JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 4)
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]
[Take $$\mu $$0 = 4$$\pi $$ × 10–7 NA–2]
3 $$\mu $$T
18 $$\mu $$T
9 $$\mu $$T
1 $$\mu $$T
Explanation
For a current carrying wire, magnetic field at
a distance r is given by
_10th_April_Evening_Slot_en_4_1.png)
$$ B=\frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) $$
Now, in given case,
_10th_April_Evening_Slot_en_4_2.png)
Due to symmetry of arrangement, net field at centre of triangle is
$$ \begin{aligned} B_{\text {net }} & =\text { Sum of fields of all wires (sides) } \\\\ & =3 \times \frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) \end{aligned} $$
Here, $\theta_1=\theta_2=60^{\circ}$
$$ \begin{aligned} & \therefore \sin \theta_1=\sin \theta_2=\frac{\sqrt{3}}{2}, i=10 \mathrm{~A}, \frac{\mu_0}{4 \pi}=10^{-7} \mathrm{NA}^{-2} \\\\ & \text { and } r=\frac{1}{3} \times \text { altitude } \\\\ & \qquad=\frac{1}{3} \times \frac{\sqrt{3}}{2} \times \text { sides length }=\frac{1}{2 \sqrt{3}} \times 1 \mathrm{~m}=\frac{1}{2 \sqrt{3}} \mathrm{~m} \end{aligned} $$
So,
$$ \begin{aligned} B_{\mathrm{net}} & =\frac{3 \times 10^{-7} \times 10 \times 2\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2 \sqrt{3}}\right)}=18 \times 10^{-6} \mathrm{~T} \\\\ \Rightarrow B_{\mathrm{net}} & =18 \mu \mathrm{T} \end{aligned} $$
$$ B=\frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) $$
Now, in given case,
Due to symmetry of arrangement, net field at centre of triangle is
$$ \begin{aligned} B_{\text {net }} & =\text { Sum of fields of all wires (sides) } \\\\ & =3 \times \frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) \end{aligned} $$
Here, $\theta_1=\theta_2=60^{\circ}$
$$ \begin{aligned} & \therefore \sin \theta_1=\sin \theta_2=\frac{\sqrt{3}}{2}, i=10 \mathrm{~A}, \frac{\mu_0}{4 \pi}=10^{-7} \mathrm{NA}^{-2} \\\\ & \text { and } r=\frac{1}{3} \times \text { altitude } \\\\ & \qquad=\frac{1}{3} \times \frac{\sqrt{3}}{2} \times \text { sides length }=\frac{1}{2 \sqrt{3}} \times 1 \mathrm{~m}=\frac{1}{2 \sqrt{3}} \mathrm{~m} \end{aligned} $$
So,
$$ \begin{aligned} B_{\mathrm{net}} & =\frac{3 \times 10^{-7} \times 10 \times 2\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2 \sqrt{3}}\right)}=18 \times 10^{-6} \mathrm{~T} \\\\ \Rightarrow B_{\mathrm{net}} & =18 \mu \mathrm{T} \end{aligned} $$
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