The magnetic field of an electromagnetic-wave obeys the relation in a certain region is `B=10^(-12) sin (5xx10^(t))`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

Options

A. `-2xx10^(-5)cos (5xx10^(-6)t)V`
B. `-3xx10^(-6)cos ( 5xx10^(6)t)V`
C. `-2.5xx10^(-6)cos(5xx10^(6)t)V`
D. `-3.3xx10^(-6)cos (5xx10^(6)t)V`

Answer

B

Explanation

Given, `B=10^(-12) sin (5xx10^(6)t)T, N=300` turns
and area, `A=20 cm^(2)=20xx10^(-4) m^(2)`
Hence, the emf induced
`e=-N(d phi)/(dt)=-N (d(BA))/(dt)=-AN (dB)/(dt)`
`rArr e=-20xx10^(-4)xx300xx(d)/(dt)(10^(-12)sin 5 xx 10^(6)t)`
`rArr e=6000xx10^(-4)xx10^(-12)xx(cos 5 xx 10^(6)t)xx5xx10^(6)`
`rArr e=-3xx10^(-6)cos (5xx10^(6)t)V`
Hence, the correct option is (b).