A square loop of edge b having M turns is rotated with a uniform angular velocity `omega` about one of its diagonals which is kept fixed in a horizontal position. A uniform magnetic field `B_(0)` exists in the vertical direction.

Find (i) the emf induced in the coil as a function of time t.
(ii) the maximum emf induced.
(iii) the average emf induced in the loop over a long period.
(iv) if resistance of loop is R, amount of charge flown in time t = 0 to t = 2T.
(v) heat produced in time t = 0 to t = 2T.

Explanation

Initially, plane of loop is `_|_^(ar)` to the field. Let at any time, normal to loop makes an angle `theta` with magnetic field `theta=omegat`.
The flux passing through loop,
`phi_(B)=MB_(0)Scos theta=MB_(0)b^(2)cosomegat`
`rArr" "e=-(dphi_(B))/(dt)=MB_(0)b^(2)omega sin omegat=e_(0)sinomegat,`
where, `e_(0)=MB_(0)b^(2)omega`
(ii) For emf to be maximum, then sin `omegat` should be equal to 1.
`rArr" "e_(max)=e_(0)`
(iii) Long period means one time period, i.e., `0rarrT`
We know that average emf is given as
`bar(e)=(int_(0)^(T)edt)/(int_(0)^(T)dt)=(1)/(T).e_(0)int_(0)^(T)sinomegatdt=0" "(becauseint_(0)^(T)sinomegatdt=0)`
(iv) Using the relation, induced current,
`i=(e)/(R)=(e_(0))/(R)sinomegat=i_(0)sinomegat`
where, `i_(0)=(e_(0))/(R)`.
We know q = idt
So, for time `Orarr2T`,
`q=int_(0)^(2T)"idt"=i_(0)int_(0)^(2T)sinomegatdt=0`
(v) Heat (H) generated in a loop during time `0rarr2T`, is given as
`H=int_(0)^(2T)(e^(2))/(R)dt=(e_(0)^(2))/(0)int_(0)^(2T)sin^(2)omegatdt`
`=(e_(0)^(2))/(R).2.(T)/(2)=(e_(0)^(2))/(R).(2pi)/(omega)=(2pie_(0)^(2))/(omegaR)" "(becauseT=(2pi)/(omega))`