JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Electromagnetic Induction

A circular conducting coil of radius 1 m being heated by the change of magnetic field

B

passing perpendicular to the plane in which the coil is laid. The resistance of the coil is

2 μ Ω

. The magnetic field is slowly switched off such that its magnitude changes in time as

B = \frac{4}{π} \times 1 0^{- 3} T (1 - \frac{t}{100})

The energy dissipated by the coil before the magnetic field is switched off completely is
E =

____

Answer: 80

Solution

The flux through the coil is

Φ = B π r^{2} = π \times \frac{4}{π} \times 1 0^{- 3} (1 - \frac{t}{100}) = 4 \times 1 0^{- 3} (1 - \frac{t}{100}) Wb

which falls linearly, so the induced emf is constant:

ε = \frac{d Φ}{d t} = \frac{4 \times 1 0 ^{- 3}}{100} = 4 \times 1 0^{- 5} V

The current is then steady at

I = \frac{ε}{R} = \frac{4 \times 1 0 ^{- 5}}{2 \times 1 0 ^{- 6}} = 20

A, and the field reaches zero at

t = 100

s. The heat developed over that interval is

E = I^{2} R t = (20)^{2} \times 2 \times 1 0^{- 6} \times 100 = 0.08 J

so the coil dissipates

80

mJ.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer