A steel wire length $ℓ$ has a magnetic moment $M$ . It is then bent into a semiconductor arc. The new magnetic moment is :

\frac{M}{π}

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\frac{2 M}{π}

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\frac{3 M}{π}

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\frac{4 M}{π}

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Solution

The correct option is C $\frac{2 M}{π}$
Let pole strength $= m$
So, $M = m l$
When wire is in form of arc, then the distance between poles $= \frac{2 l}{π}$
So, $M^{'} = \frac{m 2 l}{π} = \frac{2 M}{π}$

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A steel wire length ℓ has a magnetic moment M. It is then bent into a semiconductor arc. The new magnetic moment is :

The correct option is C 2MπLet pole strength =mSo, M=mlWhen wire is in form of arc, then the distance between poles =2lπSo, M′=m2lπ=2Mπ

A steel wire length $ℓ$ has a magnetic moment $M$ . It is then bent into a semiconductor arc. The new magnetic moment is :

The correct option is C $\frac{2 M}{π}$
Let pole strength $= m$
So, $M = m l$
When wire is in form of arc, then the distance between poles $= \frac{2 l}{π}$
So, $M^{'} = \frac{m 2 l}{π} = \frac{2 M}{π}$