Lenz Law and Conservation of Energy
Lenz Law is the direction rule of electromagnetic induction. Faraday's law gives the size of induced EMF, while Lenz Law tells the direction of induced current and explains why the current always opposes the change in magnetic flux that produced it. This page connects the statement, negative sign, diagrams, formulas and exam-style practice for NEET, JEE Main, JEE Advanced, CBSE, IB, ICSE, IGCSE and A-Level Physics.
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1. Introduction
Whenever magnetic flux linked with a coil, conducting loop, solenoid or moving rod changes, an induced EMF is produced. If the circuit is closed, induced current flows. Lenz Law decides the direction of that current by asking one question: what magnetic effect will oppose the change in flux?
Faraday gives magnitude
Use the rate of change of flux linkage to find the size of induced EMF.
Lenz gives direction
Decide whether the induced field must increase or decrease the original flux.
Energy stays conserved
External work is converted into electrical energy and heat; energy is not created for free.
2. How to Use This Page
Step 1: Find flux change
Check whether magnetic flux is increasing, decreasing or constant.
Step 2: Apply opposition
Choose the induced magnetic field that opposes that change.
Step 3: Convert field to current
Use the right-hand grip rule or Fleming's right hand rule when needed.
Step 4: Calculate magnitude
Use e = N dΦ/dt, e = Blv, I = e/R and P = I²R.
Step 5: Check energy
The magnetic force should oppose the motion or cause producing the flux change.
Step 6: Practice by exam
Use the NEET, JEE, IB, ICSE, IGCSE and A-Level sections according to your course.
3. Statement of Lenz Law
The direction of induced current is always such that it opposes the change in magnetic flux that produces it. It opposes the change, not necessarily the magnetic field itself.
e = −dΦ/dte = −N dΦ/dte = BlvI = e/RF = BIlP_mech = P_elec = I²R4. Lenz Law with Moving Magnet and Coil
For magnet-coil questions, first decide whether the pole is approaching or moving away. The coil face forms a polarity that opposes the change: repulsion for approach and attraction for recession.
5. Galvanometer Animation
Current flows only while magnetic flux changes. If the magnet is at rest relative to the coil, there is no flux change and no induced current.
6. Fleming's Right Hand Rule
Fleming's Right Hand Rule is used to determine induced current direction when a conductor moves in a magnetic field. Thumb shows motion, first finger shows magnetic field and second finger shows induced current.
Lenz Law
Explains why the induced effect opposes the change producing it.
Fleming Rule
Gives current direction directly in conductor-motion generator cases.
7. Conservation of Energy in Lenz Law
If induced current supported the change in flux, the change would reinforce itself and energy could be created without external work. Instead, the induced current opposes the cause. In moving conductor problems, an external force must do work against magnetic drag, and that work becomes electrical energy and heat.
Mechanical Energy → Electrical Energy → Heat EnergyP_mech = Fv = B²l²v²/R = I²R8. Mathematical Derivation of Energy Conservation
e = BlvI = Blv/RF = BIl = B²l²v/RP_mech = FvP_elec = I²RP_mech = P_elec9. Conducting Loop Problems
For loop problems, first decide whether flux is increasing, decreasing or constant. Then choose the induced current direction that opposes that change.
Loop enters field
Flux through the loop increases; induced current opposes the increase.
Loop fully inside
Flux is constant in a uniform field, so net induced current is zero.
Loop leaves field
Flux decreases; induced current tries to maintain the original flux.
10. Solenoid and Moving Magnet Problems
When a magnet enters or leaves a solenoid, the flux linked with the solenoid changes. Faster motion gives greater rate of flux change and larger galvanometer deflection.
11. NCERT-Style Conceptual Checks
NCERT-1What exactly is opposed by induced current?
Level: Conceptual
A strong magnetic field alone does not induce current unless the flux linkage changes.
NCERT-2Why is current zero when a magnet is stationary near a coil?
Level: Conceptual
The coil may have magnetic flux linked with it, but dΦ/dt is zero.
NCERT-3Can induced EMF exist without induced current?
Level: Conceptual
A changing flux can create EMF, but current needs a closed path.
12. Common Mistakes in Lenz Law
Opposing field instead of change
Lenz Law opposes flux change, not every magnetic field.
Ignoring open circuits
Open circuits can have induced EMF but no sustained current.
Forgetting constant flux
A loop fully inside a uniform field has no net induced current if flux is constant.
Wrong page direction
Clockwise current gives into-page field; anticlockwise gives out-of-page field.
Mixing hand rules
Use Fleming's right hand rule for generator direction, not motor force cases.
Skipping energy check
The magnetic force or polarity should oppose the motion or change causing induction.
13. Exam Strategy for NEET, JEE Main and JEE Advanced
NEET
Focus on statement, negative sign, pole-face logic, galvanometer observations, eddy currents and simple e = Blv numericals.
JEE Main
Practice formula chaining: e = Blv, I = e/R, F = BIl, P = I²R and graph slope questions.
JEE Advanced
Draw flux regions carefully, handle variable fields or resistance, and verify direction with energy conservation.
14. NEET Questions
Each NEET question has four options, the correct answer and a short explanation.
NEET-1Which statement best expresses Lenz Law?
- A) Induced current supports the flux change
- B) Induced current opposes the change in magnetic flux
- C) Induced current is always clockwise
- D) Induced current is independent of flux
Difficulty: NEET level
The direction is chosen so that the magnetic effect of induced current resists the flux change that produced it.
NEET-2The minus sign in Faraday's law mainly represents
- A) loss of charge in the circuit
- B) Lenz Law opposition to flux change
- C) the resistance of the wire
- D) the direction of external magnetic field only
Difficulty: NEET level
The negative sign is the mathematical sign of Lenz Law and does not mean that the magnitude of EMF is negative.
NEET-3A north pole of a bar magnet approaches the left face of a coil. The left face of the coil becomes
- A) north
- B) south
- C) uncharged
- D) alternately north and south without motion
Difficulty: NEET level
The approaching north pole increases flux, so the coil forms a north face to repel the approach.
NEET-4A north pole is pulled away from a nearby coil. The coil face near the magnet becomes
- A) north
- B) south
- C) east
- D) zero-pole
Difficulty: NEET level
The coil tries to maintain the decreasing flux by attracting the receding north pole, so its near face is south.
NEET-5A south pole moves away from a coil. What polarity appears on the near face of the coil?
- A) south, to repel it
- B) north, to attract it
- C) no polarity because it is moving away
- D) north only if the circuit is open
Difficulty: NEET level
For decreasing flux due to a receding south pole, the coil attracts the magnet; the near face therefore becomes north.
NEET-6Flux into the page through a closed circular loop is increasing. The induced current is
- A) clockwise
- B) anticlockwise
- C) zero
- D) alternating at every instant
Difficulty: NEET level
An increasing into-page flux is opposed by an out-of-page induced field, produced by anticlockwise current.
NEET-7Flux out of the page through a loop is decreasing. The induced current should be
- A) anticlockwise
- B) clockwise
- C) zero
- D) radially outward
Difficulty: NEET level
The loop tries to maintain the out-of-page flux; anticlockwise current produces an out-of-page magnetic field.
NEET-8A rectangular loop enters a region where magnetic field is into the page. The induced current is
- A) clockwise
- B) anticlockwise
- C) zero
- D) first zero then infinite
Difficulty: NEET level
While entering, into-page flux increases, so the induced field is out of the page and the current is anticlockwise.
NEET-9A loop leaves a region where magnetic field is into the page. The induced current while leaving is
- A) clockwise
- B) anticlockwise
- C) zero
- D) opposite on both vertical sides only
Difficulty: NEET level
The into-page flux is decreasing, so the induced current creates an into-page field, which is clockwise.
NEET-10A closed loop moves with constant velocity completely inside a uniform magnetic field. The induced current is
- A) maximum
- B) zero
- C) proportional to speed only
- D) opposite on alternate halves
Difficulty: NEET level
The flux through the complete loop is constant when field, area and orientation do not change.
NEET-11Eddy-current magnetic braking is an application of
- A) Lenz Law
- B) Coulomb's inverse-square law
- C) Snell's law
- D) Archimedes' principle
Difficulty: NEET level
Eddy currents produce magnetic effects that oppose the motion causing the flux change.
NEET-12In an open coil with changing magnetic flux, which statement is correct?
- A) Induced EMF can exist but current cannot flow
- B) Neither EMF nor current can exist
- C) Current is infinite
- D) Only current exists without EMF
Difficulty: NEET level
Faraday's law can produce EMF in an open circuit, but current requires a closed conducting path.
NEET-13For a straight conductor of length l moving perpendicular to B with speed v, motional EMF is
- A) Blv
- B) Bv/l
- C) lv/B
- D) BR/v
Difficulty: NEET level
When length, velocity and magnetic field are mutually perpendicular, the magnitude is e = Blv.
NEET-14If the resistance of a closed induction circuit is doubled while induced EMF remains the same, current becomes
- A) double
- B) half
- C) four times
- D) zero always
Difficulty: NEET level
Using I = e/R, doubling R halves the induced current for the same EMF.
NEET-15A magnet is moved faster towards a coil. The galvanometer deflection generally becomes larger because
- A) magnetic flux changes faster
- B) wire resistance becomes zero
- C) magnet mass increases
- D) charge is created in the coil
Difficulty: NEET level
Faster motion increases the rate of change of magnetic flux and therefore increases induced EMF.
NEET-16In Fleming's right hand rule, the thumb represents
- A) motion of conductor
- B) magnetic field
- C) induced current
- D) resistance
Difficulty: NEET level
Thumb gives motion, first finger gives magnetic field and second finger gives induced current.
NEET-17Why does Lenz Law agree with conservation of energy?
- A) It removes the need for external work
- B) It makes induced current oppose the cause, so external work is converted into electrical energy
- C) It makes current flow without EMF
- D) It cancels resistance
Difficulty: NEET level
The opposing magnetic effect means an external agent must do work, which becomes electrical energy and heat.
NEET-18If the number of turns of a coil is doubled for the same rate of flux change per turn, induced EMF becomes
- A) half
- B) double
- C) unchanged
- D) zero
Difficulty: NEET level
For a coil, e = N dΦ/dt in magnitude, so EMF is proportional to the number of turns.
NEET-19A loop area increases in a uniform magnetic field directed out of the page. The induced current is
- A) clockwise
- B) anticlockwise
- C) zero
- D) not determined by area change
Difficulty: NEET level
Out-of-page flux increases, so the induced field is into the page, produced by clockwise current.
NEET-20Magnetic flux through a flat surface is written as
- A) BA cos θ
- B) B/A cos θ
- C) B + A + θ
- D) IR
Difficulty: NEET level
Flux is the dot product of magnetic field and area vector: Φ = BA cos θ.
NEET-21Flux through a loop is maximum when
- A) the area vector is parallel to magnetic field
- B) the plane of loop is parallel to magnetic field
- C) the field is zero
- D) resistance is maximum
Difficulty: NEET level
Flux equals BA cos θ, so it is maximum when θ = 0 between B and the area vector.
NEET-22Induced EMF is zero when
- A) magnetic flux is changing rapidly
- B) magnetic flux linked with the circuit is constant
- C) the loop is closed
- D) the conductor has finite resistance
Difficulty: NEET level
Faraday's law gives induced EMF only when flux linkage changes with time.
NEET-23Eddy currents in transformer cores are reduced by using
- A) laminated cores
- B) thicker solid cores
- C) plastic magnets
- D) open circuits only
Difficulty: NEET level
Laminations break large current paths and reduce eddy-current heating.
NEET-24Flux into the page through a loop is decreasing. The induced current is
- A) clockwise
- B) anticlockwise
- C) zero
- D) along the radius
Difficulty: NEET level
The loop tries to maintain into-page flux; clockwise current produces a magnetic field into the page.
NEET-25A loop has induced EMF 2 V and resistance 4 Ω. The induced current is
- A) 0.5 A
- B) 2 A
- C) 6 A
- D) 8 A
Difficulty: NEET level
Ohm's law gives I = e/R = 2/4 = 0.5 A.
15. JEE Main Questions
These questions include conceptual and numerical practice on induced EMF, current, force, power and graph interpretation.
JM-1A 200-turn coil has flux per turn changing from 4.0 mWb to 1.0 mWb in 0.06 s. Find average induced EMF magnitude.
Difficulty / Format: JEE Main numerical/conceptual
e = N|ΔΦ|/Δt- Flux change per turn is 3.0 mWb = 3.0 × 10^-3 Wb.
- e = 200 × 3.0 × 10^-3 / 0.06.
- Therefore e = 10 V.
JM-2A rod of length 0.40 m moves at 6 m/s perpendicular to a 0.50 T field. Find motional EMF.
Difficulty / Format: JEE Main numerical/conceptual
e = Blv- Substitute B = 0.50 T, l = 0.40 m and v = 6 m/s.
- e = 0.50 × 0.40 × 6.
- The motional EMF is 1.2 V.
JM-3A 0.50 m rod moves at 8 m/s in a 0.30 T field and the circuit resistance is 2 Ω. Find induced current.
Difficulty / Format: JEE Main numerical/conceptual
I = Blv/R- First e = Blv = 0.30 × 0.50 × 8 = 1.2 V.
- Current I = e/R = 1.2/2.
- The current is 0.60 A.
JM-4For the current in the previous style of rod circuit, take B = 0.30 T, I = 0.60 A and l = 0.50 m. Find magnetic force on the rod.
Difficulty / Format: JEE Main numerical/conceptual
F = BIl- Use the magnetic force on a current-carrying rod.
- F = 0.30 × 0.60 × 0.50 = 0.09 N.
- By Lenz Law the force opposes the rod's motion.
JM-5A rod circuit has induced current 0.60 A and resistance 2 Ω. Find heat power.
Difficulty / Format: JEE Main numerical/conceptual
P = I^2R- Joule power is I^2R.
- P = (0.60)^2 × 2.
- The dissipated power is 0.72 W.
JM-6For B = 0.40 T, l = 0.50 m, v = 10 m/s and R = 4 Ω, find mechanical power needed to keep the rod moving uniformly.
Difficulty / Format: JEE Main numerical/conceptual
P = B^2l^2v^2/R- In steady motion, external mechanical power equals electrical power.
- P = (0.40)^2 × (0.50)^2 × (10)^2 / 4.
- The required power is 1.0 W.
JM-7A 50-turn coil has flux-time slope -0.020 Wb/s per turn. Find induced EMF magnitude.
Difficulty / Format: JEE Main numerical/conceptual
e = N|dΦ/dt|- The magnitude of slope is 0.020 Wb/s per turn.
- e = 50 × 0.020.
- The EMF magnitude is 1.0 V.
JM-8A square loop of side 0.20 m enters a 0.50 T field at 3 m/s. If R = 0.60 Ω, find induced current while entering.
Difficulty / Format: JEE Main numerical/conceptual
I = Blv/R- The effective length cutting the boundary is the side length 0.20 m.
- e = 0.50 × 0.20 × 3 = 0.30 V.
- I = 0.30/0.60 = 0.50 A.
JM-9A loop area increases at 0.020 m^2/s in a 0.40 T uniform field. With R = 2 Ω, find current magnitude.
Difficulty / Format: JEE Main numerical/conceptual
I = B(dA/dt)/R- Flux rate is dΦ/dt = B dA/dt.
- e = 0.40 × 0.020 = 0.008 V.
- I = 0.008/2 = 0.004 A.
JM-10A 100-turn coil of area 0.010 m^2 is in a field changing at 0.20 T/s perpendicular to the coil. Find EMF.
Difficulty / Format: JEE Main numerical/conceptual
e = NA(dB/dt)- Flux per turn changes at A dB/dt.
- e = 100 × 0.010 × 0.20.
- The induced EMF is 0.20 V.
JM-11A rod of length 0.75 m moving at 4 m/s produces EMF 1.5 V. Find the perpendicular magnetic field.
Difficulty / Format: JEE Main numerical/conceptual
B = e/(lv)- Rearrange e = Blv.
- B = 1.5/(0.75 × 4).
- B = 0.50 T.
JM-12In a moving rod circuit with fixed B, l and R, speed is doubled. What happens to magnetic drag force and power?
Difficulty / Format: JEE Main numerical/conceptual
F = B^2l^2v/R, P = B^2l^2v^2/R- Force is proportional to v.
- Power is proportional to v^2.
- Therefore doubling speed doubles force and quadruples power.
JM-13A 20-turn coil with total resistance 5 Ω experiences flux change 0.050 Wb per turn. Find total charge passed.
Difficulty / Format: JEE Main numerical/conceptual
q = NΔΦ/R- Total charge is independent of time for a given net flux change.
- q = 20 × 0.050 / 5.
- q = 0.20 C.
JM-14A changing flux links an open circuit. What are induced EMF, current and power in the ideal open circuit?
Difficulty / Format: JEE Main numerical/conceptual
I = e/R with R effectively infinite- Faraday's law gives EMF from changing flux.
- In an open circuit, no continuous path exists for current.
- With I = 0, electrical power in the circuit is zero.
JM-15Flux out of the page through a loop is decreasing. State induced field and current direction.
Difficulty / Format: JEE Main numerical/conceptual
Right-hand grip rule plus Lenz Law- The loop opposes the decrease by reinforcing out-of-page flux.
- An out-of-page magnetic field is produced by anticlockwise current.
- Hence the current is anticlockwise.
JM-16A rectangular loop with side length 0.20 m is pulled out of a 0.30 T field at 5 m/s. Total resistance is 1.5 Ω. Find current magnitude.
Difficulty / Format: JEE Main numerical/conceptual
I = Blv/R- Only the side crossing the field boundary contributes to changing area.
- e = 0.30 × 0.20 × 5 = 0.30 V.
- I = 0.30/1.5 = 0.20 A.
JM-17Why does magnetic braking force approximately increase with speed at low speeds?
Difficulty / Format: JEE Main numerical/conceptual
e = Blv, I = e/R, F = BIl- Motional EMF is proportional to speed.
- For fixed resistance, current is also proportional to speed.
- Magnetic drag force is therefore proportional to speed.
JM-18Flux through a 30-turn coil is Φ = 0.010 cos(100t) Wb per turn. Find maximum EMF.
Difficulty / Format: JEE Main numerical/conceptual
e_max = NΦ0ω- For sinusoidal flux, maximum |dΦ/dt| is Φ0ω.
- e_max = 30 × 0.010 × 100.
- Maximum EMF is 30 V.
JM-19A 100-turn coil of area 0.020 m^2 rotates from area vector parallel to B to 60° in 0.50 s. B = 0.20 T. Find average EMF magnitude.
Difficulty / Format: JEE Main numerical/conceptual
e_avg = N|BA(cosθ2 - cosθ1)|/Δt- Initial cosθ is 1 and final cosθ is cos60° = 0.5.
- Flux change per turn is 0.20 × 0.020 × 0.5 = 0.002 Wb.
- e = 100 × 0.002/0.50 = 0.40 V.
JM-20For B = 0.20 T, l = 0.50 m, v = 10 m/s and R = 0.50 Ω, find external force for constant speed.
Difficulty / Format: JEE Main numerical/conceptual
F = B^2l^2v/R- The external force balances magnetic drag.
- F = (0.20)^2 × (0.50)^2 × 10 / 0.50.
- F = 0.20 N.
JM-21An induced current of 0.50 A flows through 4 Ω for 3 s. Find heat produced.
Difficulty / Format: JEE Main numerical/conceptual
H = I^2Rt- Use Joule heating.
- H = (0.50)^2 × 4 × 3.
- Heat produced is 3 J.
JM-22A loop moves entirely inside a uniform magnetic field. Explain why no net EMF appears in the loop.
Difficulty / Format: JEE Main numerical/conceptual
e_loop = -dΦ/dt- The loop area, orientation and field are unchanged.
- Therefore dΦ/dt = 0 for the closed loop.
- Individual side effects cancel around the complete loop.
JM-23If induced EMF is fixed and resistance is doubled, what happens to heat power?
Difficulty / Format: JEE Main numerical/conceptual
P = e^2/R- For a fixed EMF, power is inversely proportional to resistance.
- Doubling R gives P/2.
- This is different from the case of fixed current.
JM-24A circular loop of radius 0.10 m and resistance 0.50 Ω is in a field into the page increasing at 2 T/s. Find current and direction.
Difficulty / Format: JEE Main numerical/conceptual
I = πr^2(dB/dt)/R- Area is π(0.10)^2 = 0.0314 m^2.
- e = 0.0314 × 2 = 0.0628 V and I = 0.0628/0.50 = 0.126 A.
- The increasing into-page flux is opposed by an out-of-page field, so current is anticlockwise.
JM-25External mechanical power supplied to a rod circuit is 8 W and resistance is 2 Ω. Find induced current and EMF.
Difficulty / Format: JEE Main numerical/conceptual
P = I^2R and e = IR- I = sqrt(P/R) = sqrt(8/2) = 2 A.
- e = IR = 2 × 2 = 4 V.
- The mechanical power is converted into electrical heating.
16. JEE Advanced Style Original Questions
These are original practice questions in multiple-correct, integer-type, assertion-reason, matching-type, paragraph-based, graph-based and multi-step numerical formats.
JA-1Multiple-correct: A rod of length l moves on rails in uniform B with v = kt and resistance R. Choose all correct relations.
Difficulty tag: Hard, multiple-correct
Key concept tested: Variable motional EMF and power
- Motional EMF is e = Blv = Blkt.
- Current is I = e/R = Blkt/R.
- The magnetic force is BIl = B^2l^2kt/R and by Lenz Law it opposes motion.
- Heat is integral of I^2R dt from 0 to T, giving B^2l^2k^2T^3/(3R).
JA-2Integer-type: B = 0.5 T, l = 0.4 m, R = 0.2 Ω and v = 5t. Find magnetic drag at t = 2 s in newton.
Difficulty tag: Hard, integer-type
Key concept tested: Drag force in variable-speed rod
- At t = 2 s, v = 10 m/s.
- F = B^2l^2v/R.
- F = (0.5)^2 × (0.4)^2 × 10 / 0.2 = 2 N.
JA-3Assertion-reason: Assertion: Lenz Law is necessary for conservation of energy. Reason: If induced current helped the flux change, the change would reinforce itself without external work.
Difficulty tag: Medium, assertion-reason
Key concept tested: Energy logic behind the negative sign
- A helping induced current would increase the original flux change.
- That would permit self-amplifying current and energy creation without work.
- The observed opposing direction prevents this and requires external work.
JA-4Matching-type: Match flux change cases with induced current direction as seen by the observer.
Difficulty tag: Hard, matching-type
Key concept tested: Direction traps for increasing and decreasing flux
- Let option 1 be clockwise and option 2 be anticlockwise.
- P: into-page flux increasing needs out-of-page induced field, so anticlockwise.
- Q: into-page flux decreasing needs into-page induced field, so clockwise.
- R: out-of-page flux increasing needs into-page induced field, so clockwise.
- S: out-of-page flux decreasing needs out-of-page induced field, so anticlockwise.
JA-5Paragraph-based: A square loop of side 0.20 m enters a wide 1.0 T field at 3 m/s. R = 0.60 Ω. Find current, drag force and heat during entry.
Difficulty tag: Hard, paragraph-based
Key concept tested: Loop entering finite magnetic field
- During entry, e = Blv = 1.0 × 0.20 × 3 = 0.60 V.
- I = e/R = 0.60/0.60 = 1 A.
- Magnetic drag on the active side is F = BIl = 1.0 × 1 × 0.20 = 0.20 N.
- Entry time is side/speed = 0.20/3 s. Heat is I^2Rt = 1 × 0.60 × 0.20/3 = 0.040 J.
JA-6Graph-based: Flux per turn rises linearly from 0 to 0.060 Wb in 0.030 s for a 10-turn coil, then remains constant. R = 5 Ω. Find current in both intervals.
Difficulty tag: Medium, graph-based
Key concept tested: Flux-time slope and zero-current plateau
- During the rising part, dΦ/dt = 0.060/0.030 = 2 Wb/s.
- e = N dΦ/dt = 10 × 2 = 20 V.
- I = 20/5 = 4 A.
- On the flat part, dΦ/dt = 0, so current is zero.
JA-7Non-uniform field: A rod of length 0.50 m moves at 4 m/s through B(x) = B0x/a, with B0 = 0.80 T. Find EMF at x = a/2.
Difficulty tag: Hard, numerical
Key concept tested: Motional EMF in position-dependent field
- At x = a/2, B = B0/2 = 0.40 T.
- Use e = B(x)lv for the rod at that position.
- e = 0.40 × 0.50 × 4 = 0.80 V.
JA-8Time-varying field with changing area: B = 3t into the page and loop area A = 0.01(1 + t). Find EMF magnitude and direction at t = 2 s.
Difficulty tag: Hard, multi-step numerical
Key concept tested: Product rule for flux
- Flux is Φ = BA, so dΦ/dt = A dB/dt + B dA/dt.
- At t = 2 s, A = 0.03 m^2, dB/dt = 3 T/s, B = 6 T and dA/dt = 0.01 m^2/s.
- dΦ/dt = 0.03 × 3 + 6 × 0.01 = 0.15 Wb/s.
- Into-page flux is increasing, so induced field is out of page and current is anticlockwise.
JA-9Variable resistance: A rod moves with constant speed in B. The circuit resistance is R0(1 + αt). Describe I(t) and power.
Difficulty tag: Hard, reasoning
Key concept tested: Changing resistance in a motional EMF circuit
- The motional EMF Blv is constant because B, l and v are constant.
- Current follows Ohm's law with time-dependent resistance.
- Power is e^2/R(t), so it decreases as resistance increases.
JA-10Magnetic braking model: If drag force is F = -kv on a mass m initially at speed v0, find v(t) and stopping distance limit.
Difficulty tag: Hard, derivation
Key concept tested: Lenz-law damping as velocity-proportional drag
- Newton's law gives m dv/dt = -kv.
- Solving the differential equation gives v = v0e^(-kt/m).
- The distance is integral of v dt from 0 to infinity, which equals mv0/k.
JA-11Mutual induction direction trap: Primary current in a nearby loop is anticlockwise and increasing. What is the induced current in a coaxial secondary loop?
Difficulty tag: Medium, conceptual
Key concept tested: Mutual induction and Lenz Law
- An anticlockwise primary current produces out-of-page magnetic flux through the secondary.
- Since the primary current is increasing, out-of-page flux is increasing.
- The secondary opposes this increase by producing into-page field, requiring clockwise current.
JA-12Vertical rails: A conducting rod of mass m slides down under gravity in a perpendicular field B. Find terminal speed if rail separation is l and resistance is R.
Difficulty tag: Hard, derivation
Key concept tested: Energy conversion at terminal speed
- At speed v, current is I = Blv/R.
- Magnetic drag is F = BIl = B^2l^2v/R.
- At terminal speed, mg = B^2l^2v/R.
- Therefore v_t = mgR/(B^2l^2).
JA-13Integer-type terminal speed: m = 0.10 kg, R = 2 Ω, B = 0.50 T, l = 0.40 m and g = 10 m/s^2. Find terminal speed in m/s.
Difficulty tag: Hard, integer-type
Key concept tested: Terminal speed of a falling conducting rod
- Use v_t = mgR/(B^2l^2).
- mgR = 0.10 × 10 × 2 = 2.
- B^2l^2 = 0.25 × 0.16 = 0.04.
- v_t = 2/0.04 = 50 m/s.
JA-14Finite-field direction: A loop enters, then is fully inside, then leaves a region where B is out of the page. State current direction in each stage.
Difficulty tag: Medium, direction-trap
Key concept tested: Entering, constant-flux and leaving cases
- While entering, out-of-page flux increases, so induced current is clockwise.
- When fully inside a uniform field, flux is constant and induced current is zero.
- While leaving, out-of-page flux decreases, so induced current is anticlockwise.
JA-15Graph/numerical: Flux through each turn is Φ = 0.020t^2 Wb for a 50-turn coil of resistance 10 Ω. Find EMF, current and power at t = 3 s.
Difficulty tag: Hard, graph-based numerical
Key concept tested: Differentiating flux and calculating power
- dΦ/dt = 0.040t, so at t = 3 s it is 0.12 Wb/s.
- e = N dΦ/dt = 50 × 0.12 = 6 V.
- I = e/R = 6/10 = 0.60 A.
- P = I^2R = 0.36 × 10 = 3.6 W.
17. IB Physics Questions
IB-1Explain Lenz Law using the idea of energy conservation.
Level: IB Physics structured
If it supported the change, electrical energy could appear without work. Opposition means mechanical work is converted into electrical energy and thermal energy.
IB-2A magnet is moved into a coil connected to a galvanometer, then held still. Describe the observations.
Level: IB Physics structured
Only changing flux induces EMF. A stationary magnet may produce flux, but not a changing flux.
IB-3A flux-time graph is a straight line with positive gradient. What does the gradient represent?
Level: IB Physics structured
For N turns, multiply the magnitude of the gradient by N. The sign is interpreted using Lenz Law.
IB-4State two assumptions in a simple e = Blv rod experiment.
Level: IB Physics structured
These assumptions make the calculated EMF and power match the ideal equations.
IB-5Why does timing uncertainty matter in average EMF experiments?
Level: IB Physics structured
A smaller timing uncertainty or repeated trials improve reliability.
IB-6Compare induced EMF in open and closed circuits.
Level: IB Physics structured
An open circuit has no continuous path, so charge separation may exist without current flow.
IB-7Explain magnetic braking in terms of eddy currents.
Level: IB Physics structured
The kinetic energy lost by the object becomes thermal energy in the conductor.
IB-8A 40-turn coil has flux per turn changing at 0.025 Wb/s. Calculate EMF magnitude.
Level: IB Physics structured
Use e = N dΦ/dt = 40 × 0.025 = 1.0 V.
IB-9A north pole approaches a coil. Explain the polarity at the near face.
Level: IB Physics structured
A north face repels the approaching north pole and opposes the increase in flux.
IB-10Suggest one improvement to a classroom Lenz Law demonstration.
Level: IB Physics structured
This reduces subjective observation and helps relate deflection to rate of flux change.
18. ICSE / IGCSE Questions
ICSE-Level Questions
ICSE-1State Lenz Law in one sentence.
Level: ICSE level
Mention opposition to change in magnetic flux, not simply opposition to magnetic field.
ICSE-2What happens to galvanometer deflection when a magnet is held stationary inside a coil?
Level: ICSE level
There is flux, but no change of flux, so induced current is zero.
ICSE-3A north pole approaches a coil. What pole is produced on the near face?
Level: ICSE level
The coil repels the approaching north pole to oppose the motion.
ICSE-4Why are transformer cores laminated?
Level: ICSE level
Thin insulated sheets break large circulating current paths.
ICSE-5Write the formula for magnetic flux through a flat coil.
Level: ICSE level
θ is the angle between magnetic field and the area vector.
ICSE-6Flux changes by 0.12 Wb in 0.40 s through a single loop. Find average EMF.
Level: ICSE level
Average EMF magnitude is ΔΦ/Δt = 0.12/0.40 = 0.30 V.
ICSE-7A magnet is withdrawn quickly from a coil. How does deflection compare with slow withdrawal?
Level: ICSE level
A faster motion gives a larger rate of flux change.
ICSE-8Why is a closed circuit needed for induced current?
Level: ICSE level
Changing flux can create EMF, but charges cannot circulate unless the path is closed.
ICSE-9A loop enters a magnetic field region. Is flux through it increasing or decreasing?
Level: ICSE level
A larger part of the loop becomes linked with the field as it enters.
ICSE-10Differentiate Faraday's law and Lenz Law.
Level: ICSE level
Together they form e = -dΦ/dt.
IGCSE-Level Questions
IGCSE-1What condition is required for electromagnetic induction?
Level: IGCSE level
This can happen by moving a magnet, moving a conductor or changing field strength.
IGCSE-2Why does moving a magnet faster increase induced voltage?
Level: IGCSE level
Induced voltage depends on how quickly flux linkage changes.
IGCSE-3What is the effect of increasing the number of turns in a coil?
Level: IGCSE level
More turns mean greater flux linkage change for the same motion.
IGCSE-4Which hand rule is commonly used for generator current direction?
Level: IGCSE level
It connects motion, magnetic field and induced current direction.
IGCSE-5Flux changes by 0.060 Wb in 0.30 s. Find average induced voltage in one turn.
Level: IGCSE level
e = ΔΦ/Δt = 0.060/0.30 = 0.20 V.
IGCSE-6Why does Lenz Law support conservation of energy?
Level: IGCSE level
The work done becomes electrical or thermal energy.
IGCSE-7A galvanometer deflects one way when a magnet enters a coil. What happens when it leaves by the same end?
Level: IGCSE level
The flux change reverses, so the induced current direction reverses.
IGCSE-8Why does a copper plate slow down between strong magnets?
Level: IGCSE level
The plate's kinetic energy is converted into heat.
IGCSE-9How does an iron core inside a solenoid affect induction?
Level: IGCSE level
Greater linked flux change gives greater induced EMF.
IGCSE-10Can there be induced current in an open circuit?
Level: IGCSE level
An induced voltage may exist, but a complete path is needed for current.
19. A-Level Physics Questions
AL-1Define flux linkage for an N-turn coil.
Level: A-Level Physics
Flux linkage is the total magnetic flux linked with all turns of the coil.
AL-2A coil has flux linkage changing from 0.80 Wb-turn to 0.20 Wb-turn in 0.15 s. Find average EMF.
Level: A-Level Physics
Average EMF magnitude is change in flux linkage divided by time: 0.60/0.15 = 4.0 V.
AL-3A coil rotates so that its area vector moves from parallel to B to perpendicular to B. What happens to flux?
Level: A-Level Physics
Because Φ = BA cos θ and θ changes from 0° to 90°.
AL-4A rod circuit dissipates 18 W as heat while moving at constant speed. What mechanical power is supplied ideally?
Level: A-Level Physics
At constant speed in an ideal circuit, supplied mechanical power equals electrical power dissipated.
AL-5In an experiment, EMF is plotted against rod speed. What graph is expected?
Level: A-Level Physics
Since e = Blv, EMF is directly proportional to speed for fixed B and l.
AL-6Why is an external force required to keep a rod moving in a magnetic field?
Level: A-Level Physics
This is the mechanical expression of Lenz Law.
AL-7A flux linkage-time graph has gradient -3 Wb-turn/s. What is induced EMF?
Level: A-Level Physics
The equation is e = -d(NΦ)/dt.
AL-8The percentage uncertainties in B, l and v are 3%, 2% and 4%. Estimate percentage uncertainty in e = Blv.
Level: A-Level Physics
For multiplication, percentage uncertainties add: 3 + 2 + 4 = 9%.
AL-9A secondary coil experiences increasing flux produced by a primary coil. How does Lenz Law determine secondary current?
Level: A-Level Physics
The exact clockwise or anticlockwise direction depends on the direction of the primary flux.
AL-10Derive the terminal speed of a falling rod on rails in a magnetic field.
Level: A-Level Physics
At terminal speed, mg equals magnetic drag BIl. With I = Blv/R, mg = B^2l^2v/R.
20. FAQs on Lenz Law and Conservation of Energy
FAQ-1What is Lenz Law?
FAQ-2What does the negative sign in Faraday's law mean?
FAQ-3Does induced current oppose magnetic field or change in flux?
FAQ-4Why does Lenz Law obey conservation of energy?
FAQ-5Can induced EMF exist in an open circuit?
FAQ-6When is induced current zero in a moving loop?
21. Final Revision Sheet
Induced current opposes flux changee = −dΦ/dte = −N dΦ/dte = BlvI = Blv/RF = BIlP = I²R = Fve = 0Need Personal Help?
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