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QJEEM19Ph3a58: JEE Main 2019 Physics - Class 12 JEEM19Ph3a58 Solution: The magnitude of the magnetic field at the centre of an…

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 μ\mu 0 = 4 π\pi × 10–7 NA–2]

Multiple Choice Options

  • A 3 μ\mu T
  • B 18 μ\mu T
  • C 9 μ\mu T
  • D 1 μ\mu T

Step-by-Step Verified Solution

For a current carrying wire, magnetic field at a distance r is given by

JEE Main 2019 (Online) 10th April Evening Slot Physics - Magnetic Effect of Current Question 148 English Explanation 1
B=μ0i4πr(sin⁡θ1+sin⁡θ2)B=\frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right)

Now, in given case,

JEE Main 2019 (Online) 10th April Evening Slot Physics - Magnetic Effect of Current Question 148 English Explanation 2

Due to symmetry of arrangement, net field at centre of triangle is

Bnet amp;= Sum of fields of all wires (sides) amp;=3×μ0i4πr(sin⁡θ1+sin⁡θ2)\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, θ1=θ2=60∘\theta_1=\theta_2=60^{\circ}

amp;∴sin⁡θ1=sin⁡θ2=32,i=10 A,μ04π=10−7NA−2amp; and r=13× altitude amp;=13×32× sides length =123×1 m=123 m\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,

Bnetamp;=3×10−7×10×2(32)(123)=18×10−6 T⇒Bnetamp;=18μT\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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