A straight conducting wire of length $0.30, m$ moves with a constant velocity of $4, m s^{- 1}$ perpendicular to a uniform magnetic field of $0.15, T$ directed into the page. Find the magnitude of the induced emf across the ends of the wire.

easy 2 marks Class 12 · Physics · Electromagnetic Induction

Answer

1.8 \times 1 0^{- 1}, V

The correct answer is $1.8 \times 1 0^{- 1}, V$ — Step 5: The direction of induced emf is such that it opposes the motion (Lenz’s law).

Step-by-step solution

Step 1: Identify the formula for motional emf: $E = B, l, v$ .
Step 2: Plug in the given values: $B = 0.15, T$ , $l = 0.30, m$ , $v = 4, m s^{- 1}$ .
Step 3: Calculate: $0.15 \times 0.30 \times 4 = 0.18$ .
Step 4: Write the result: $E = 0.18, V = 1.8 \times 1 0^{- 1}, V$ .
Step 5: The direction of induced emf is such that it opposes the motion (Lenz’s law).

Common mistake: Using the wrong formula or forgetting to multiply by the wire length

Concept tested: Motional emf in a moving conductor

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked