Question

What will be the magnitude of e.m.f. induced in a 200 turns coil with cross section area 0.16 $m^2$ if the magnetic field through the coil changes from 0.10 Wb $m^{-2}$ to 0.30 $m^{-2}$ in 0.05 second.

• A. 128V
• B. 130V
• C. 118V
• D. 132V

Answer 1

The correct option is A 128V

Answer is A.
The magnetic flux $\varphi =B\times A$ depends on the magnetic field, the area of the loop, and their relative orientation.
In this case, The magnetic field through the coil changes from $0.10Wb{m}^{-2}$ to $0.30Wb{m}^{-2}$, at a uniform rate over a period of 0.05 s.
The change in magnetic feild is $0.10Wb{m}^{-2}$ - $0.30Wb{m}^{-2}$ = $0.20Wb{m}^{-2}$.
The magnetic flux in one turn of the coil = $\varphi =B\times A=0.20\times 0.16=0.032T$
Therefore, the Magnetic flux in 200 turns of coil = 0.032 * 200 = 6.4 T.
Therefore, the e.m.f induced in it, if this coil is removed from the field in 0.05 second is d$\varphi$/dt = 6.4/0.05 = 128 V.
The magnitude of the e.m.f induced is 128 V.