Start learning →

A horizontal conducting rod 0.5 m long moves at 10 m/s perpendicular to a 0.4 T magnetic field. The induced emf is

A0.5 V
B4.0 V
C1.0 V
D2.0 V
Answer & Solution
Correct answer: D. 2.0 V
ε = Blv = 0.4 × 0.5 × 10 = 2.0 V.
Practise this question with Quill — Karnataka PUC II →

Keep practising Electromagnetic Induction

Browse all Electromagnetic Induction questions for Karnataka PUC II, or explore more Physics topics.

Start a Electromagnetic Induction lesson →