A conducting wire of length 50 cm moves perpendicular to a uniform magnetic field of 0.2 T with a constant velocity of 10 m s⁻¹ (the wire, its velocity and the field are mutually perpendicular). The emf induced between the two ends of the wire is V.

Question

Accepted Answer

(1) 1.0. Given: ℓ=50 cm=0.5 m, B=0.2 T, v=10 m s⁻¹, all mutually perpendicular. Idea: the magnetic force on the free electrons pushes them along the wire until its ends differ in potential by ε=Bℓ v. ε=0.2×0.5×10. So ε=1.0 V.

A conducting wire of length 50 cm moves perpendicular to a uniform magnetic field of 0.2 T with a constant velocity of 10 m s⁻¹ (the wire, its velocity and the field are mutually perpendicular). The emf induced between the two ends of the wire is ______ V.

Step-by-step solution

Why the other options are wrong