A bar magnet of magnetic moment 5 ,Am^{2} is | vedclass

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(C) The force $F$ experienced by a magnetic pole of strength $m$ in a uniform magnetic field $B$ is given by $F = mB$.Given $F = 2.5 	imes 10^{-4} \,

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$0.6$

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(C) The force $F$ experienced by a magnetic pole of strength $m$ in a uniform magnetic field $B$ is given by $F = mB$.Given $F = 2.5 	imes 10^{-4} \,N$ and $B = 3 	imes 10^{-5} \,T$.Calculating the pole strength $m$:$m = \frac{F}{B} = \frac{2.5 	imes 10^{-4}}{3 	imes 10^{-5}} = \frac{25}{3} \,Am$.The magnetic moment $M$ is defined as the product of pole strength $m$ and magnetic length $L$, i.e., $M = mL$.Given $M = 5 \,Am^{2}$.Therefore, the magnetic length $L$ is:$L = \frac{M}{m} = \frac{5}{25/3} = \frac{5 	imes 3}{25} = \frac{15}{25} = 0.6 \,m$.

$A$ bar magnet of magnetic moment $5 \,Am^{2}$ is placed in a uniform magnetic induction $3 \times 10^{-5} \,T$. If each pole of a magnet experiences a force of $2.5 \times 10^{-4} \,N$, then the magnetic length of the magnet is: (in $\,m$)

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