First slide

Magnetic Dipole moment

Question

A thin circular wire carrying a current I has a magnetic moment M. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment

Moderate

A

M
B

$\frac{4}{π^{2}} M$
C

$\frac{4}{π} M$
D

$\frac{π}{4} M$

Solution

$I ({πr}^{2}) = M 2 πr = l$
$M = \frac{{Iπl}^{2}}{4 π^{2}} = \frac{{Il}^{2}}{4 π} r = \frac{1}{2 π}$
When it is made a square l = 4a.
$M^{1} = {Ia}^{2} = I \frac{l^{2}}{16} a = \frac{1}{4}$
$\frac{M}{{M^{1}}_{}} = \frac{{Il}^{2} / 4 π}{{Il}^{2} / 16} = \frac{16}{4 π} = \frac{4}{π}$
$M^{1} = \frac{πM}{4}$

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