Question

Question 5.12: A short bar magnet has a magnetic moment of 0.48 J T −1 . Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-201

Answer 1

Given, Magnetic moment, M = 0.48J/T
distance of observation point from centre of the magnet, r = 10cm = 0.1m

(a) we know, magnetic field on axial line due to magnetic dipole is given by ,
$\bf{B_{axial}=\frac{\mu_0}{4\pi}\frac{2M}{r^3}}$

= (10^-7 × 2 × 0.48)/(0.1)³ = 9.6 × 10^-5T along S-N direction.

(b) magnetic field on equatorial line due to magnetic dipole is given by ,
$\bf{B_{equatorial}=\frac{\mu_0}{4\pi}\frac{M}{r^3}}$

= (10^-7 × 0.48)/(0.1)³ = 4.8 × 10^-5 T along N-S direction.

Answer 2

Answer:

Magnetic moment = 0.48 J T^-1

M= 0.48 J T^-1

Distance d= 10 cm =0.1 m

We've a relation for magnetic field produced by magnets as;

Magnetic field B= (μ₀*2*M)/(2π*d³)  

B= 0.48*(0.1)³ *10^−7

B= 0.48*10^−4 T

Since the direction of magnetic field is opposite to the direction of magnetic moment.

B= 0.48*10^−4 which is opposite to the direction of magnetic moment which is along North-South direction.