Q. A rectangular loop of sides 10 cm and 5 cm with a cut is stationary between the pole pieces of an electromagnet. The magnetic field of the magnet is normal to the loop. The current feeding the electromagnet is reduced so that the field decreased from its initial value of 0.3 T at the rate of 0.02$\ \Omega$. If the cut is joined and the loop has a resistance of 2.0$\ \Omega$, the power dissipated by the loop as heat is
A) 5 nW
B) 4 nW
C) 3 nW
D) 2 nW
Explanation
(a) Here, $A = 10 \times 5 = 50cm^{2} = 50 \times 10^{- 4}m^{2}$
$\frac{dB}{dt} = 0.2\ {Ts}^{- 1}$
$R = 2\Omega$
$E = \frac{d\phi}{dt} = A.\frac{dB}{dt} = 50 \times 10^{- 4} \times 0.02 = 10^{- 4}V$
Power dissipated in the form of heat
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{E^{2}}{R} = \frac{10^{- 4} \times 10 - 4\ }{2} = 0.5 \times 10^{- 8}W$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5 \times 10^{- 9}W = 5nW$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.