A magnetized wire of magnetic moment M and le | vedclass

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Question

(A) The initial magnetic moment of the wire is $M = m \cdot l$, where $m$ is the pole strength.When the wire of length $l$ is bent into a semicircle,

Vedclass · Accepted Answer

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

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(A) The initial magnetic moment of the wire is $M = m \cdot l$, where $m$ is the pole strength.When the wire of length $l$ is bent into a semicircle, the length $l$ becomes the arc length of the semicircle.Thus, $l = \pi r$, which implies $r = \frac{l}{\pi}$.The new magnetic moment $M^{\prime}$ is the product of the pole strength $m$ and the straight-line distance between the poles (the diameter of the semicircle).$M^{\prime} = m \cdot (2r) = m \cdot \left( \frac{2l}{\pi} \right)$.Since $M = m \cdot l$, we can substitute $m \cdot l = M$ into the equation:$M^{\prime} = \frac{2}{\pi} (m \cdot l) = \frac{2 M}{\pi}$.

$A$ magnetized wire of magnetic moment $M$ and length $l$ is bent in the form of a semicircle of radius $r$. Then its new magnetic moment is:

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