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.

Answer

A, B, C

Explanation

Let us consider a small strip of thickness dx as shown in the figure.
The megnetic field at this strip
`B=(B_A)+(B_B)`
(perpendiculare to the plane of paper directed upwards)
`=(mu_(0))/(2 pi)(I)/(x)+(mu_(0))/(2 pi)(1)/(3a-x)`
`B_(A)`= Magnetic feild due to current in wire A
`(mu_(0))/(2 pi)(1/x+1/(3a-x)]`
`B_(B)`= Magnetic feild due to current in wire B

Small amount of magnetic flux passing through the stip of thickness dx is
`d(phi)=Bxxadx =(mu_(0)Iaxx3a dx)/(2 pi x(3a-x))`
Total fulc through the square loop
(phi)=int_(0)^(2a)(mu_(0)I xx 3a^(2))/(2 pi) (dx)/(x(3a-x))=(mu_(0)Ia)/(pi)In 2`
`=(mu_(0)aIn(2))/(pi) (I_(0) sin omega t)`
The emf produced
`e=|-(d phi)/(dt)| = (mu_(0)aI_(0)omega)/(pi) In (2) cos omega t`
Charge stored in the capacitor
`q=C xx e=Cxx(mu_(0)aI_(0)omega)/(pi) In (2) cos omega t ...(i)
:. Current in the loop
`i=(dq)/(dt)=(C xx mu_(0)aI_(0)omega^(2))/(pi) In (2) cos omega t`
`:. I_(max)=(mu_(0)aI_(0)omega^(2)C In(2))/(pi) `.
(b) From (i) the graph between charge and time is

Here, `(q_0)=(C xx mu_(0)aI_(0)omega In(2))/(pi) `