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Q.
An equilateral triangular loop of wire of side $2l$ carries a current $I$. The magnetic field produced at the centre of the loop is
UP CPMTUP CPMT 2011Moving Charges and Magnetism
Solution:
The magnetic field at the centre $O$ due to current $I$ through one side $BC$ of the triangle is
$B = \frac{\mu_{0}I}{4\pi r} \left[ sin\phi_{1} + sin \phi_{2}\right] $
Here,$ r = OD = \frac{BD}{tan \,60^{\circ}}$
$\frac{l}{tan \,60^{\circ}} = \frac{l}{\sqrt{3}} $
$ \phi_{1}=\phi_{2}= 60^{\circ} $
$ \therefore B= \frac{\mu_{0}}{4\pi} \frac{I}{\left(l \sqrt{3}\right)} \left(sin\, 60^{\circ} +sin \,60^{\circ}\right) $
$= \frac{\mu_{0}}{4\pi} \frac{3I}{l} $
Since the direction of magnetic field at $O$ due to the current through all the three sides is same in magnitude and direction, hence total magnetic field at $O$ due to current through the triangle is
$B_O = 3B$
$ = \frac{\mu_{0}}{4\pi} \frac{9I}{l} $
