JEE Main 2019 Physics question
A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is:
Answer: (B)
Solution
V = Bvl = 0.1 × 6 × 0.02 = 12 mV.