Question 193: Magnetism and Matter

Question

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is
carrying a current of 10 A is : [Take $$\mu $$0 = 4$$\pi $$ × 10–7
 NA–2]

Accepted Answer

For a current carrying wire, magnetic field at
a distance r is given by
$$
B=\frac{\mu_0 i}{4 \pi r}\left(\sin 	heta_1+\sin 	heta_2ight)
$$
Now, in given case,
Due to symmetry of arrangement, net field at centre of triangle is
$$
\begin{aligned}
B_{	ext {net }} & =	ext { Sum of fields of all wires (sides) } \\
& =3 	imes \frac{\mu_0 i}{4 \pi r}\left(\sin 	heta_1+\sin 	heta_2ight)
\end{aligned}
$$
Here, $	heta_1=	heta_2=60^{\circ}$
$$
\begin{aligned}
& 	herefore \sin 	heta_1=\sin 	heta_2=\frac{\sqrt{3}}{2}, i=10 \mathrm{~A}, \frac{\mu_0}{4 \pi}=10^{-7} \mathrm{NA}^{-2} \\
& 	ext { and } r=\frac{1}{3} 	imes 	ext { altitude } \\
& \qquad=\frac{1}{3} 	imes \frac{\sqrt{3}}{2} 	imes 	ext { sides length }=\frac{1}{2 \sqrt{3}} 	imes 1 \mathrm{~m}=\frac{1}{2 \sqrt{3}} \mathrm{~m}
\end{aligned}
$$
So,
$$
\begin{aligned}
B_{\mathrm{net}} & =\frac{3 	imes 10^{-7} 	imes 10 	imes 2\left(\frac{\sqrt{3}}{2}ight)}{\left(\frac{1}{2 \sqrt{3}}ight)}=18 	imes 10^{-6} \mathrm{~T} \\
\Rightarrow B_{\mathrm{net}} & =18 \mu \mathrm{T}
\end{aligned}
$$

Answer

3 $$\mu $$T

Answer

18 $$\mu $$T

Answer

9 $$\mu $$T

Answer

1 $$\mu $$T

Magnetism and Matter

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is : [Take $\mu 0 = 4\pi $ × 10–7 NA–2]

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