Question 79 of 276

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side $1m$ which is carrying a current of $10A$ is:

[Take ${\mu }_{0}=4\pi \times {10}^{-7}N{A}^{-2}$ ]

A$3\mu T$B$1\mu T$C$18\mu T$D$9\mu T$

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Answer: (C) $18\mu T$

The perpendicular distance of centroid from any side, $\dfrac{1}{2\sqrt{3}}m$
Magnetic field due to one side.
${B}_{1}=\dfrac{{µ}_{o}I}{4\pi r}[cos{\theta }_{1}+cos{\theta }_{2}]$
$=\dfrac{{\mu }_{0}(10)}{4\pi r\left(\dfrac{1}{2\sqrt{3}}\right)}\left[cos{30}^{^{\circ}}+cos{30}^{^{\circ}}\right]$
$=\dfrac{5\sqrt{3}}{\pi }{\mu }_{0}\left[\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\right]=\dfrac{15{\mu }_{0}}{\pi }$
Net magnetic field $B=3{B}_{1}=\dfrac{45}{\pi }{\mu }_{0}=\dfrac{45}{\pi }\times 4\pi \times {10}^{-7}$
$=180\times {10}^{-7}T$
$=18\mu T$