A bar magnet of length $6 \mathrm{~cm}$ has a magnetic moment of $4 \mathrm{~J} \mathrm{~T}^{-1}$. Find the…
A bar magnet of length $6 \mathrm{~cm}$ has a magnetic moment of $4 \mathrm{~J} \mathrm{~T}^{-1}$. Find the strength of magnetic field at a distance of $200 \mathrm{~cm}$ from the centre of the magnet along its equatorial line.
$4 \times 10^{-8}$ tesla
$3.5 \times 10^{-8}$ tesla
$5 \times 10^{-8}$ tesla
$3 \times 10^{-8}$ tesla
Solution
Along the equatorial line, magnetic field strength
$
\left.B=\frac{\mu_0}{4 \pi} \frac{M}{\left(r^2+\ell^2{ }^{3 / 2}\right.}\right)
$
Given: $M=4 \mathrm{~J} T^{-1}$
$
r=200 \mathrm{~cm}=2 \mathrm{~m}
$
$\ell=\frac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$
$
\therefore B=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{4}{\left[2^2+\left(3 \times 10^{-2}\right)^2\right]^{3 / 2}}
$
Solving we get, $B=5 \times 10^{-8}$ tesla