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 $20\, cm ^{2}$ 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, $A=20 \,cm ^{2}=20 \times 10^{-4} \,m ^{2}$
Hence, the emf induced
$e=-N \frac{d \phi}{d t}=-N \frac{d(B A)}{d t}=-A N \frac{d B}{d t}$
$\Rightarrow e=-20 \times 10^{-4} \times 300 \times \frac{d}{d t}\left(10^{-12} \sin 5 \times 10^{6} t\right)$
$\Rightarrow e=6000 \times 10^{-4} \times 10^{-12} \times\left(\cos 5 \times 10^{6} t\right) \times 5 \times 10^{6}$
$\Rightarrow \, e=-3 \times 10^{-6} \cos \left(5 \times 10^{6} t\right) V$