JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Electromagnetic Induction

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.

Answer: 150

Solution

The motional emf induced across the cube is given by

Δ V = (v \times B) \cdot d

. With

v = 2

m/s along y-axis,

B = 0.5

T along z-axis, and the cube side

d = 0.15

m along x-axis, we get

Δ V = v B d = 2 \times 0.5 \times 0.15 = 0.15

= 150

mV.

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer

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