Two long parallel horizontal rails a, a distance d aprt and each having a risistance `lambda` per unit length are joing at one end by a resistance R. A perfectly conduction rod MN of mass m is free to slide along the rails without friction (see figure). There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that, as the rod moves a constant current flows through R.

(i) Find the velocity of the rod and the applied force F as function of the distance x of the rod from R.
(ii) What fraction of the work done per second by F is converted into heat?

Answer

A, B, C, D

Explanation

(i) A variable force F is applied to the rod MN such that as the rod moves in the uniform magnetic field a constant current flows through R. Consider the loop MPQN. Lenght of rails in loop =2x.

Resistance of rails in loop `=2x(lambda)`
Total resistance of loop`=R+2(lambda)x`
Induced emf `=Bvd`
Induced current `(I)=(Bvd)/(R+2 lambda x)`
so for constant I,
`v=(R+2 lambda x)/(Bd)I`
Furthermore, as due to induced current I the wire will experience force `F_(M)=BId` opposite to its motion, the equation of motion of the wire will be
`F-(F_M)=ma i.e., F=(F_M)+ma`
But as here `(F_m)`=Bld and from equation (i)
`a=(dv)/(dt)=(2lambdal)/(Bd) (dx)/(dt) =(2 lambda i_(v))/(Bd0=(2 lambdaI^(2))/(Bd^2)(R+2lambdax)`
so, `F=Bld+(2lambda mI^(2))/((Bd)^(2))(R+2lambdax)`
(ii) As the work done by force F per sec.
`(dW)/(dt)=P=Fv=[bld+(2lambda mI^(2))/((Bd)^(2))(R+2 lambda x)][(R+2 lambda x)/(Bd).I]`
i,e.,`P=[(I62)(R+2lambda x)=(2lambda mI^(3))/(b^(3)d^(3))(R+2 lambda x)^(2)]`
and heat produced per second, i.e., joule heat
`H=(I^2)(R2 lambda x)`
so, `f=H/P=[1+(2 lambda mI(R+2 lambdax))/(B^(3)d^(3))]^(-1)`.