A magnetic rod of length L has a magnetic dip | vedclass

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(B) The initial magnetic dipole moment is $M = m \cdot L$,where $m$ is the pole strength.When the rod is bent into a semicircle of radius $R$,its leng

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$\frac{2M}{\pi}$

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(B) The initial magnetic dipole moment is $M = m \cdot L$,where $m$ is the pole strength.When the rod is bent into a semicircle of radius $R$,its length $L$ becomes the arc length of the semicircle,so $L = \pi R$,which gives $R = \frac{L}{\pi}$.The new magnetic dipole moment $M'$ is the product of the pole strength $m$ and the straight-line distance between the poles (the diameter $2R$).$M' = m \cdot (2R) = m \cdot \left( \frac{2L}{\pi} \right)$.Since $M = m \cdot L$,we substitute $M$ into the equation:$M' = \frac{2M}{\pi}$.

$A$ magnetic rod of length $L$ has a magnetic dipole moment $M$. If it is bent into a semicircle,what will be its new magnetic dipole moment?

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