In figure the switch is closed at t = 0, with the capacitor of capacity $1μF$ and having initial charge of $20μC$ (the polarity shown). The left square loop has very large dimensions as compared to the distance between loops and a resistance of $10 Ω$ is connected in series with capacitor as shown. A wire of length 8 mm and resistance per unit length of $0.5Ω/mm$ is bent in the form of square loop ‘A’ and placed at a distance x = 1 m from left loop.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The mutual inductance of the system will be $8 \times 10^{- 13} H$

The induced current in the loop A will be clockwise

The induced current in the loop A will be anticlockwise

The induced current in the loop A at time $t = 2 RC l n2$ will be $10^{- 8} A$

answer is A, B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$M= {\frac{_{2}}{i_{1}}}_{2} =BA$
Where $B= \frac{μ_{0} i_{1}}{2πx}$
$A=4 \times 10^{- 6} m^{2} M = \frac{μ_{0}}{2πx} .A = \frac{2 \times 10^{- 7}}{1} \times 4 \times 10^{- 6} = 8 \times 10^{- 3} H i_{1} = i_{0} e^{- t/RC} e_{2} = \frac{-_{2}}{dt} & i_{2} = \frac{e_{2}}{(8 \times 0.5)} \Rightarrow i_{2} = \frac{\frac{M}{RC} . \frac{q}{RC} e^{- t/RC}}{4} = \frac{8 \times 10^{- 13} \times 20 \times 10^{- 6}}{{(10 \times 10^{- 6})}^{2} \times 4} e^{\frac{2 RC l n2}{RC}} = 10^{- 8} A$