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There exists a long conductor along z-ax...

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

A

zero

B

`(mu_0VI_0)/(piRb)`

C

`(mu_0VI_0)/(2piRb)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
(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.)
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Knowledge Check

  • 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 total torque acting on the loop is nearly

    A
    `(mu_0VI_0)/(piR)(b-a) hati`
    B
    `-2(mu_0VI_0)/(piR)(b-a) hati`
    C
    `(mu_0VI_0)/(2piR)(b-a) hati`
    D
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  • A long straight wire along the z-axis carries a current I in the negative z-direction. The magnetic vector field vec(B) at a point having coordinnates (x, y) in the z = 0 plane is

    A
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  • A long straight wire along the z- axis carries a current I in the negative z- direction . The magnetic vector field vec(B) at a point having coordinates (x,y) in the Z = 0 plane is

    A
    ` (mu_(0)I(y hat(i)) - x hat(j))/ ( 2pi( x^(2) + y^(2))`
    B
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    D
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