Question

A bar magnet of magnetic moment 'M' has a magnetic length '2d'. Find magnetic induction on its equatorial line at a distance ${}^{\prime }{\sqrt{13d}}^{\prime }$.

• A. $\frac{{\mu }_{0}M}{4\pi \left(d^3\right)(17{)}^{\frac{3}{2}}}$
• B. $\frac{2{\mu }_{0}M}{4\pi \left(d^3\right)(17{)}^{\frac{3}{2}}}$
• C. $\frac{4{\mu }_{0}M}{4\pi \left(d^3\right)(17{)}^{\frac{3}{2}}}$
• D. $\frac{8{\mu }_{0}M}{4\pi \left(d^3\right)(17{)}^{\frac{3}{2}}}$

Answer 1

The correct option is A $\frac{{\mu }_{0}M}{4\pi \left(d^3\right)(17{)}^{\frac{3}{2}}}$

$r=\sqrt{13}d;2l=2d$
$B_{e}q=\frac{{\mu }_0}{4\pi }\times \frac{\mu }{(r^2+1^2{)}^{\frac{3}{2}}}$
$=\frac{{\mu }_0}{4\pi }\times \frac{M}{{\left(\right(\sqrt{13}d{)}^2+\left(2d{)}^2\right)}^{\frac{3}{2}}}$
$=\frac{{\mu }_0}{4\pi }\times \frac{M}{(17d^2{)}^{\frac{3}{2}}}=\frac{{\mu }_{0}M}{4\pi \left(d^3\right)(17{)}^{\frac{3}{2}}}$