Question 10: Magnetism and Matter

Question

The current passing through a conducting loop in the form of equilateral triangle of side $4 \sqrt{3} \mathrm{~cm}$ is 2 A . The magnetic field at its centroid is $\alpha 	imes 10^{-5} \mathrm{~T}$. The value of $\alpha$ is $\_\_\_\_$ .
(Given : $\mu_{\mathrm{o}}=4 \pi 	imes 10^{-7}$ SI units)

Accepted Answer

Let L be the side of the equilateral triangle, where $\mathrm{L}=4 \sqrt{3} \mathrm{~cm}=4 \sqrt{3} 	imes 10^{-2} \mathrm{~m}$.
The centroid $O$ is equidistant from all three sides. Let $r$ be the perpendicular distance from the centroid to any side.
In an equilateral triangle :
$$ 	an 60^{\circ}=\sqrt{3}=\frac{L / 2}{r} \Rightarrow r=\frac{L}{2 \sqrt{3}} $$
Substituting the value of L :
$$ \mathrm{r}=\frac{4 \sqrt{3} 	imes 10^{-2}}{2 \sqrt{3}}=2 	imes 10^{-2} \mathrm{~m} $$
The magnetic field $B_1$ due to a straight wire of finite length at a distance $r$ is given by :
$$ \mathrm{B}_1=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{r}}\left(\sin 	heta_1+\sin 	heta_2ight) $$
Here, $	heta_1=	heta_2=60^{\circ}$
$$ B_1=\frac{\mu_0 I}{4 \pi r}\left(\sin 60^{\circ}+\sin 60^{\circ}ight)=\frac{\mu_0 I}{4 \pi r}\left(2 \sin 60^{\circ}ight) $$
$$\Rightarrow $$ $B_1=\frac{\mu_0 I}{4 \pi r}\left(2 	imes \frac{\sqrt{3}}{2}ight)=\frac{\mu_0 I \sqrt{3}}{4 \pi r}$
Since there are three sides and the current flows in a way that all three fields at the centroid point in the same direction,
$$ B_{n e t}=3 	imes B_1=\frac{3 \sqrt{3} \mu_0 I}{4 \pi r} $$
Substituting the given values :
1. $\mu_0=4 \pi 	imes 10^{-7}$
2. $\mathrm{I}=2 \mathrm{~A}$
3. $\mathrm{r}=2 	imes 10^{-2} \mathrm{~m}$
$\mathrm{B}_{\mathrm{net}}=\frac{3 \sqrt{3} 	imes\left(4 \pi 	imes 10^{-7}ight) 	imes 2}{4 \pi 	imes\left(2 	imes 10^{-2}ight)} \mathrm{T}$
$$\Rightarrow $$ $\mathrm{B}_{\mathrm{net}}=\frac{3 \sqrt{3} 	imes 10^{-7} 	imes 2}{2 	imes 10^{-2}} \mathrm{~T}$
$$\Rightarrow $$ $\mathrm{B}_{\mathrm{net}}=3 \sqrt{3} 	imes 10^{-5} \mathrm{~T}=\alpha 	imes 10^{-5} \mathrm{~T} $
$\Rightarrow \alpha=3 \sqrt{3}$

Answer

$3 \sqrt{3}$

Answer

$2 \sqrt{3}$

Answer

$\sqrt{3}$

Answer

$\frac{\sqrt{3}}{2}$

Magnetism and Matter

The current passing through a conducting loop in the form of equilateral triangle of side 4 \sqrt{3} \mathrm{~cm} is 2 A . The magnetic field at its centroid is \alpha \times 10^{-5} \mathrm{~T}. The value of \alpha is \_\_\_\_ . (Given : \mu_{\mathrm{o}}=4 \pi \times 10^{-7} SI units)

Choose the correct answer: