Question

A short bar magnet of magnetic movement 5.25 × 10–2 J T–1 is placedwith its axis perpendicular to the earth’s field direction. At whatdistance from the centre of the magnet, the resultant field is inclinedat 45° with earth’s field on (a) its normal bisector and (b) its axis.Magnitude of the earth’s field at the place is given to be 0.42 G.Ignore the length of the magnet in comparison to the distances involved

Answer 1

Given, the magnetic moment of the bar magnet is 5.25× 10 −2  J/T and the magnitude of the Earth’s magnetic field is 0.42× 10 −4  T.

a)

The formula to calculate the magnitude of the magnetic field at a distance r on the axis of the magnet is,

B= μ 0 4π ( m r 3 ) r 3 = μ 0 m 4πB

Here, the permeability of the free space is μ 0 , the magnetic moment is m, the magnitude of the magnetic field is B and the distance from the bar magnet is r.

Substituting the values in the above equation, we get:

r 3 = ( 4π× 10 −7 )( 5.25× 10 −2 ) 4π( 0.42× 10 −4 ) r= 12.5× 10 −5 3 =0.05 m× 100 cm 1 m =5.0 cm

Thus, the distance from the center is 5.0 cm.

b)

The formula to calculate the magnitude of the magnetic field at a distance r 1 on the axis of the magnet is,

B 1 = μ 0 4π ( 2m r 1 3 ) r 1 3 = μ 0 ( 2m ) 4π B 1

Here, the permeability of the free space is μ 0 , the magnetic moment is m, the magnitude of the magnetic field is B 1 and the distance from the bar magnet is r 1 .

Substituting the values in the above equation, we get:

r 1 3 = ( 4π× 10 −7 )×2×( 5.25× 10 −2 ) 4π( 0.42× 10 −4 ) r 1 = 25× 10 −5 3 =0.063 m× 100 cm 1 m =6.3 cm

Thus, the distance from the center is 6.3 cm.