The horizontal component of the earth's magnetic field at a place is 3 × 10-4 T and the dip is tan -1((4/3)) . A metal rod of length 0.25 m placed in the north-south position and is moved at a constant speed of 10 cm / s towards the east. The emf induced in the rod will be

Question

The horizontal component of the earth's magnetic field at a place is 3 × 10-4 T and the dip is tan -1((4/3)) . A metal rod of length 0.25 m placed in the north-south position and is moved at a constant speed of 10 cm / s towards the east. The emf induced in the rod will be (A) Zero (B) 1 μ V (C) 5 μ V (D) 10 μ V

Tardigrade Admin · Accepted Answer

Rod is moving towards east, so induced emf across its end will be e=BV v l=(BH tan δ) v l ∴ e=3 × 10-4 × (4/3) ×(10 × 10-2) × 0.25 =10-5 V =10 μ V

Zero

$1 μ V$

$5 μ V$

$10 μ V$

Rod is moving towards east, so induced emf across its end will be
$e = B_{V} v l = (B_{H} tan δ) v l$
$∴ e = 3 \times 1 0^{- 4} \times \frac{4}{3} \times (10 \times 1 0^{- 2}) \times 0.25$
$= 1 0^{- 5} V = 10 μ V$

Q. The horizontal component of the earth's magnetic field at a place is 3×10−4T and the dip is tan−1(34​). A metal rod of length 0.25m placed in the north-south position and is moved at a constant speed of 10cm/s towards the east. The emf induced in the rod will be

Zero

1μV

5μV

10μV

Rod is moving towards east, so induced emf across its end will be e=BV​vl=(BH​tanδ)vl ∴e=3×10−4×34​×(10×10−2)×0.25 =10−5V=10μV

$1 μ V$

$5 μ V$

$10 μ V$

Rod is moving towards east, so induced emf across its end will be
$e = B_{V} v l = (B_{H} tan δ) v l$
$∴ e = 3 \times 1 0^{- 4} \times \frac{4}{3} \times (10 \times 1 0^{- 2}) \times 0.25$
$= 1 0^{- 5} V = 10 μ V$