Question 202: 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 ]

Choose the correct answer: