A conducting circular loop of radius a and resistance per unit length R is moving with a constant velocity `v_0` , parallel to an infinite conducting wire carrying current `i_0`. A conducting rod of length `2a` is approaching the centre of the loop with a constant velocity `v_0/2` along the direction 2 of the current. At the instant `t = 0`, the rod comes in contact with the loop at `A` and starts sliding on the loop with the constant velocity. Neglecting the resistance of the rod and the self-inductance of the circuit, find the following when the rod slides on the loop.

(a) The current through the rod when it is at a distance of `(a/2)` from the point `A` of the loop.
(b) Force required to maintain the velocity of the rod at that instant.

Answer

B, C

Explanation

a. At the given instant

`AC=a/2, OC=a/2`
and `costheta=(a/2)/a=1/2, theta=pi/3`
`:.` Velocity of rod
`=(+v_0/2)` along the direction of current.
Emf induced across the ends `M` and `N`
`E_(rod)=int_(asqrt(3)-(asqrt(3))/(2))^(asqrt(3)+(asqrt(3))/(2))v_(rod) B dx`
`=v_(rod)int_(a(sqrt(3))/(2))^(3a(sqrt(3))/(2))(mu_(0)i_(0))/(2pix) dx`
`(v_0/2)(mu_0i_0)/(2pi) ln (3/1)`
with end `M` at higher potential.
Since, the effective length of both the arcs `MAN` and `MBN` is `MN`.

`:. E_(MAN)=E_(MBN)+V_("loop") (mu_0i_0)/(2pi) ln3`
`=v_0 (mu_0i_0)/(2pi) ln e`
with point `M` at higher potential.
Resistance of arc `MAN`
`implies R_1=(R)2(atheta)=2aRpi/3`
`implies` Resistance of arc `MBN`
`implies R_2=(R)a(2pi-2theta)=4aRpi/3`
Equivalent circuit at the given instant is shown in figure.

Current through the rod `MN`,
`i=(i_1+i_2)=((E_(MAN)-E_(rod))/R_1)+((E_(MBN)-R_(rod))/R_2)`
`i=(E_(MAN)-E_(rod))[1/R_1+1/R_1]`
`=(v_0mu_0i_0)(ln 30/(4pi)) [91/2+1/4][3/(aRpi)]`
`(9v_0i_0mu_0)/(16aRpi^2)ln (3)`
b. Force on the rod

`F_(rod)=int_((asqrt3)/2)^((3asqrt3)/2) idxB`
`=(imu_0i_0)/(2pi) ln 3=(9mu_0^2i_0^2v_0)/(32aRpi^3) (ln 3)^2`