The Total solution for NCERT class 6-12
A short bar magnet ofmagnetic moment 5.25 × 10−2 J T−1 is placedwith its axis perpendicular to the earth’s field direction. At what distancefrom the centre of the magnet, the resultant field is inclined at 45º withearth’s field on
(a) its normal bisectorand (b) its axis. Magnitude of the earth’s field at the place is given to be0.42 G. Ignore the length of the magnet in comparison to the distancesinvolved.
Magnetic moment of thebar magnet, M = 5.25 × 10−2 J T−1
Magnitude of earth’s magneticfield at a place, H = 0.42 G = 0.42 × 10−4 T
(a) The magnetic field at adistance R from the centre of the magnet on the normalbisector is given by the relation:
Where,
= Permeability of freespace = 4π × 10−7 Tm A−1
When the resultant fieldis inclined at 45° with earth’s field, B = H
(b) The magnetic field at adistanced from the centre of themagnet on its axis is given as:
The resultant fieldis inclined at 45° with earth’s field.