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अभ्यासक्रम

Motional Emf: A Conceptual Approach Based on Lenz's Law and Dynamic Flux Analysis

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Estimated time: 7 minutes
CISCE: Class 12

Definition: Motional EMF

The electromotive force induced in a conductor when it moves through a magnetic field, causing a change in magnetic flux through the circuit it completes.

CISCE: Class 12

Concept Setup

  • A conducting rod MN rests on two parallel rails and slides with velocity v.
  • A uniform magnetic field B acts perpendicular to the plane and points into the page.
  • The moving rod effectively divides the circuit into two loops: a left subloop (increasing area) and a right subloop (decreasing area).
CISCE: Class 12

Two-Subloop Flux Analysis

Subloop Area Change Flux Change Induced Current Direction
Left loop Increasing Increasing (into page) Anticlockwise (opposes increase)
Right loop Decreasing Decreasing (into page) Clockwise (opposes decrease)

Why do the currents add up at the rod?
Both subloop currents flow through the shared rod MN in the same direction (from N to M), because one loop's anticlockwise sense and the other's clockwise sense both push current the same way through the shared branch. They combine rather than cancel.

Analogy: Think of two connected water tanks — one filling, one draining, joined by a single pipe (the rod). Water pushed by the filling tank and water pulled by the draining tank both move through the shared pipe in the same direction, reinforcing the flow rather than opposing it.

CISCE: Class 12

Derivation of Motional EMF

Step 1: Flux through the circuit changes as the rod moves:

  • Φ = B ⋅ l ⋅ x

where x is the distance of the rod from a reference end.

Step 2: Induced emf by Faraday's law:

  • ε = −\[\frac {dΦ}{dt}\] = −Bl\[\frac {dx}{dt}\] = −Blv

Step 3: Induced current in the circuit:

  • I = \[\frac {ε}{R}\] = \[\frac {Blv}{R}\]

Step 4: Power dissipated (equal to power delivered by the external agent moving the rod):

  • P = I2R = \[\frac {B^2l^2v^2}{R}\]

Key Formula Box:  ε = Blv | I = \[\frac {Blv}{R}\]​ | P = \[\frac {B^2l^2v^2}{R}\]

CISCE: Class 12

Rod Sliding Down an Inclined Plane

At terminal velocity, the net force is zero — gravity's component along the incline balances the magnetic retarding force:

  • vT = \[\frac{mgR\sin\theta}{B^2l^2\cos^2\theta}\]
CISCE: Class 12

Example

Given: l = 0.5 m, B = 0.15 T, R = 3 Ω, v = 2 m/s

To Find: External force required to maintain constant velocity; power delivered

Solution:

  • I = \[\frac {Blv}{R}\] = \[\frac {0.15×0.5×2}{3}\] = 0.05 A
  • F = BIl = 0.15 × 0.05 × 0.5 = 3.75 × 10−3 N
  • P = Fv = 3.75 × 10−3 × 2 = 7.5 × 10−3 W

Answer: External force = 3.75 × 10−3 N; Power delivered = 7.5 × 10−3 W

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