A closed coil having 100 turns is rotated in a uniform magnetic field `B = 4.0 xx 10^(-4)` T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is `25 cm^2` and its resistance is `4.0 Omega.` Find (a) the average emf developed in the half a turn form a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part (a).

Explanation

`N=100` , `B=400muT=400xx10^(-6)T`
`omega=(2pixx300)/(60)=10pi rad/sec, A=25cm^(2)`
`=25xx10^(-4)m^(-2)`
`R=4Omega`
(a) `phi_(1)=NBAcos0=NBA`
`phi_(2)=NBAcos180^(@)=-NBA`
`Deltat(T)/(2)=(2pi//omega)/(2)=pi//omega`
`|bar(e)|=(Deltaphi)/(Deltat)=(2NBA)/(pi//omega)=(2NBAomega)/(pi)`
`=(2xx100xx4xx10^(-4)xx25xx10^(-4)xx10pi)/(pi)=2xx10^(-3)V`
(b) `phi_(1)=NBAcos0=NBA` ,
`phi_(2)=NBAcos360^(@)=NBA` , `Deltaphi=0`
`bar(e)=0`
(c) `Deltaq=(Deltaphi)/(R)=(2NBA)/(R)=(2xx10^(-4))/(4)`
`=5xx10^(-5)C`