JEE MainPhysicsElectromagnetic Induction
A metallic cube of side 20 cm is moving at a constant speed v along the positive x -axis. It is moving in a region of uniform magnetic field of magnitude 0.4 T directed along the positive y -axis. If the potential difference developed between the faces of the cube perpendicular to the z -axis is 240 mV, the value of v in m/s is
Correct answer
3
Step-by-step solution
The motional electromotive force (EMF) developed across a conductor moving in a magnetic field is given by the relation E = v B . Here, the velocity v = v i and the magnetic field B = 0.4 j . The induced electric field is E = (v i ) (0.4 j ) = 0.4v k V/m. The potential difference V is developed along the z -axis, which corresponds to the faces perpendicular to the z -axis. The distance between these faces is the side length of the cube, L = 20 cm = 0.2 m. Using V = | E | L , we have: V = (0.4v) 0.2 = 0.08v Given th