A rectangular loop of sides 8 cm and 2 cm having resistance of 1.6 $Ω$ is placed in a magnetic field of 0.3 T directed normal to the loop. The magnetic field is gradually reduced at the rate of 0.02 T/s. How much power is dissipated by the loop as heat ?

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

12.8 x 10^-8 W

6.4 x 10^-10 W

3.2 x 10^-8 W

1.6 x 10^-4 W

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

We know that induced e.m.f. (e) is equal to the product of rate of change of magnetic field and area. Hence
$e = (0 \cdot 02) \times (8 \times 2 \times 10^{- 4}) = 3 \cdot 2 \times 10^{- 5} V Now i = \frac{e}{R} = \frac{3 \cdot 2 \times 10^{- 5}}{1 \cdot 6} = 2 \cdot 0 \times 10^{- 5} amp ∴ power loss = i^{2} R = {(2 \cdot 0 \times 10^{- 5})}^{2} \times 1 \cdot 6 = 6 \cdot 4 \times 10^{- 10} W$