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Magnetism And Matter

Question
CBSEENPH12037813

A small magnet of magnetic moment π x 10–10 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 0.05 m 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.

Solution
Given,
Magnetic moment, M =  π x 10–10 Am2
Distance of magnet from origin at y-axis = 0.1 m
Number of tuns of coil, N = 169
Radius of the coil, r = 0.05 m
Distance of coil from origin at x-axis, x = 0.12 m

 

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

Magnetic field, at the origin, due to the magnet is given by
                   Bm = μ04πMr3

                   Bm = 10-7×π×10-100.1 × 0.1 × 0.1T

                   Bm = π×10-14T  (equitorial line) 

Now,
The magnetic field B
c due to the circular coil should be opposite to Bm. 
For this, the current through the coil should flow anticlockwise as seen from the origin.

Hence,    BC = μ0 NIr22r2+x23/2   (axial line)

              BC = 4π×10-7×169×I×(0.05)22(0.05)2+(0.12)23/2T

              BC = 0.845π × 10-7 × I0.0025+0.01443/2

              BC = 0.845 π × 10-7× I0.01693

              BC =0.845π × 10-7 × I0.13 × 0.13 × 0.13

i.e.,            = 3.84π × 10-5 × I 

For needle to point in north-south direction 

3.846 π × 10-5 × I = π × 10-14 

                      I = 10-143.846 × 10-5A 

                           = 2.6 × 10-10A  

Hence, as viewed from origin, the current flows in anti-clockwise direction.