A bar magnet with a magnetic moment of 3.0 , | vedclass

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(D) The magnetic moment $M$ of a bar magnet is given by $M = m \cdot L$,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 $M = m \cdot L$,where $m$ is the pole strength and $L$ is the length of the magnet.The force $F$ acting on a pole in a magnetic field $B$ is given by $F = m \cdot B$.From the first equation,we have $m = \frac{M}{L}$.Substituting this into the force equation: $F = \frac{M}{L} \cdot B$.Given $M = 3.0 \, A \cdot m^2$,$B = 2 	imes 10^{-5} \, T$,and $F = 6 	imes 10^{-4} \, N$.Substituting the values: $6 	imes 10^{-4} = \frac{3.0}{L} 	imes 2 	imes 10^{-5}$.$L = \frac{3.0 	imes 2 	imes 10^{-5}}{6 	imes 10^{-4}} = \frac{6 	imes 10^{-5}}{6 	imes 10^{-4}} = 10^{-1} = 0.1 \, m$.

$A$ bar magnet with a magnetic moment of $3.0 \, A \cdot m^2$ is placed in a magnetic field of $2 \times 10^{-5} \, T$. If the force acting on each pole is $6 \times 10^{-4} \, N$,what is the length of the magnet in $m$?

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