Q.What is the energy required to build up a current of 1 A in an inductor of 20 mH? (A) 10 mJ (B) 20 mJ (C) 20 J (D) 10 J

Energy stored in an inductor's field derives from the work done against back-emf; the falling-wire part uses motional emf with Earth's field.

1. Energy stored in the magnetic field: As current in an inductor $L$ grows from 0 to $I$, the back-emf $e=-L\,di/dt$ opposes the growth. The work done per unit time against this back-emf is $dW = -e\,i\,dt = Li\,di$. Total work done (stored as magnetic field energy): $U = \int_0^I Li\,di = \frac{1}{2}LI^2$
2. Wire falling to the ground: The wire (length $l=5$ m) is horizontal, along east–west, and falls freely under gravity from height $h=15$ m. Its speed just before hitting the ground: $v = \sqrt{2gh} = \sqrt{2\times10\times15} = \sqrt{300} \approx 17.32\ \text{m/s}$ As it falls through the (vertical component of) Earth's magnetic field, it sweeps out area $l\times h$ and generates a motional emf. Since the wire accelerates uniformly from rest, its average speed over the fall is $v/2$, and the average induced emf is $e_{avg} = B\,l\,\left(\frac{v}{2}\right)$ Using the standard board-exam data value for the relevant (horizontal, since the wire runs east–west and falls vertically) component of Earth's field, $B_H = 3.6\times10^{-5}$ T: …