First slide

Magnetic field due to a straight current carrying wire

Question

The magnetic field at the centre of an equilateral triangular loop of side 2L and carrying a current i is

Moderate

A

$\frac{9 μ_{0} i}{4 π L}$
B

$\frac{3 \sqrt{3} μ_{0} i}{4 π L}$
C

$\frac{2 \sqrt{3} μ_{0} i}{π L}$
D

$\frac{3 μ_{0} i}{4 π L}$

Solution

Magnetic field at O is
$B = 3 [\frac{μ_{0}}{4 π} \frac{i}{r} (\sin 60^{0} + \sin 60^{0})]$
$\begin{array}{l} = 3 (\frac{μ_{0}}{4 π}) (i) (\frac{\sqrt{3}}{L}) (\sqrt{3}) \\ = \frac{9 μ_{0} i}{4 πL} \end{array}$

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