A small magnet of magnetic moment π x10–10 Am2 is placed on the y-axis at a distance of 0.1 m from
the origin with its axis parallel to the x-axis. A coil having 169 turns and radius 0.05m is placed on the
x-axis at a distance of 0.12 m from the origin with the axis of the coil coinciding with x-axis. Find the
magnitude and direction of the current in the coil for a compass needle placed at the origin to point in
the north-south direction

Dear student,
The compass needle placed at origin will point north-south direction if the magnetic field at the origin,produced by the magnet and coil are equal in magnitude and opposite in direction.

The origin O is in bisectorial position with respect to small magnet having magnetic dipole moment M,hence magnetic field Bm produced at O due to magnet is 
Bm=μ04π×Mr3=10-7×π×10-100.13=π×10-14 Tesla

The magnetic field Bc produced by the coil must be opposite Bm .For this the current through the coil should flow in opposite direction when seen from the origin.Radius of coil =0.05m,x=0.12 m.
Bc=μ0NIr22r2+x232Bc =4π×10-7×169×I×0.05220.052+0.12232Bc =0.845π×10-7×I0.01693Bc =3.846π ×10-5×I T
For the needle to point in north south direction
3.846π×10-5×I =π×10-14 TestlaI=π×10-143.846π×10-5I=2.6×10-10A

When viewed from origin the current will flow in anticlockwise direction.

Regards.
 

  • 4
6*10^-9A
  • -1
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