A) \[8\]
B) \[\frac{1}{2\sqrt{2}}\]
C) \[3\]
D) \[4\]
Correct Answer: A
Solution :
Magnetic field due to a bar magnet at a distance\[r\] from the centre of magnet on axial position \[B=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2M}{{{r}^{3}}}\]\[\Rightarrow \]\[\frac{{{B}_{1}}}{{{B}_{2}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{3}}={{\left( \frac{48}{24} \right)}^{3}}=\frac{8}{1}=8\]You need to login to perform this action.
You will be redirected in
3 sec