Two capacitors are connected by the ideal wire forming a square with side d the central wire

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Text Solution

Answer : A::B::C

Solution : Let us consider a small strip of thickness dx as shown in the figure. <br> The megnetic field at this strip <br> `B=(B_A)+(B_B)` <br> (perpendiculare to the plane of paper directed upwards) <br> `=(mu_(0))/(2 pi)(I)/(x)+(mu_(0))/(2 pi)(1)/(3a-x)` <br> `B_(A)`= Magnetic feild due to current in wire A <br> `(mu_(0))/(2 pi)(1/x+1/(3a-x)]` <br> `B_(B)`= Magnetic feild due to current in wire B <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/JMA_EIA_C15_050_S01.png" width="80%"> <br> Small amount of magnetic flux passing through the stip of thickness dx is <br> `d(phi)=Bxxadx =(mu_(0)Iaxx3a dx)/(2 pi x(3a-x))` <br> Total fulc through the square loop <br> (phi)=int_(0)^(2a)(mu_(0)I xx 3a^(2))/(2 pi) (dx)/(x(3a-x))=(mu_(0)Ia)/(pi)In 2` <br> `=(mu_(0)aIn(2))/(pi) (I_(0) sin omega t)` <br> The emf produced <br> `e=|-(d phi)/(dt)| = (mu_(0)aI_(0)omega)/(pi) In (2) cos omega t` <br> Charge stored in the capacitor <br> `q=C xx e=Cxx(mu_(0)aI_(0)omega)/(pi) In (2) cos omega t ...(i) <br> :. Current in the loop <br> `i=(dq)/(dt)=(C xx mu_(0)aI_(0)omega^(2))/(pi) In (2) cos omega t` <br> `:. I_(max)=(mu_(0)aI_(0)omega^(2)C In(2))/(pi) `. <br> (b) From (i) the graph between charge and time is <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/JMA_EIA_C15_050_S02.png" width="80%"> <br> Here, `(q_0)=(C xx mu_(0)aI_(0)omega In(2))/(pi) `