A circular ring of radius a is made from a wire having resistance `(lambda)` per unit length. The ring is mounted on a car such that ring remains vertical. The care moves along a horizontal circle of radius R and completes n revolutions per minute. The horizontal component of earth's magnetic field be H,
Calculate average rate at which heat is produced in the ring

Options

A. `(pi^(3)a^(3)H^(2)n^(2))/(1800 lambda)J s^(-1)`
B. `(pi^(3)a^(3)H^(2)n^(2))/(3600 lambda)J s^(-1)`
C. `(pi^(3)a^(3)H^(2)n^(2))/(1200 lambda)J s^(-1)`
D. None of these

Answer

B

Explanation

Path of the car is a circle of radius R as shown in fig. Let at instant `t=0`, the car be at A when
the axis of ring was parallel to horizontal component H of earth's magnetism. Since, car completes n revolutions per mintues, therefore, angular velocity of car is `omega = ((2pi n)/(60))`
Hence, at time t, its position B is given by,
`theta = omega t = (2 pi nt)/(60) = (pi n t)/(30)`
Axis of the ring has also turned through the same angle `theta`. Therefore, component of H along ring axis is now equal to H cos `theta`.
Hence, Flux linked with the ring at this instant is ,
`phi = pi a^(2)(H cos theta)=pi a^(2)H cos ((pi n t)/(30))`
Induced emf,
`e=-(d phi)/(dt) = (pi^(2)a^(2)nH)/(30) sin ((pi nt )/(30))`
Resistance of the ring = `2 pi a l`.
Due to induced emf, a current starts to flowing through the ring and due to this current heat is produced in the ring. At time t, thermal power.
`P=((e^2)/(R e s i s t a n c e))`
Perio of one complete revolution of the car is ,
`P_(avg) = (int_(0)^(t)pdt)/(T) = (pi^(3)a^(2)H^(2)n^(2))/(3600 lambda) js^(-1)`.