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  • A current I is flowing through the sides of an equilateral triangle of side a. The magnitude of the magnetic field at the centroid of the triangle isOption: 1</s

A current I is flowing through the sides of an equilateral triangle of side a. The magnitude of the magnetic field at the centroid of the triangle is

Option: 1

\mathrm{\frac{3 \mu_0 I}{2 \pi a} }


Option: 2

\mathrm{\frac{9 \mu_0 I}{2 \pi a} }


Option: 3

\mathrm{\frac{3 \sqrt{3} \mu_0 I}{2 \pi a} }


Option: 4

zero


Answers (1)

best_answer

The magnetic field at centroid O due to current I flowing in side A B of the triangle is given by
\mathrm{B_{A B}=\frac{\mu_0 I}{4 \pi r}(\sin \alpha+\sin \beta) }
It is clear that  \mathrm{\alpha=\beta=60^{\circ} } and

\mathrm{\quad O D =r=\frac{A D}{\tan \alpha}=\frac{a / 2}{\tan 60^{\circ}}=\frac{a / 2}{\sqrt{3}}=\frac{a}{2 \sqrt{3}} \\ }
\mathrm{\therefore \quad \quad B_{A B} =\frac{\mu_0 I}{4 \pi} \times \frac{2 \sqrt{3}}{a} \times\left(\sin 60^{\circ}+\sin 60^{\circ}\right) \\ }
\mathrm{ =\frac{3 \mu_0 I}{2 \pi a} }

By symmetry, the magnetic fields due to current in sides B C and A C= that due to side A B. Hence, the magnetic field at $O$ due to the current in the three sides of triangle A B C is

\mathrm{ B=B_{A B}+B_{B C}+B_{C A}=3 B_{A B} }

Hence the correct choice is (b).

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