A metallic cube of side 15 cm moving along y-axis at a uniform velocity of 2 ms⁻¹. In a region of uniform magnetic field of magnitude 0.5 T directed along z-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be _________ mV.

Solution

Physics sir · 3 steps

Electrons get pushed sideways until their own electric field stops them, and that balance gives V = Bvl.

1. Direction of push. The force on a moving charge is q v⃗ × B⃗, and y × z = x. So the free electrons, being negative, are pushed towards -x. The two faces perpendicular to the x-axis become charged: the face at smaller x turns negative and the face at larger x turns positive.
2. Equilibrium. Charge piles up until the electric field E between these two faces cancels the magnetic push on each electron. At that point eE = evB, so E = vB. The field is uniform across the side l, so V = El.

   V = Bvl = 0.5 × 2 × 0.15 = 0.15 V

3. So. The potential difference between the higher-potential and lower-potential faces is 0.15 V, which is 150 mV.

The common slip: 0.15. Stopping at the answer in volts gives 0.15, but the blank asks for millivolts, so multiply by 1000 to get 150.

The idea to keep. For a conductor moving through a field, find the direction of v⃗ × B⃗ to see which faces charge up. Then the magnetic push balances the electric pull, giving V = Bvl, with l in metres.

Answer from the NTA final answer key. Solution by PrepWiser.