) = π r^{2} \frac{d B}{d t}$ , so $E = \frac{r}{2} \frac{d B}{d t}$ .
Force on the ring $= q E$ , so the torque is $τ = q E r = \frac{q r ^{2}}{2} \frac{d B}{d t}$ .
Angular impulse: $\displaystyle\int \tau\,dt = \dfrac{qr^{2}}{2}\int_{0}^{B} dB = \dfrac{qr^{2}B}{2}$.
This equals the change in angular momentum $I ω$ with $I = m r^{2}$ :
$m r^{2} ω = \frac{q r ^{2} B}{2}$

Answer: $ω = \frac{q B}{2 m}$ . Neither $r$ nor the switching time appears, and the ring turns the way that makes its own magnetic moment oppose $B$ , which is Lenz's law again.

Solved Example 11

The flux through a single-turn coil changes with time as in the flux-time graph of Section 3: it rises from

0

0.6 Wb

in the first

2 s

, stays at

0.6 Wb

until

t = 4 s

and falls to zero at

t = 5 s

. The magnitude of the induced emf is largest during
(A)

0

2 s

(B)

2

4 s

(C)

4

5 s

(D) it is the same throughout

Solution:

The emf is minus the slope of each straight piece.

$0$ to $2 s$ : slope $= \frac{0.6}{2} = 0.3 Wb s^{- 1}$ , so $∣ ε ∣ = 0.3 V$ .

$2$ to $4 s$ : the flux is constant, so $ε = 0$ , even though the flux is at its largest.

$4$ to $5 s$ : slope $= \frac{0.6}{1} = 0.6 Wb s^{- 1}$ , so $∣ ε ∣ = 0.6 V$ .

Answer: (C). The steepest part of the graph, not the highest, gives the largest emf. (B) is the trap.

Solved Example 12

A coil of

100

turns, each of area

0.010 m^{2}

, has resistance

5.0 Ω

. It lies with its plane perpendicular to a field that increases uniformly from

0.20 T

0.80 T

0.10 s

. Find the induced emf, the current and the charge that flows.

Solution:

Given: $N = 100$ , $A = 0.010 m^{2}$ , $Δ B = 0.60 T$ , $Δ t = 0.10 s$ , $R = 5.0 Ω$ .

emf: $∣ ε ∣ = N A \frac{Δ B}{Δ t} = (100) (0.010) \frac{0.60}{0.10} = 6.0 V$ .

Current: $i = \frac{6.0}{5.0} = 1.2 A$ .

Charge: $q = \frac{N A Δ B}{R} = \frac{( 100 ) ( 0.010 ) ( 0.60 )}{5.0} = 0.12 C$ , which checks with $i Δ t = (1.2) (0.10)$ .

Answer: $6.0 V$ , $1.2 A$ , $0.12 C$ .

Solved Example 13

The angular speed of the coil of an ac generator is doubled, everything else being unchanged. The peak emf and the average power delivered to a fixed resistive load become
(A)

2

times and

2

times
(B)

2

times and

4

times
(C)

4

times and

4

times
(D)

4

times and

2

times

Solution:

Peak emf $ε_{0} = N B A ω$ , so it doubles.

Average power $P = \frac{ε _{0}^{2}}{2 R} \propto ω^{2}$ , so it becomes four times.

Answer: (B). Power goes as the square of the emf.

Solved Example 14

A long solenoid has

1000

turns per metre and radius

4.0 cm

. A circular loop of radius

10 cm

is placed round it, coaxially, near its middle. The solenoid current rises uniformly from

0

5.0 A

0.10 s

. Find the emf induced in the loop.

Solution:

Key idea. The field of a long solenoid is $B = μ_{0} ni$ inside it and almost zero outside. The loop is bigger than the solenoid, so only the solenoid's cross-section $π r_{s}^{2}$ carries flux.

ϕ = μ_{0} ni π r_{s}^{2} \Rightarrow ∣ ε ∣ = μ_{0} n π r_{s}^{2} \frac{d i}{d t}

Substitute: $\frac{d i}{d t} = \frac{5.0}{0.10} = 50 A s^{- 1}$ , $r_{s} = 0.040 m$ :

∣ ε ∣ = (4 π \times 1 0^{- 7}) (1000) π (0.040)^{2} (50) = 3.2 \times 1 0^{- 4} V

Answer: about $0.32 mV$ . The loop's own radius never enters. Only if the loop were smaller than the solenoid (say $2 cm$ ) would its own area be used, giving $μ_{0} n π (0.020)^{2} (50) = 7.9 \times 1 0^{- 5} V$ .

Practice Questions

A coil of $200$ turns, each of area $50 cm^{2}$ , lies with its plane perpendicular to a field of $0.25 T$ . The field is reduced to zero uniformly in $0.10 s$ . Find the average induced emf.Answer: $2.5 V$ (use $ε = N A Δ B /Δ t$ ).
The flux through a coil varies as $ϕ = 5 t^{3} + 4 t^{2} + 2 t + 5$ weber. Find the induced emf at $t = 2 s$ .Answer: $78 V$ , since $∣ ε ∣ = 15 t^{2} + 8 t + 2$ .
A coil of $100$ turns has resistance $10 Ω$ . The flux through each turn changes by $2 mWb$ . Find the charge that flows.Answer: $0.02 C$ , from $q = N ∣Δ ϕ ∣/ R$ . The time taken is not needed.
A square loop of side $10 cm$ and resistance $0.5 Ω$ sits normal to a field of $0.20 T$ , which falls uniformly to zero in $0.50 s$ . Find the induced current.Answer: $∣ ε ∣ = 4.0 \times 1 0^{- 3} V$ , so $i = 8.0 mA$ .
Show that when the flux through a circuit of total resistance $R$ changes from $ϕ_{1}$ to $ϕ_{2}$ , the charge that flows is $q = \frac{∣ ϕ _{2} - ϕ _{1} ∣}{R}$ .Answer: Put $i = d q / d t$ in $i = ∣ d ϕ / d t ∣/ R$ and integrate once with respect to time; the time variable cancels.
A field confined to a cylinder of radius $20 cm$ grows at $0.50 T s^{- 1}$ . Find the induced electric field at $r = 5 cm$ and at $r = 50 cm$ from the axis.Answer: $0.0125 V m^{- 1}$ inside and $0.020 V m^{- 1}$ outside.
A $50$ -turn coil of area $0.020 m^{2}$ rotates at $30 rad s^{- 1}$ about an axis perpendicular to a field of $0.50 T$ . Find the peak emf and the flux through the coil at the instant the emf is peak.Answer: $ε_{0} = N B A ω = 15 V$ ; the flux is zero at that instant (coil plane parallel to $B$ ).

Common Mistakes to Avoid

Watch out

Measuring $θ$ from the plane of the loop instead of from the normal. If the field lies in the plane, $θ = 9 0^{\circ}$ and the flux is zero.
Assuming a large flux means a large emf. A huge but steady flux gives zero emf; only $d ϕ / d t$ matters.
Dropping the factor $N$ for a coil of many turns, in both $ε = N ∣ d ϕ / d t ∣$ and $q = N ∣Δ ϕ ∣/ R$ .
Forgetting unit conversions: $20 cm^{2}$ is $20 \times 1 0^{- 4} m^{2}$ , not $20 \times 1 0^{- 2} m^{2}$ .
Carrying the minus sign of Faraday's law into a magnitude calculation. Use magnitudes for the number and Lenz's law for the direction.
Thinking the induced charge depends on how fast the flux changed. It depends only on the total change $∣Δ ϕ ∣$ and on $R$ .
Placing the peak emf of a rotating coil where the flux is largest. The emf is largest when the flux is zero (coil plane parallel to $B$ ) and zero when the flux is largest.
Substituting $t$ into $ϕ (t)$ when the question asks for the emf. Differentiate first, then substitute.

Frequently Asked Questions

What is Faraday's law of induction in simple words?

Faraday's law of induction says that an emf appears in a circuit whenever the magnetic flux through it changes, and the emf equals how fast the flux changes: $ε = - N d ϕ / d t$ . A steady field produces nothing. It is the change, not the field itself, that drives the current.

What is the difference between magnetic field and magnetic flux?

Magnetic field $B$ is a vector defined at each point and measured in tesla. Magnetic flux $ϕ$ is a scalar that counts how much of that field crosses a chosen surface, $ϕ = B A cos θ$ , measured in weber. A strong field can still give zero flux if it runs parallel to the surface.

Why is there a minus sign in Faraday's law?

The minus sign is Lenz's law built into the equation. It says the induced emf always acts to oppose the change in flux that created it. Without it, an induced current would reinforce its own cause and energy would be created from nothing, which is impossible.

Does the induced emf depend on the resistance of the loop?

No. The induced emf depends only on how fast the flux changes and on the number of turns. Resistance decides the induced current, $i = ∣ ε ∣/ R$ , and the induced charge, but not the emf. An open loop of infinite resistance still has an emf across its ends.

Can an emf be induced where there is no wire at all?

Yes. A changing magnetic field sets up closed loops of induced electric field in empty space, described by $\oint E \cdot d l = - d ϕ / d t$ . A wire only makes the effect visible as a current. This induced field is non-conservative, so no potential can be defined for it.

Is an emf induced in a coil at rest in a non-uniform magnetic field?

No. The field being non-uniform in space does not matter; what matters is whether the flux through the coil changes with time. A coil held still in a steady field, however uneven that field is, has constant flux and therefore no induced emf and no induced current.

How is Faraday's law tested in NEET?

NEET usually asks single-step numericals: emf from a given $ϕ (t)$ , average emf when a field collapses in a given time, induced charge, or the peak emf of a rotating coil. Statement questions on where the emf peaks and on the non-conservative induced electric field also appear.

What kind of Faraday's law questions come in JEE Main and Advanced?

JEE Main favours flux integrals near a long current-carrying wire and rotating-coil emf. JEE Advanced pushes further: induced electric field inside and outside a cylindrical region, torque and power for a generator, and problems that mix induced $E$ with mechanics, such as a charged ring set spinning.

Previous year questions on Faraday’s Law of Induction

22 questions from past papers, each with a step-by-step solution.

Show all 22 questions

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1. Magnetic Flux: the Quantity That Matters

1.1 Rules for the area vector

2. What Faraday Actually Observed

3. Faraday's Law of Induction

3.1 Induced current

3.2 Induced charge: the time drops out

4. Three Ways to Change the Flux

4.1 Changing the field: ϕ=AB(t)

4.2 Changing the area: ϕ=BA(t)

4.3 Changing the angle: the a.c. generator

5. The Induced Electric Field

5.1 The standard cylindrical case

6. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is Faraday's law of induction in simple words?

What is the difference between magnetic field and magnetic flux?

Why is there a minus sign in Faraday's law?

Does the induced emf depend on the resistance of the loop?

Can an emf be induced where there is no wire at all?

Is an emf induced in a coil at rest in a non-uniform magnetic field?

How is Faraday's law tested in NEET?

What kind of Faraday's law questions come in JEE Main and Advanced?

Previous year questions on Faraday’s Law of Induction

Ready to master Electromagnetic Induction?

4.1 Changing the field: $ϕ = A B (t)$

4.2 Changing the area: $ϕ = B A (t)$