Two circular rings each of radius `a` are joined together such that their planes are perpendicular to each other as shown in the figure. The resistance of each half part of the ring is indicated. A very small loop of mass `m` and radius `r` carrying a current `I_(0)` is placed in the plane of the paper at the common centre of each ring. The loop can freely rotate about any of its diametric axes. If the loop is slightly dispalced, find the time period of its oscillations.
(Given `ma=(2pimu_(0)I_(0))/(pi^(2))`

Explanation

The equivalent resistance of the circuit is
`(1)/(R_(eq))=3+(1)/(2)`
`rArr R_(eq)=(2)/(7)Omega`
`:. I=`total current `=(E)/(R_(eq))=7"amp"`.
The magnetic field at the centre of the loop is
`B=(mu_(0)(2))/(4a)+(mu_(0)(2))/(4a)-(mu_(0)(2))/(4a)-(mu_(0)(I))/(4a)`
`rArr B=(mu_(0))/(4a)` (inward)
The magnetic moments of the tiny ring is `M=I_(0)A=piI_(0)r^(2)`
If the ring is rotated through angle `theta`, then restoring torque is
`tau=-MB sintheta= -piI_(0)r^(2)(mu_(0))/(4a)theta` (for small angle)
Using `II`nd law of motion
`tau=(mr^(2))/(2)(d^(2)theta)/(dt^(2))= -piI_(0)r^(2)(mu_(0))/(4a)theta`
`:. T=2pisqrt((2ma)/(pimu_(0)I_(0))`