question_answer

A bar magnet of length 3 cm has points A and B along its axis at distances of 24 cm and 48 cm on the opposite sides. Ratio of magnetic fields at these points will be:

A)  8             

B)  \[1/2\sqrt{2}\]

C)  3           

D)  4

Correct Answer: A

Solution :

 Key Idea: The given position is also known as end-on or tan A position. When magnet is small compared to distance of points B and A, then magnetic field B is given by \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\,M}{{{r}^{3}}}\] where M is magnetic moment. \[\therefore \] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{r_{2}^{3}}{r_{1}^{3}}\] Given, \[{{r}_{2}}=48\,cm,\,{{r}_{1}}=24\,cm\] \[\therefore \] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{(48)}^{3}}}{{{(24)}^{3}}}={{(2)}^{3}}=8\]