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=100turns, B=4xx10^(-4)`
`A=25cm^2=25xx10^(-4)m^2`
`(a) when the coil is perpendicular to the field `
`phi_1=nBA`
`When coil goes through half turn`
`phi_2=nBA cos 180^(@)=-nBA`
`:. D phi =2nBA`
`The coil tmdergoes 300 rev. in 1 min `
`300xx2 pi rad//min =10 pi rad//sec.`
`10 pi rad is swept in 1 sec`
` pi rad is swept in (1/10 pi)pi =10 sec`
`e= (d phi)/(dt)=(2nBA)/(dt)`
`=(2xx100xx4xx10^(-4)xx25xx10^(-4))/(1//10)`
`=2xx10^(-3)V`
`(b) phi_1= nBA, phi_2=nBA( theta =360^(@))`
`d phi=0`
`(c ) i=(e/R)=(2xx10^(-3))/4 =1/2xx10^(-3)A.`
`=0.5xx10^(-3)=5xx10^(-4)A`
`Q=idt=5xx10^(-4)xx(1/10)`
`=5xx10^(-5)C`.