Comment
Share
Q)

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

$\begin{array}{1 1} 0.4 \;Am^2 \\ 0.04 \;Am^2 \\ 0.4 \;A/m^2 \\ 0.04 \;A/m^2 \end{array}$

Comment
A)
Solution:
$L= 31.4 \;cm$
$m= 0.2 \;Am$
$L= \pi r$
=> $31.4 =3.14 (r)$
$r= 10\;cm$
Distance between the two ends is $2r$
$\qquad= 2r$
$\qquad= 20\;cm=0.2 \;m$
$M= m \times 2l$
$\qquad= 0.2 \times 0.2 =0.04\;Am^2$
Answer : $0.04 \;Am^2$