StreakPeaked· Practice

ExamsJEE MainPhysics

A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth’s magnetic field at that place is 4 × 10⁻⁵ T and the angle of dip is 45°. The emf induced in the rod is ________ mV.

  1. 16.00
  2. 1.60
  3. 0.16
  4. 0.016

Correct answer: 0.16

Solution

The rod lies N-S and moves East, so the field component perpendicular to the plane of v and L is the vertical component B_V = B_H tan(45) = 4e-5 T. emf = B_V*v*L = (4e-5)(20)(0.2) = 1.6e-4 V = 0.16 mV.

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →