JEE Main202325 Jan 2023Evening ShiftPhysicsElectromagnetic InductionActual
A wire of length 1 m moving with velocity 8 m s - 1 at right angles to a magnetic field of 2 T . The magnitude of induced emf, between the ends of wire will be ___________.
Options
- A20   V
- B8   V
- C12   V
- D16   V
Correct answer
D. 16   V
Step-by-step solution
The expression of motional emf is e = B L v sin θ , where, θ is angle between magnetic field and direction of motion. Here, θ = 90 ° , So, induced emf across the ends of wire is e =   B L v sin 90 ° = B L v =   2   ×   1   ×   8   =   16   V