A short bar magnet of magnetic moment 255 ${JT}^{- 1}$ is placed with its axis perpendicular to earth’s field direction. At what distance from the center of the magnet, the resultant field is inclined at $45{^\circ}$ with earth’s field, $H = 0.4 imes 10^{- 4}\ T\ ?$

Question

A short bar magnet of magnetic moment 255 ${JT}^{- 1}$ is placed with its axis perpendicular to earth’s field direction. At what distance from the center of the magnet, the resultant field is inclined at $45{^\circ}$ with earth’s field, $H = 0.4 	imes 10^{- 4}\ T\ ?$

Dr. Abubacker · Accepted Answer

(b) Since, B and H are perpendicular to each other and the resultant field is inclined at an angle $45{^\circ}$ with.
So, B = H
$\frac{\mu_{0}}{4\pi}\frac{2M}{r^{3}} = H$
$	herefore r^{3} = \frac{\mu_{0}}{4\pi}\frac{2M}{H} = 0.5\ m$

Q. A short bar magnet of magnetic moment 255 ${JT}^{- 1}$ is placed with its axis perpendicular to earth’s field direction. At what distance from the center of the magnet, the resultant field is inclined at $45{^\circ}$ with earth’s field, $H = 0.4 \times 10^{- 4}\ T\ ?$

Explanation

Reviewed by Dr. Abubacker

Want more practice?