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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.
- 16.00
- 1.60
- 0.16
- 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.
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