Question

Question 5.15: A short bar magnet of magnetic moment 5.25 × 10 −2 J T −1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45º 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.

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-202

Answer 1

(a) at normal bisector,
$\bf{tan\theta=\frac{B_H}{B}}$
here, according to question resultant field is inclined at 45° with earth's field.
so, $\theta$=45°
so, tan45°= 1 = $\frac{B_H}{B}$
so, $B_H=B=\frac{\mu_0}{4\pi}\frac{M}{r^3}$
Given, B =0.42 × 10^-4 T , M = 5.25 × 10^-2J/T
so, 0.42 × 10^-4 = 10^-7 × 5.25 × 10^-2/r³
r³ = 12.5 × 10^-5 = 125 × 10^-6
r = 5 × 10^-2 m = 5cm

(b) at axis of magnet,
tan45° = 1 = B/$B_H$
$B_H=B=\frac{\mu_0}{4\pi}\frac{2M}{r^3}$
so, 0.42 × 10^-4 = 10^-7 × 2 × 5.25 × 10^-2/r³
r³ = 25 × 10^-5 = 250 × 10^-6
r = 6.3 cm