Question

a magnetised wire of moment 3.14Am^2 is bent in the form of a semi circle , then the new magnetic moment will be

Answer 1

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Answer 2

The new magnetic moment formed in the wire is 2.

Solution:

The pole strength of wire is assumed to be p and the length is assumed L, then

$P \times L=3.14 \mathrm{Am}^{2}$

The wire is then bent in the form of a semicircle

We know L = 3.14R

So $R=\frac{L}{3.14}$

So the semicircle’s effective length = 2 R $= \frac{2 L}{3.14}$

So the new magnetic moment formed in wire $=\mathrm{P} \times(2 \mathrm{R})=\left(\frac{2 L}{3.14}\right) \times \mathrm{P}$

$\{\mathrm{P} \times \mathrm{L}=3.14\}$

$\left(\frac{2}{3.14}\right) \times(3.14)=2$

The magnetic moment is 2.