Magnetic field due to currentMedium
Question
The current passing through a conducting loop in the form of equilateral triangle of side $4\sqrt{3}\text{ }cm$ is 2A. The magnetic field at its centroid is $\alpha \times 10^{- 5}\text{ }T$. The value of $\alpha$ is $\_\_\_\_$ .
(Given : $\mu_{o} = 4\pi \times 10^{- 7}$ SI units)
Options
A.$2\sqrt{3}$
B.$\sqrt{3}$
C.$3\sqrt{3}$
D.$\frac{\sqrt{3}}{2}$
Solution
$${B = \frac{\mu_{0}}{4\pi} \times \frac{I}{d}\left\lbrack sin60^{\circ} + sin60^{\circ} \right\rbrack \times 3 }{B = 10^{- 7} \times \frac{2}{2 \times 10^{- 2}}\left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right) \times 3 }{= \sqrt{3} \times 10^{- 5} \times 3 = 3\sqrt{3} \times 10^{- 5}}$$
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