A square loop of side 'a' with a capacitor of capacitance C is located between two current carrying long parallel wires as shown. The value of I in the wires in given as `I=(I_0) sin omega t`.

(a) Calculate maximum current in the square loop.
(b) Draw a graph between charges on the upper plates of the capacitor vs time.

Explanation

Let us first find the flux through the loop due to left wire.

`phi_(L) = int_(a)^(2a) (mu_(0)I)/(2pi)(log_(e)2)`
This flux will be in outward direction because magnetic field due to wire is in outward direction because anticlockwise direction is positive. Now flux due to right wire also be same because loop is placed symmtrically w.r.t. both wires. So net flux through the loop is
`phi = 2phi_(L) = (mu_(0)Ia)/(pi)log_(e)2` ltbr. Induced emf in the loop:
`e = -(dphi)/(dt) = -(mu_(0)a(log_(e)2))/(pi)((dI)/(dt))`
`= -(mu_(0)I_(0)omega a)/(pi)(log_(e)2)cosomegat` ltbr. Negative sign indicates that induced emf will be in clockwise sence because anticlockwise sence is positive. Finally, induced emf in the loop is `e = (mu_(0)omega aI_(0)(log_(e)2))/(pi)cosomegat` as shown in Fig. 3.93.
let `q` be the charge on the capacitor at any time, then
`q = Ce = (mu_(0)I_(0)omega aClog_(e)2cos omegat)/(pi)`
Current `I` in the circuit: `I = (dq)/(dt) = (mu_(0)I_(0)omega^(2) aClog_(e)2)/(pi)sin omegat`
Hence, maximum current in the loop is ltbr. `I_(max) = (mu_(0)I_(0)omega^(2)aClog_(e)2)/(pi)`
Charge on the upper plate: `q_(up) = -q = -q_(0)cos omegat`
where `q_(0) = (mu_(0)I_(0)omega aC(log_(e)2))/(pi)`
The variation of `q_(up)` with time as shown in Fig. 3.94.