Lenz’s Law

Q: What is Lenz's law in simple words?

Lenz's law says the induced current always flows in the direction that fights the change producing it. Push a magnet towards a coil and the coil pushes back; pull it away and the coil holds on. It is the minus sign in ε = -N dφ/dt.

Q: Does the induced current oppose the magnetic flux or the change in flux?

The change, always. If the flux through a loop is increasing, the induced current makes flux the opposite way. If it is decreasing, the induced current makes flux the same way as the original field, trying to keep it alive. Confusing the two flips every answer.

Q: Is Lenz's law a consequence of conservation of energy?

Yes. If the induced current helped the change instead of opposing it, a magnet nudged towards a coil would be pulled in faster and faster while heat poured out of the coil, creating energy from nothing. Lenz's law is exactly the sign that prevents that.

Q: How do I decide clockwise or anticlockwise quickly?

Learn one case: flux into the page and increasing gives an anticlockwise current. Flip the answer if the field is out of the page instead, and flip again if the flux is decreasing instead. Two flips return you to the original answer.

Q: Why does a magnet fall slowly through a copper pipe?

The pipe acts like a stack of closed rings. Eddy currents induced in the walls oppose the magnet's motion, giving a retarding force that grows with speed until it balances the weight. The magnet then drifts down at a constant terminal velocity.

Q: What are eddy currents and are they useful or harmful?

Eddy currents are closed loops of induced current inside a solid conductor. They are harmful in transformer and motor cores, where they waste energy as heat, so cores are laminated. They are useful in magnetic braking, induction furnaces, induction cooktops and dead-beat galvanometers.

Q: How is Lenz's law asked in NEET?

Almost always as a quick reasoning question: the direction of current in a coil for a given magnet motion, the pole formed on the near face, whether a falling magnet has acceleration less than g, or an assertion and reason pair linking Lenz's law to conservation of energy.

Q: What kind of Lenz's law questions come in JEE Main and Advanced?

JEE Main asks for current directions near a long wire and for the force on a loop entering a field. JEE Advanced combines Lenz's law with mechanics: terminal velocity of a falling loop, exponential decay of speed on rails, and energy balance between work done and heat produced.

PhysicsElectromagnetic InductionFor NEET aspirants

Lenz's law fixes the direction of an induced current: it always flows in the sense that opposes the change in magnetic flux which produced it. That single sentence is the minus sign in $ε = - N \frac{d ϕ}{d t}$ , and it is really conservation of energy in disguise, because the induced current can only appear if something does work against an opposing force. Lenz's law is a guaranteed question in NEET and JEE Main, usually as a one-line "clockwise or anticlockwise" check.

On this page1Statement2Magnet and coil3Three-step method4Energy conservation5Loop crossing a field6Formal sign method7Eddy currents

Key Formulas - Quick Reference

★ Must learn Lenz's law: the induced emf and current oppose the change in flux that causes them, which is the minus sign in $ε = - N \frac{d ϕ}{d t}$
★ Must learn Flux into the page increasing $\Rightarrow$ induced current anticlockwise; decreasing $\Rightarrow$ clockwise (reverse both for flux out of the page)
Magnetic moment of the loop $M = N i A$ , found by the right-hand grip rule; the induced $M$ always fights the change
★ Must learn Magnet approaching a coil: the near face becomes the same pole (repulsion). Magnet receding: the near face becomes the opposite pole (attraction)
★ Must learn Retarding force on a loop of side $l$ and resistance $R$ entering a field at speed $v$ : $F = \frac{B ^{2} l ^{2} v}{R}$
Magnet dropped through a closed conducting ring: $a = g - \frac{F}{m} < g$ . Through a cut ring or a plastic ring: $a = g$
★ Must learn Energy balance: work done against the opposing force $=$ heat produced, $\int F v d t = i^{2} R t$
Terminal speed of a loop falling out of a field under gravity: $v_{T} = \frac{m g R}{B ^{2} l ^{2}}$
Induced charge $q = \frac{N ∣Δ ϕ ∣}{R}$ ; Lenz's law fixes its direction, not its size
Eddy-current loss grows with the square of the rate of change of flux, so cores are laminated and plates are slotted

1. What Lenz's Law Says

Faraday's law gives the size of the induced emf but not its sense. Lenz's law supplies that missing half.

The induced emf drives a current in the direction that opposes the change in flux that produced it. Written into Faraday's law:

ε = - N \frac{d ϕ}{d t}

the minus sign is Lenz's law. Note the word change: the induced current does not oppose the flux, it opposes whatever the flux is doing.

That distinction is where most marks are lost. If the flux through a loop is increasing, the induced current sets up flux in the opposite sense, trying to hold it down. If the flux is decreasing, the induced current sets up flux in the same sense as the original, trying to prop it up. A loop never simply fights the external field; it fights the change.

Flux increasing

The induced current makes flux opposite to the original, to hold the growth down. A magnet approaching a coil is repelled.

Flux decreasing

The induced current makes flux in the same direction as the original, to prop it up. A magnet moving away is attracted back.

Figure 1: The coil always forms the pole that fights the motion: a like pole to push an approaching magnet back, an unlike pole to hold a receding magnet back. Seen from the magnet, an anticlockwise current makes the near face N, a clockwise one makes it S.

1.1 The face rule for a magnet and a coil

For magnet-and-coil questions there is a one-line shortcut that follows from the magnet-and-coil figure:

A pole moving towards the coil sees the same pole appear on the near face, so the coil pushes the magnet back.
A pole moving away from the coil sees the opposite pole appear on the near face, so the coil pulls the magnet back.
Either way the coil resists the motion. Nothing the magnet does can ever be helped along by the coil.

Once the face pole is known, the direction of the current follows from the right-hand grip rule: curl the fingers of your right hand in the sense of the current and the thumb points towards the face that behaves as the north pole.

Figure 2: Curl the fingers of the right hand along the current: the thumb gives the field inside the loop. Seen from the page, anticlockwise = out of the page = N face; clockwise = into the page = S face.

Key idea

The induced current opposes the change in flux, never the flux itself: approaching poles are repelled, receding poles are attracted back.

2. Finding the Direction in Three Steps

For a flat loop drawn on the page, the whole job reduces to three questions:

Which way is the flux? Decide whether the field through the loop points into the page or out of the page.
Is it growing or shrinking? The loop may be moving, the field may be changing, or the area may be changing, but only the trend matters.
Oppose the trend. If the flux is growing, the induced current must make flux the other way; if it is shrinking, the induced current must make flux the same way. Then use the right-hand grip rule: current anticlockwise on the page gives flux out of the page, clockwise gives flux into the page.

Figure 3: The four cases you will ever need. Change one word (into or out of, increasing or decreasing) and the answer flips; change both and it flips back.

Exam Trick

Memorise one case only: flux into the page and increasing gives an anticlockwise current. Every other case is obtained by flipping. Change exactly one word in the question (into $\to$ out of, or increasing $\to$ decreasing) and the answer flips. Change both words and it flips back to anticlockwise.

Figure 4: Direction in three steps. The only decision is whether the flux is growing (oppose it) or shrinking (support it).

Quick Recall: tap to check

The flux out of the page through a loop is increasing. Which way is the induced current?

Clockwise: it must make flux into the page.

A loop in a field into the page is squeezed so its area shrinks. Which way is the current?

Clockwise: the flux into the page is falling, so the current supports it.

The S pole of a magnet is pushed towards a coil. Which pole forms on the near face?

S, a like pole, so the magnet is repelled.

Does the induced current oppose the flux or the change in flux?

Always the change in flux.

3. Lenz's Law is Conservation of Energy

Suppose Lenz's law ran the other way, so that the induced current helped the change. Push a magnet a little towards a coil and the coil would pull it in; that would speed the magnet up, which would increase the rate of change of flux, which would increase the current, which would pull harder still. The magnet would accelerate for ever while the coil poured out heat, and energy would appear from nothing. Lenz's law is exactly the condition that stops this.

Figure 5: Lenz's law is conservation of energy. The ring can take electrical energy from the magnet only by slowing it, so

a < g

. Cut the ring and the whole effect disappears:

a = g

Figure 6: Computed emf as a magnet falls from rest through a ring. The pulses have opposite signs (flux rising, then falling),

ε = 0

as the magnet passes the ring (flux maximum), and the second pulse is about

1.2

times taller and narrower because the magnet is faster. The two areas are equal: the same

∣Δ ϕ ∣

in and out.

Exam Trick

Falling-magnet questions: closed ring, metal pipe or coil with a closed circuit gives $a < g$ throughout the fall (repelled while approaching, attracted while leaving). A cut ring, a plastic ring or an open coil gives $a = g$ . The emf-time graph is two pulses of opposite sign, the second taller and narrower because the magnet is faster.

So the electrical energy has to be paid for. Whoever moves the magnet (or the loop, or turns up the current) does work against the opposing force, and that work reappears as heat in the resistance:

work done against the induced force = electrical energy generated = heat produced

\int F v d t = \int i^{2} R d t

JEE Advanced

For a loop of side $l$ and resistance $R$ moving at speed $v$ with one arm in a field $B$ , the emf is $ε = B v l$ , the current is $i = \frac{B v l}{R}$ , and the magnetic force on that arm is

F = B i l = \frac{B ^{2} l ^{2} v}{R}

which points backwards, against $v$ . Two standard results follow at once. A loop released so that it falls out of a field region reaches a terminal speed when $m g = \frac{B ^{2} l ^{2} v _{T}}{R}$ , that is

v_{T} = \frac{m g R}{B ^{2} l ^{2}}

and a loop given an initial speed $u$ on frictionless rails slows down exponentially, $v = u e^{- B^{2} l^{2} t / m R}$ , because the retarding force is proportional to $v$ itself.

3.1 A loop entering and leaving a field region

This is the single most common figure in the chapter. Watch what switches on and off:

Figure 7: The current reverses between entering (anticlockwise) and leaving (clockwise), but the force on the loop points backwards both times. Fully inside, the flux is constant and everything switches off.

Entering: flux into the page is growing, so the current is anticlockwise and the force on the loop points backwards.
Fully inside: the flux is constant, because as much field enters on one side as leaves on the other. No emf, no current, no force. The loop coasts.
Leaving: flux into the page is shrinking, so the current reverses to clockwise, but the force still points backwards.

Figure 8: For a loop of side

L

crossing a region of width

D > L

at constant

v

ϕ

is a trapezium,

ε = - \frac{d ϕ}{d t}

is two opposite steps of size

B Lv

, and the force

\frac{B ^{2} L ^{2} v}{R}

points backwards in both steps.

The current reverses between entering and leaving, yet the force does not. That is Lenz's law being consistent: the force opposes the motion in both cases, never the current direction for its own sake.

Key idea

Lenz's law is energy conservation: the magnetic force always opposes the motion, so the work done against it reappears as heat,

\int F v d t = \int i^{2} R d t

4. The Formal Sign Method

The three-step method is fast but needs a picture. When a problem gives you flux as a function of time and asks for a sign, use the formal route instead:

Choose a direction to walk round the loop, and use the right-hand grip rule to fix the area vector $A$ that goes with it.
Work out $ϕ = B \cdot A$ with that sign convention, then find $\frac{d ϕ}{d t}$ .
Compute $ε = - \frac{d ϕ}{d t}$ . If $ε$ comes out positive the current flows the way you chose to walk; if negative, it flows the other way.

Worked in one line. A loop lies beside a long wire whose current is increasing. Walk round the loop clockwise, so $A$ points into the page, the same way as $B$ at the loop. Then $ϕ$ is positive and growing, so $\frac{d ϕ}{d t} > 0$ , so $ε < 0$ : the current runs opposite to the clockwise sense you chose, that is, anticlockwise. The three-step method gives the same answer in half the time, but this route never depends on reading a picture correctly.

Figure 9: Falling flux into the page gives a clockwise current in both cases. The near arm then carries current parallel to the wire and is attracted: the loop is pulled towards the stronger field, trying to keep its flux.

Exam Trick

Loop beside a long wire: if the wire current is decreasing, or the loop moves away, the loop is attracted towards the wire; if the current is increasing, or the loop moves closer, it is repelled. The loop always moves the way that would keep its flux constant.

5. Eddy Currents

Nothing in Lenz's law requires a wire. Move a solid block of metal through a changing flux and induced currents swirl around inside it in closed loops. These are eddy currents, named after the whirlpools they resemble, and they oppose the motion exactly as a loop current would.

Figure 10: Eddy currents are Lenz's law inside solid metal. Slots (or laminations) break the big current path into narrow high-resistance ones, so the braking force and the

i^{2} R

heating both fall.

Because a solid plate offers very low resistance, the currents are large and the damping is strong: a copper plate swinging between magnet poles stops within a swing or two. Cut slots in the plate and the large loops are broken into small ones, the resistance of each path rises, the currents fall and the plate swings almost freely. The same trick, applied to a transformer core, is called lamination.

Figure 11: Laminating a transformer or motor core. The field runs along the sheets, the varnish blocks the current from crossing between them, so each eddy loop is thin and carries little current.

Where it is used	What Lenz's law is doing
Electromagnetic braking in trains and roller coasters	Eddy currents in a metal disc or rail oppose the motion and bring it to a smooth, contact-free stop
Induction furnace	Strong eddy currents are deliberately produced in the metal charge, and their $i^{2} R$ heating melts it
Induction cooktop	Eddy currents heat the steel pan directly while the glass top stays cool
Dead-beat galvanometer	The coil is wound on a metal frame; eddy currents damp the swing so the pointer settles without oscillating
Analogue energy meter	The shiny aluminium disc turns because of eddy currents, and a braking magnet gives it a steady speed
Transformer and motor cores (the harmful side)	Eddy currents waste energy as heat, so the core is built from thin laminations separated by varnish

Figure 12: In copper the retarding eddy-current force grows with speed until it equals

m g

, so

v = v_{T} (1 - e^{- g t / v_{T}})

levels off (illustrative

v_{T} = 0.6 m s^{- 1}

, reached in about

184 ms

). In plastic there is no current and

v = g t

Quick Recall: tap to check

A magnet falls through a ring that has a cut in it. What is its acceleration?

g

: an emf appears, but no current can flow, so there is no opposing force.

A closed loop moves at constant velocity entirely inside a uniform field. Is there a current?

No. The flux through it is constant.

Why is a transformer core laminated?

To break up eddy-current paths, which cuts the

i^{2} R

heat loss.

A loop is leaving a field region. Which way does the magnetic force on it point?

Backwards, opposite to its velocity, exactly as while entering.

Key idea

Eddy currents are Lenz's law inside solid metal: useful for braking, damping and induction heating, and kept small in cores by lamination.

Figure 13: Mind map of Lenz's law. Every branch is the same idea: the induced current opposes the change.

6. Solved Examples

Solved Example 1

A magnetic field perpendicular to the plane of a circular coil points into the page and is increasing steadily with time. Find the direction of the induced current in the coil.

Solution:

Step 1: which way is the flux? Into the page.

Step 2: growing or shrinking? Growing, because $B$ is increasing.

Step 3: oppose the change. The induced current must set up flux out of the page inside the loop, and by the right-hand grip rule that needs an anticlockwise current.

Answer: anticlockwise. Note that the coil is not fighting the field, which is into the page; it is fighting the growth of that field.

Solved Example 2

A long straight wire carries a constant current. Two rectangular loops lie in the same plane as the wire: loop (i) slides parallel to the wire, keeping the same distance from it, and loop (ii) moves directly away from the wire. Find the direction of the induced current in each.

Solution:

Loop (i). Moving parallel to the wire, every point of the loop stays at the same distance, so the field pattern through it does not change and $\frac{d ϕ}{d t} = 0$ .

Answer for (i): no induced current. Motion by itself is not enough; the flux must change.

Loop (ii). The wire's field falls off as $\frac{μ _{0} i}{2 π x}$ , so as the loop moves away the flux through it (into the page, say) decreases. To prop it up, the induced current must add flux into the page.

Answer for (ii): clockwise (the sense that gives flux into the page). The loop is also pulled back towards the wire, as Lenz's law demands.

Solved Example 3

A copper ring is held horizontally and a bar magnet is dropped through it along the axis of the ring. Is the acceleration of the magnet equal to, greater than or less than

g

? What changes if the ring is cut at one point?

Solution:

While the magnet approaches, the flux through the ring grows, so an induced current flows and the near face of the ring becomes a like pole, repelling the magnet. The force on the magnet is upwards.

While the magnet recedes below the ring, the flux shrinks, the current reverses, the near face becomes an unlike pole, and it attracts the magnet, which is again upwards.

So throughout the fall

ma = m g - F \Rightarrow a = g - \frac{F}{m} < g

Answer: the acceleration is less than $g$ at every stage. If the ring is cut, no current can circulate, there is no opposing force, and the magnet falls with $a = g$ . The lost gravitational energy in the closed-ring case has gone into heating the ring.

Solved Example 4

Two identical coaxial circular loops carry equal currents circulating in the same direction. The loops are now moved towards each other. What happens to the current in each loop?

Solution:

Flux bookkeeping. The currents circulate the same way, so the field each loop produces at the other threads it in the same sense as its own field. Bringing them closer increases the mutual flux through each loop.

Apply Lenz's law. Each loop must oppose this increase, so the induced emf drives current in the sense that reduces the existing current.

Answer: the current in each loop decreases. A useful cross-check: same-direction currents attract, so the loops are being pushed together with help from the magnetic force, and the system must give something back, which it does by weakening the currents.

Solved Example 5

A square loop of side

10 cm

and resistance

0.50 Ω

enters a region of uniform field

0.40 T

directed into the page, moving at a steady

5.0 m s^{- 1}

. Find the induced current and its direction, the force needed to keep the speed steady, and the power supplied.

Solution:

Given: $l = 0.10 m$ , $R = 0.50 Ω$ , $B = 0.40 T$ , $v = 5.0 m s^{- 1}$ .

emf and current:

ε = B v l = (0.40) (5.0) (0.10) = 0.20 V i = \frac{ε}{R} = \frac{0.20}{0.50} = 0.40 A

The flux into the page is increasing, so the current is anticlockwise.

Force: only the arm inside the field feels a force,

F = B i l = (0.40) (0.40) (0.10) = 0.016 N

directed backwards, so an equal forward force must be applied.

Power: $P = F v = (0.016) (5.0) = 0.080 W$ , which matches $i^{2} R = (0.40)^{2} (0.50) = 0.080 W$ .

Answer: $0.40 A$ anticlockwise, $F = 0.016 N$ , $P = 0.080 W$ . The two ways of computing the power agreeing is the check that Lenz's law balances the energy books.

Solved Example 6

The magnetic flux through a loop is directed out of the page and is decreasing steadily. The induced current in the loop is
(A) clockwise
(B) anticlockwise
(C) zero
(D) clockwise at first and then anticlockwise

Solution:

The flux out of the page is being lost, so the loop must replace it: the induced current has to produce flux out of the page inside the loop.

By the right-hand grip rule, flux out of the page needs an anticlockwise current.

Answer: (B). Option (A) is the trap for students who think the induced current always opposes the field itself rather than the change. Since the flux decreases steadily, nothing reverses, which rules out (D).

Solved Example 7

A flexible circular loop lies in a uniform magnetic field directed into the page. The loop is pulled from opposite sides so that its radius shrinks steadily. Find the direction of the induced current.

Solution:

Flux: $ϕ = B π r^{2}$ , into the page. With $B$ fixed and $r$ shrinking, the flux into the page is decreasing.

Oppose the change: the loop must try to keep the into-the-page flux up, so the induced current has to produce flux into the page inside itself.

Answer: clockwise. Compare this with Solved Example 1: the same field direction gives the opposite current sense, purely because the flux is falling instead of rising.

Solved Example 8

A bar magnet is dropped down a long vertical copper pipe and an identical magnet is dropped down a plastic pipe of the same size. Which takes longer to fall through, and where does the difference in energy go?

Solution:

Copper pipe. The pipe behaves like a stack of closed rings. As the magnet falls, each ring in turn sees the flux rise and then fall, so eddy currents circulate in the pipe wall. By Lenz's law these currents oppose the motion, producing a retarding force that grows with speed.

The magnet quickly reaches a terminal velocity, when the retarding force equals $m g$ , and then drifts down slowly at constant speed.

Plastic pipe. Plastic is an insulator, no current can flow, no opposing force appears, and the magnet is in free fall.

Answer: the magnet takes far longer in the copper pipe. The gravitational energy it does not gain as kinetic energy has been dissipated as heat in the copper, exactly accounting for the difference.

Solved Example 9

A long vertical wire carries a steady current upwards. A rectangular loop lies to the right of the wire, in the same plane, with its long sides parallel to the wire. The current in the wire is now switched off. Find the direction of the induced current in the loop and the direction of the force on the loop while the current dies away.

Solution:

Field at the loop. For an upward current, the field to the right of the wire points into the page.

What is changing. Switching off means the flux into the page through the loop is falling to zero.

Direction of current. To hold that flux up, the induced current must add flux into the page inside the loop, which by the grip rule means clockwise.

Force. Going clockwise, the near (left) arm of the loop carries current upwards, the same direction as the wire current, and parallel currents attract.

Answer: clockwise, and the loop is pulled towards the wire. This is Lenz's law again: by moving into the stronger-field region the loop is trying to keep its flux from falling.

Solved Example 10

A conducting rod of length

l

slides at constant speed

v

on frictionless rails closed by a resistance

R

, in a field

B

perpendicular to the plane of the rails. Show that the work done per second by the agent pulling the rod equals the heat produced per second in

R

Solution:

Induced emf: $ε = B v l$ , so the current is $i = \frac{B v l}{R}$ .
Magnetic force on the rod: $F_{mag} = B i l = \frac{B ^{2} l ^{2} v}{R}$ , directed against the motion (Lenz's law).
Since the speed is constant, the agent must apply an equal forward force, so the power it supplies is
$P_{agent} = F_{mag} v = \frac{B ^{2} l ^{2} v ^{2}}{R}$
Heat produced per second:
$P_{heat} = i^{2} R = (\frac{B v l}{R})^{2} R = \frac{B ^{2} l ^{2} v ^{2}}{R}$

Answer: the two are equal, so all the mechanical work becomes heat. If Lenz's law had the opposite sign, $F_{mag}$ would push the rod along and the same heat would still be produced, creating energy from nothing.

Solved Example 11

A bar magnet is released from rest above a horizontal closed copper ring and falls through it along its axis. The graph of the emf induced in the ring against time is best described as
(A) two equal pulses of the same sign
(B) two pulses of opposite sign, the second taller and narrower
(C) two pulses of opposite sign and equal height
(D) a single pulse while the magnet is inside the ring

Solution:

While the magnet approaches, the flux through the ring rises; while it leaves, the flux falls. So the two pulses have opposite signs, and the emf is zero at the instant the magnet is in the plane of the ring, where the flux is largest.

The magnet speeds up as it falls, so it leaves faster than it arrived: the same flux change happens in less time, giving a taller and narrower second pulse. The areas under the two pulses are equal, because $∣Δ ϕ ∣$ is the same.

Answer: (B). See the computed graph in Section 3.

Solved Example 12

A square loop is pulled at constant velocity through a region of uniform magnetic field that is wider than the loop. The external force that must be applied to the loop is
(A) along

v

at all times
(B) along

v

while entering and while leaving, zero while fully inside
(C) along

v

while entering, opposite to

v

while leaving
(D) zero at all times

Solution:

While entering and while leaving, a current flows and the magnetic force on the loop points backwards (Lenz's law). To keep the velocity constant the applied force must balance it, so it points along $v$ in both stages.

Fully inside, the flux is constant, there is no current and no magnetic force, so no applied force is needed.

Answer: (B). (C) is the trap for students who think the force reverses when the current reverses.

Solved Example 13

A coil of

50

turns, each of area

20 cm^{2}

, and resistance

10 Ω

lies in the plane of the page. A magnetic field out of the page through it increases uniformly from

0

0.50 T

0.20 s

. Find the induced emf, the current, and its direction as seen by the reader.

Solution:

emf: $∣ ε ∣ = N A \frac{Δ B}{Δ t} = (50) (20 \times 1 0^{- 4}) \frac{0.50}{0.20} = 0.25 V$ .

Current: $i = \frac{0.25}{10} = 0.025 A = 25 mA$ .

Direction: the flux out of the page is increasing, so the induced current must make flux into the page, which needs a clockwise current.

Answer: $0.25 V$ , $25 mA$ , clockwise.

Solved Example 14

A square loop of side

10 cm

, mass

20 g

and resistance

0.50 Ω

hangs in a vertical plane with its upper arm inside a horizontal uniform field of

2.0 T

(into the page) and its lower part below the field region. It is released and falls out of the field. Find the terminal speed it reaches, the current at that speed and its direction, and check the energy balance (

g = 9.8 m s^{- 2}

Solution:

Terminal speed: the retarding force on the upper arm, $\frac{B ^{2} l ^{2} v}{R}$ , balances the weight:

v_{T} = \frac{m g R}{B ^{2} l ^{2}} = \frac{( 0.020 ) ( 9.8 ) ( 0.50 )}{( 2.0 ) ^{2} ( 0.10 ) ^{2}} = 2.45 m s^{- 1}

Current: $i = \frac{B v _{T} l}{R} = \frac{( 2.0 ) ( 2.45 ) ( 0.10 )}{0.50} = 0.98 A$ . The flux into the page is falling, so the current is clockwise; the force $B i l = (2.0) (0.98) (0.10) = 0.196 N$ on the upper arm points up and equals $m g$ .

Energy: power lost by gravity $m g v_{T} = (0.196) (2.45) = 0.48 W$ ; heat produced $i^{2} R = (0.98)^{2} (0.50) = 0.48 W$ .

Answer: $2.45 m s^{- 1}$ , $0.98 A$ clockwise, and the two powers agree, as Lenz's law requires.

Practice Questions

A bar magnet is dropped with its south pole downwards towards a horizontal copper ring. Which pole appears on the top face of the ring, and is the acceleration more or less than $g$ ?Answer: South pole on the top face (it repels the approaching south pole), and the acceleration is less than $g$ .
A closed loop moves with constant velocity in a uniform magnetic field, staying entirely inside the field region. Is there an induced current?Answer: No. The flux through the loop is constant, so $d ϕ / d t = 0$ even though the loop is moving.
Why is the core of a transformer made from thin laminated sheets rather than one solid block of iron?Answer: To break up eddy-current paths. Each lamination has a much higher resistance, so the induced currents and the $i^{2} R$ heat loss fall sharply.
A circular loop of radius $8.0 cm$ and resistance $2.0 Ω$ lies in a field of $0.50 T$ out of the page, which falls uniformly to zero in $0.10 s$ . Find the induced current and its direction.Answer: $A = 2.01 \times 1 0^{- 2} m^{2}$ , so $∣ ε ∣ = 0.10 V$ and $i = 5.0 \times 1 0^{- 2} A$ , flowing anticlockwise.
A square loop of side $20 cm$ and resistance $0.10 Ω$ is pulled out of a field of $0.30 T$ at a steady $1.0 m s^{- 1}$ . Find the force that must be applied.Answer: $ε = 0.060 V$ , $i = 0.60 A$ , so $F = B i l = 0.036 N$ in the direction of motion.
A metal plate swinging between the poles of a strong magnet comes to rest within a couple of swings. What happens if several slots are cut in the plate, and why?Answer: It swings almost freely, because the slots break the large eddy-current loops into small high-resistance ones, so the opposing force becomes small.
A square loop is pulled out of a field region at a steady speed. If a loop of the same size but twice the resistance is used, how do the force needed and the power supplied change?Answer: both halve, since $F = \frac{B ^{2} l ^{2} v}{R}$ and $P = \frac{B ^{2} l ^{2} v ^{2}}{R}$ .

Common Mistakes to Avoid

Watch out

Saying the induced current opposes the field. It opposes the change in flux; when the flux is falling, the induced current supports the original field.
Forgetting that a loop moving inside a uniform field region has constant flux, so no emf at all, even though it is clearly moving.
Using the left hand, or reading the loop from behind the page, when applying the grip rule. Always decide clockwise or anticlockwise as seen by the reader.
Assuming a magnet dropped through any ring falls with $a = g$ . That is only true for a cut ring or an insulating ring.
Thinking the retarding force reverses when the induced current reverses. The current flips as a loop goes from entering to leaving, but the force still opposes the motion.
Treating Lenz's law as a separate rule from energy conservation. It is the statement that induction cannot create energy.
Believing eddy currents are always a nuisance. They are deliberately used in braking, induction heating and damping.
Mixing up the pole on the near face with the pole of the approaching magnet. The near face copies an approaching pole and opposes a receding one.

Frequently Asked Questions

What is Lenz's law in simple words?

Lenz's law says the induced current always flows in the direction that fights the change producing it. Push a magnet towards a coil and the coil pushes back; pull it away and the coil holds on. It is the minus sign in $ε = - N d ϕ / d t$ .

Does the induced current oppose the magnetic flux or the change in flux?

The change, always. If the flux through a loop is increasing, the induced current makes flux the opposite way. If it is decreasing, the induced current makes flux the same way as the original field, trying to keep it alive. Confusing the two flips every answer.

Is Lenz's law a consequence of conservation of energy?

Yes. If the induced current helped the change instead of opposing it, a magnet nudged towards a coil would be pulled in faster and faster while heat poured out of the coil, creating energy from nothing. Lenz's law is exactly the sign that prevents that.

How do I decide clockwise or anticlockwise quickly?

Learn one case: flux into the page and increasing gives an anticlockwise current. Flip the answer if the field is out of the page instead, and flip again if the flux is decreasing instead. Two flips return you to the original answer.

Why does a magnet fall slowly through a copper pipe?

The pipe acts like a stack of closed rings. Eddy currents induced in the walls oppose the magnet's motion, giving a retarding force that grows with speed until it balances the weight. The magnet then drifts down at a constant terminal velocity.

What are eddy currents and are they useful or harmful?

Eddy currents are closed loops of induced current inside a solid conductor. They are harmful in transformer and motor cores, where they waste energy as heat, so cores are laminated. They are useful in magnetic braking, induction furnaces, induction cooktops and dead-beat galvanometers.

How is Lenz's law asked in NEET?

Almost always as a quick reasoning question: the direction of current in a coil for a given magnet motion, the pole formed on the near face, whether a falling magnet has acceleration less than $g$ , or an assertion and reason pair linking Lenz's law to conservation of energy.

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

JEE Main asks for current directions near a long wire and for the force on a loop entering a field. JEE Advanced combines Lenz's law with mechanics: terminal velocity of a falling loop, exponential decay of speed on rails, and energy balance between work done and heat produced.

Previous year questions on Lenz’s Law

1 question from past papers, each with a step-by-step solution.

NEET 2024, Physics Q18

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