Question

A square loop of side 1m is placed in a perpendicular magnetic field. Half of the area of the loop lies inside the magnetic field. A battery of emf 10V and negligible internal resistance is connected in the loop. The magnetic field change with time according to the relation B = (0.01 –2t) tesla. The total emf of the battery will be -

• A. 11V
• B. 9 V
• C. 12 V
• D. 6 V

Answer 1

Answer: (B)

9 V

Answer 2

emf induced is,

e = – $\frac{d\varphi}{dt}$ = $- \frac{AdB}{dt}$ = $- \frac{\mathcal{l}^{2}}{2}$ × (–2)

= l² = 1 Volt

\ Resultant emf = 10 – 1 = 9V