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:
(A) 12 mV (B) 6 mV
(C) 1 mV (D) 2 mV

Answer 1

Given:

Length $l = 2 \times 10^{-2}$ m

Speed $v = 6$ $\frac{m}{s}$

Magnetic field $B = 0.1$ T

To Find: Potential difference between two faces,

According to the faraday's law,

   $\epsilon = Blv$

Where $\epsilon =$ potential difference between two faces.

Put the values in above equation,

   $\epsilon = 0.1 \times 6 \times 2 \times 10^{-2}$

   $\epsilon = 1.2 \times 10^{-2}$

   $\epsilon = 12$ mV

Therefore, the potential difference between the two faces is 12 mV

Option (A) is correct