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Lenzs Law Motional Emf

🎓 Class 12 Physics CBSE Theory Ch 6 – Electromagnetic Induction ⏱ ~14 min
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Lenzs Law Motional Emf

6.5 Lenz's Law and Conservation of Energy

Faraday's law tells us how big the EMF is, but the negative sign requires interpretation. The German physicist Heinrich Lenz (1834) gave the rule for the direction of the induced current.

Lenz's law: The polarity of the induced EMF is such that the induced current opposes the change in flux that produced it.

In short: nature opposes the change. If flux is increasing, the induced current circulates so as to set up a field opposing it; if flux is decreasing, the induced current tries to maintain the original flux.

Lenz's law is a direct consequence of conservation of energy: if the induced current aided the change in flux, you could extract energy from the coil for free, in violation of the first law of thermodynamics.

Coil (loop seen on edge) N S v → Induced current N (face)
Fig 6.3 As the bar magnet's N-pole approaches, the coil face nearer the magnet becomes a "north" face by induction; the induced current circulates anticlockwise (as seen from the magnet) to repel the approaching N - exactly opposing the increase in flux.

Worked Example 6.4 - Direction of induced current

Example 6.4 L4 Analyse

A bar magnet with its N-pole pointing right is moved away from a circular loop on its right. Use Lenz's law to find the direction of the induced current as seen from the magnet's side.

Flux through the loop (pointing right because of the magnet's N) is decreasing as the magnet moves away. The induced current must therefore create a flux pointing right to oppose the decrease.

By the right-hand rule, an anticlockwise current (as seen from the right, i.e. from the magnet) creates flux pointing right? No - clockwise from the magnet's viewpoint creates flux pointing right (curl fingers clockwise, thumb points to the right). Hence the induced current flows clockwise as seen from the magnet's side.

6.6 Motional Electromotive Force

Consider a straight rod of length L moving with velocity \(\vec v\) perpendicular to a uniform field \(\vec B\). Each free electron in the rod experiences a magnetic force \(\vec F = -e\,\vec v \times \vec B\). This force pushes electrons toward one end of the rod, leaving a positive charge at the other end. A potential difference (motional EMF) appears across the ends.

\[\varepsilon = BLv\]

This is the motional EMF.

L v B (out) R induced current loop
Fig 6.4 A rod of length L slides with velocity v on a U-rail in a perpendicular field B; motional EMF = BLv drives a current through resistor R.

6.6.1 Motional EMF from Faraday's law

The flux through the closed circuit at time t is Φ = B × (area enclosed). If the moving rod is at position x, area = Lx, so Φ = BLx. Then

\[\varepsilon = -\dfrac{d\Phi}{dt} = -BL\dfrac{dx}{dt} = -BLv\]

So Faraday's law and the Lorentz-force picture agree.

6.7 Energy Considerations - A Quantitative Study

The induced current I = ε/R = BLv/R flows through the rod. The rod itself, carrying current I in field B, experiences a magnetic force \(F_{mag} = BIL = B²L²v/R\) opposing its motion (Lenz!). To keep the rod moving at constant v, an external agent must apply force F = B²L²v/R, doing work at the rate:

\[P_{ext} = Fv = \dfrac{B^2L^2v^2}{R}\]

This is exactly equal to the electrical power dissipated as heat in R:

\[P_R = I^2 R = \left(\dfrac{BLv}{R}\right)^2 R = \dfrac{B^2L^2v^2}{R}\]

Mechanical work done = electrical heat dissipated. Energy is conserved exactly - and Lenz's law is the rule that enforces this balance.

Worked Example 6.5 - Sliding rod power balance

Example 6.5 L3 Apply

A rod of length L = 0.50 m slides on parallel rails in a perpendicular field B = 0.40 T at v = 5.0 m/s. The total resistance of the circuit is R = 2.0 Ω. Find (a) ε, (b) I, (c) the force needed to keep the rod moving, (d) the power dissipated in R.

(a) ε = BLv = 0.40 × 0.50 × 5.0 = 1.0 V.

(b) I = ε/R = 1.0/2.0 = 0.50 A.

(c) F = BIL = 0.40 × 0.50 × 0.50 = 0.10 N.

(d) PR = I²R = (0.50)² × 2.0 = 0.50 W. Check: Pext = Fv = 0.10 × 5.0 = 0.50 W ✓.

Worked Example 6.6 - Rotating rod

Example 6.6 L4 Analyse

A copper rod of length 0.40 m rotates in a horizontal plane about one end with angular speed ω = 30 rad/s, in a vertical magnetic field B = 0.50 T. Find the EMF between the centre and the tip.

For a rotating rod, ε = ½ B ω L².

= 0.5 × 0.50 × 30 × (0.40)² = 0.5 × 0.50 × 30 × 0.16 = 1.20 V.

Interactive: Motional EMF Predictor L3 Apply

Adjust v, B, L and R to see EMF, current, force and power update.

ε = 1.00 V  |  I = 0.50 A  |  F = 0.100 N  |  P = 0.50 W
Activity 6.2 - The dropped magnet testL5 Evaluate
  1. Take a long, vertical copper or aluminium tube (a few cm in diameter).
  2. Drop a small neodymium magnet down the tube.
  3. Time its fall. Compare with a non-magnetic object (e.g. a similar piece of plastic) dropped at the same time.
Predict: Will the magnet fall faster, slower, or the same as the plastic? Why?

The magnet falls dramatically slower. As it falls, the flux in the tube (a continuous conductor) changes; eddy currents are induced. By Lenz's law these currents oppose the magnet's motion - producing an upward magnetic braking force. Mechanical PE of the magnet ⇒ heat in the copper. The plastic, with no induced currents, falls in regular free-fall time.

Competency-Based Questions L1-L6

A horizontal rod 80 cm long is moved at 4 m/s perpendicular to a uniform magnetic field of 0.5 T. The rod is part of a closed circuit of total resistance 0.4 Ω.
1. The motional EMF developed across the rod is: L3 Apply
  • (a) 0.40 V
  • (b) 1.6 V
  • (c) 2.0 V
  • (d) 4.0 V
(b) ε = BLv = 0.5 × 0.8 × 4 = 1.6 V.
2. The mechanical power required to keep the rod moving uniformly equals: L3 Apply
  • (a) 1.6 W
  • (b) 4.0 W
  • (c) 6.4 W
  • (d) 0.4 W
(c) P = ε²/R = (1.6)²/0.4 = 6.4 W.
3. State Lenz's law and explain how it follows from energy conservation. L2 Understand
The induced current always opposes the change of flux that produces it. If it aided the change, mechanical motion would generate ever-growing electric currents and dissipate increasing heat without external work - violating energy conservation.
4. Why is it easier to push a magnet through a plastic tube than through a thick copper tube? L4 Analyse
In the copper tube, the changing flux induces eddy currents that oppose the magnet's motion (Lenz). The induced current circles in the copper wall, dissipating mechanical energy as heat. Plastic is non-conducting; no eddy currents; no opposing force.
5. Design a "drop test" lab experiment that quantitatively verifies Lenz's law using a copper tube and timing equipment. L6 Create
Materials: copper tube (~50 cm), comparable plastic tube, neodymium magnet, similar non-magnetic mass, photogate or stopwatch, ruler. Procedure: time the fall of magnet through copper, plastic, and time the non-magnetic mass through copper. Compare. Vary tube wall thickness/conductivity. Plot terminal velocity vs (1/thickness) - linear relation per Lenz/Joule heating equation. Quantitative check: equate magnet's PE drop to I²R heat - confirms energy conservation.

Assertion-Reason Pairs L4 Analyse

Options: (A) Both true, R correct explanation. (B) Both true, R not the explanation. (C) A true, R false. (D) A false, R true.

Assertion: When the N-pole of a magnet approaches a coil, the face of the coil nearer the magnet acts as a north pole.
Reason: By Lenz's law, the induced current must oppose the increase in flux due to the approaching N-pole.
(A). Reason directly explains the assertion.
Assertion: A rod sliding on rails in a magnetic field generates electricity for free.
Reason: The induced EMF causes a current with no work needed.
(D). Both statements are false: the rod experiences a retarding magnetic force = BIL; an external agent must do work, exactly equal to the heat dissipated.
Assertion: The motional EMF for a rod of length L moving with velocity v perpendicular to B is ε = BLv.
Reason: The Lorentz force qv × B on free electrons in the rod separates charge until the field across the rod balances it.
(A). Both true; reason explains the assertion.

Frequently Asked Questions - Lenzs Law Motional Emf

What is the main concept covered in Lenzs Law Motional Emf?
In NCERT Class 12 Physics Chapter 6 (Electromagnetic Induction), "Lenzs Law Motional Emf" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Lenzs Law Motional Emf useful in real-life applications?
Real-life applications of "Lenzs Law Motional Emf" from NCERT Class 12 Physics Chapter 6 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Lenzs Law Motional Emf?
Key formulas in "Lenzs Law Motional Emf" (NCERT Class 12 Physics Chapter 6 Electromagnetic Induction) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 6?
NCERT Class 12 Physics Chapter 6 (Electromagnetic Induction) is structured so each part builds on the previous one. "Lenzs Law Motional Emf" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Lenzs Law Motional Emf?
CBSE board questions from "Lenzs Law Motional Emf" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Lenzs Law Motional Emf" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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