A bar magnet of magnetic moment $5 \mathrm{Am}^{2}$ is placed in a uniform magnetic induction $3 \times…
A bar magnet of magnetic moment $5 \mathrm{Am}^{2}$ is placed in a uniform magnetic
induction $3 \times 10^{-5} \mathrm{~T}$. If each pole of a magnet experiences a force of $2 \cdot 5 \times 10^{-4} \mathrm{~N}$
then the magnetic length of the magnet is
$0.8 \mathrm{~m}$
$0 \cdot 2 \mathrm{~m}$
$0.6 \mathrm{~m}$
$0 \cdot 4 \mathrm{~m}$
Solution
$\mathrm{F}=\mathrm{mB} \quad$ where $\mathrm{m}$ is pole strength.
$\therefore \mathrm{m}=\frac{\mathrm{F}}{\mathrm{B}}=\frac{2.5 \times 10^{-4}}{3 \times 10^{-5}}=\frac{25}{3} \mathrm{Am}$
Magnetic moment $\mathrm{M}=\mathrm{mL}$
$\therefore \mathrm{L}=\frac{\mathrm{M}}{\mathrm{m}}=\frac{5 \times 3}{25}=0.6 \mathrm{~m}$