Question

The magnetic field of an electromagnetic-wave obeys the relation in a certain region is $B=10^{-12}\sin \left(5\times 10^{6}t\right)T$, where $t$ is the time. Then, the induced emf, in a $300$ turns in coil of area $20c{m}^2$ oriented perpendicular to the field is

• A. $-7\times 10^{-5}\cos \left(5\times 10^{6}t\right)V$
• B. $-3\times 10^{-6}\cos \left(5\times 10^{6}t\right)V$
• C. $-2.5\times 10^{-6}\cos \left(5\times 10^{6}t\right)V$
• D. $-3.3\times 10^{-6}\cos \left(5\times 10^{6}t\right)V$

Answer 1

Answer: (B)

Given, $B=10^{-12}\sin \left(5\times 10^{6}t\right)T,N=300$ turns
and area, $A=20c{m}^2=20\times 10^{-4}{m}^2$
Hence, the emf induced
$e=-N\frac{d\varphi }{dt}=-N\frac{d(BA)}{dt}=-AN\frac{dB}{dt}$
$\Rightarrow e=-20\times 10^{-4}\times 300\times \frac{d}{dt}\left(10^{-12}\sin 5\times 10^{6}t\right)$
$\Rightarrow e=6000\times 10^{-4}\times 10^{-12}\times \left(\cos 5\times 10^{6}t\right)\times 5\times 10^6$
$\Rightarrow e=-3\times 10^{-6}\cos \left(5\times 10^{6}t\right)V$