Area vector of a coil of `5 xx 10^(-3)m^2` makes an angle of `0^@` with a uniform magnetic field. If from this position the coil is rotated in 0.5 sec., so that the angle made by area vector with the field becomes `90^@`. Find the average emf induced in the coil. The magnetic field intensity is 0.3 T. Number of turns in the coil is 500.

Options

A. 1.5 V
B. 2.5 V
C. 1.4 V
D. 1.3 V

Answer

A

Explanation

Here, N=500
`A=5xx10^(-3) m^2`
`Deltat` = 0.5 s
`theta_1=0^@`
`theta_2=90^@`
B=0.3 T
`rArr` Flux in coil
`phi_1=AB cos theta_1` and `phi_2=AB cos theta_2`
Change in flux `Deltaphi=phi_2-phi_1`
`=AB cos theta_2-AB cos theta_1`
`=AB[cos theta_2-cos theta_1]`
`=AB [cos 90^@ - cos 0^@]`
=AB[0-1]=-AB
`therefore` Induced emf in coil,
`E=-N(Deltaphi)/(Deltat)`
`=-N((-AB))/(Deltat)`
`=(NAB)/(Deltat)`
`=(500xx5xx10^(-3)xx0.3)/0.5`
`therefore` E=1.5 V