1.

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»

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.


Discussion

No Comment Found

Related InterviewSolutions