A bar magnet of magnetic moment 3.0 , A-m^2 i | vedclass

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(D) The magnetic moment $M$ of a bar magnet is given by $M = m 	imes L$,where $m$ is the pole strength and $L$ is the length of the magnet.The force

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$0.1$

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(D) The magnetic moment $M$ of a bar magnet is given by $M = m 	imes L$,where $m$ is the pole strength and $L$ is the length of the magnet.The force $F$ experienced by each pole in a uniform magnetic field $B$ is given by $F = m 	imes B$.From this,the pole strength is $m = \frac{F}{B}$.Substituting $m$ into the magnetic moment formula: $M = \left( \frac{F}{B} ight) 	imes L$.Rearranging for $L$: $L = \frac{M 	imes B}{F}$.Given: $M = 3.0 \, A-m^2$,$B = 2 	imes 10^{-5} \, T$,and $F = 6 	imes 10^{-4} \, N$.$L = \frac{3.0 	imes 2 	imes 10^{-5}}{6 	imes 10^{-4}} = \frac{6 	imes 10^{-5}}{6 	imes 10^{-4}} = 0.1 \, m$.Therefore,the length of the magnet is $0.1 \, m$.

$A$ bar magnet of magnetic moment $3.0 \, A-m^2$ is placed in a uniform magnetic induction field of $2 \times 10^{-5} \, T$. If each pole of the magnet experiences a force of $6 \times 10^{-4} \, N$,the length of the magnet is.....$m$

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