question_answer

Magnetic field along the axis at a point distance d from a short bar magnet, is :

A)  \[\frac{{{\mu }_{0}}}{3N}\times \frac{M}{{{d}^{3}}}\]

B)  \[\frac{{{\mu }_{0}}}{N}\times \frac{M}{{{d}^{3}}}\]

C)  \[\frac{{{\mu }_{0}}}{2\pi }\times \frac{M}{{{d}^{3}}}\]

D)  \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{M}{{{d}^{3}}}\]

Correct Answer: C

Solution :

Key Idea: Along axial point the position is known as tan A position. When magnetic moment of magnet is M, then the magnitude of \[\overrightarrow{B}\], at an axial distance d is given by \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\,Md}{{{({{d}^{2}}-{{t}^{2}})}^{2}}}\] where \[2\,l\] is length of magnet. Since, magnet is small compared to d, \[{{l}^{2}}\]can be ignored, then \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{d}^{3}}}=\frac{{{\mu }_{0}}}{2\pi }\frac{M}{{{d}^{3}}}\] Note: In \[\tan B\]or broadside on position, magnitude of magnetic field is half of \[\tan A\]position.