Question

A magnetised steel wire $31.4cm$ long has a pole strength of $0.2Am$. It is then bent in the form of a semicircle. Calculate the magnetic moment of this needle.

• A. $0.01A{m}^2$
• B. $0.02A{m}^2$
• C. $0.03A{m}^2$
• D. $0.04A{m}^2$

Answer 1

The correct option is D $0.04A{m}^2$

Given length of wire $l=31.4cm=0.314m$

Pole strength $m=0.2Am$

Magnetic moment $=m\times r=0.2\times 0.314=0.628A{m}^2$

For semicircle,$l=\pi r⟹r^{{}^{\prime }}=\frac{l}{\pi }$ and

$M^{{}^{\prime }}=m\times 2r^{{}^{\prime }}$

$=m\times \frac{2\times l}{\pi }$

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

$=\frac{0.628}{3.14}\times 2$

$=0.04A{m}^2$