There exists a long conductor along z-axis carrying a current `I_0` along positive z-direction. A loop having total resistance R is placed symmetrical about x-axis and y-axes in x-y plane as shown in Fig. potential difference `V_(BA)=V` is applied. Radii of arcs are a and b, respectively, as shown in the figure.

The magnitude of force experienced by the arc MN is

Options

A. `zero`
B. `(mu_0VI_0)/(piRb)`
C. `(mu_0VI_0)/(2piRb)`
D. none of these

Answer

A

Explanation

(a)
Magnetic field produced by the conductor along z-axis will
be parallel to MN. Hence, force on MN is zero.
Current from B to A (through NM or QP):
`I=V/(R//2)=(2V)/R`
Force due to `I_0` on N'N and Q'Q will be out of the paper and on
M'M and P'P will be in the paper. This will produce the torque
Torque on M'M (along negative x-axis):
`tau =int_a^b((mu_0I_0)/(2pir) Idr) r sin(pi/6)=(mu_0I_0)/(4pi)I(b-a)`
`=(mu_0I_02V)/(4Rpi)(b-a)`
Total torque =`4tau=2(mu_0I_0V)/(piR)(b-a)` (along negative x-axis.)