A metallic rod of length 20 cm is palced in NorthSouth 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 4×10−3 T and the angle of dip is 45∘. The emf induced in the rod is ______ mV.
Answer: 16
Solution
The motional emf is given by:
ε=(v×B)⋅ℓ
Since the angle of dip is 45∘, the vertical component equals the horizontal component:
BV=BH=4×10−3 T
The rod moves East with velocity 20 m/s and is oriented North-South. The vertical component of Earth's field contributes to the emf:
ε=BV⋅v⋅ℓ=(4×10−3)×20×10020
ε=4×10−3×20×0.2=16×10−3 V=16 mV
JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer
