Physics - Magnetism Question with Solution | TestHub

The magnetic field on the axis of a bar magnet of length $2\mathrm{l}$, having a pole strength $\mathrm{m}$, at a distance $\mathrm{d}$ is

$\mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{m}}{{\left[\mathrm{d}-\mathrm{l}\right]}^2}-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{m}}{{\left[\mathrm{d}+\mathrm{l}\right]}^2}$

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{m}}{4\mathrm{\pi }}\frac{{\left[\mathrm{d}+\mathrm{l}\right]}^2-{\left[\mathrm{d}-\mathrm{l}\right]}^2}{{\left[{\mathrm{d}}^2-{\mathrm{l}}^2\right]}^2}=\frac{{\mathrm{\mu }}_0\mathrm{mld}}{\mathrm{\pi }{\left[{\mathrm{d}}^2-{\mathrm{l}}^2\right]}^2}$

$\Rightarrow \frac{{\mathrm{B}}_1}{{\mathrm{B}}_2}=\mathrm{ }\frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}\mathrm{ }{\left[\frac{{\mathrm{d}}_2^2-{\mathrm{l}}^2}{{\mathrm{d}}_1^2-{\mathrm{l}}^2}\right]}^2$

$\Rightarrow \frac{12.5}{1}=\frac{10}{20}\mathrm{ }{\left[\frac{400-{\mathrm{l}}^2}{100-{\mathrm{l}}^2}\right]}^2$

$\Rightarrow \mathrm{l}=\; 5\mathrm{cm}$

Hence, length of magnet $=2\mathrm{l}=\; 10\mathrm{cm}$