A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of $5.0 \mathrm{~ms}^{-1}$, at right angles to the horizontal component of the earth's magnetic field, $0.3 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}$. The value of the induced emf in wire is :

(B) $1.1 \times 10^{-3} \mathrm{~V}$

(A) $1.5 \times 10^{-3} \mathrm{~V}$

(C) $2.5 \times 10^{-3} \mathrm{~V}$

(D) $0.3 \times 10^{-3} \mathrm{~V}$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Electromagnetic Induction

A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of

5.0 ms^{- 1}

, at right angles to the horizontal component of the earth's magnetic field,

0.3 \times 1 0^{- 4} Wb / m^{2}

. The value of the induced emf in wire is :

Answer: (B)

Only the part of the field perpendicular to both the wire and its velocity drives an emf,

E = B ℓ v sin θ

. The wire runs north-east to south-west, at

4 5^{\circ}

to the horizontal field it sweeps through as it falls, so

E = (0.3 \times 1 0^{- 4}) (10) (5) sin 4 5^{\circ} = \frac{1.5 \times 1 0 ^{- 3}}{2} \approx 1.1 \times 1 0^{- 3}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option