MHT CET202617 April 2026Evening ShiftPhysicsElectromagnetic InductionActual
A straight line conductor of length 0.4 m is moved with a speed of 7.0 ms ⁻¹ perpendicular to magnetic field of intensity 0.8 Wb m ⁻² . The induced e.m.f. across the conductor is
Options
- A2 24 V
- B2 80 V
- C3 20 V
- D5 60 V
Correct answer
A. 2 24 V
Step-by-step solution
Given l = 0.4 m , v = 7.0 ms ⁻¹ , and B = 0.8 Wb m ⁻² . The induced e.m.f. across a straight conductor moving perpendicular to a magnetic field is given by e = B l v . Substituting the given values: e = 0.8 0.4 7.0 e = 2.24 V Answer: 2 24 V