The magnitude of magnetic field at $O$ due to a current carrying loop as shown in the figure is, where $O$ is the centre of two circular portions with radii $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively is (Take, the value of current $I=\frac{1.2}{\pi} \mathrm{A}$ )

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Aicharya · Accepted Answer

Wire lengths $a b$ and $c d$ does not produces any field at $O$ as this point lies an axis of wires.

Arc $b c$ produces an inward $\otimes$ magnetic field given by $B_1=\frac{\mu_0 I}{2 R_1} \times \frac{\theta}{360^{\circ}}$ $\begin{aligned} & =\frac{4 \pi \times 10^{-7}}{2 \times 2 \times 10^{-2}} \times \frac{1.2}{\pi} \times \frac{30}{360} \\ & =\frac{10^{-7}}{10^{-2} \times 10}=10^{-6} \mathrm{~T} \\ & =10^3 \times 10^{-9} \mathrm{~T}=1000 \mathrm{nT}\end{aligned}$ Arc ad produces an outward $\odot$ magnetic field given by $B_2=\frac{\mu_0 I}{2 R_2} \times \frac{\theta}{360^{\circ}}$ $=\frac{4 \pi \times 10^{-7} \times \frac{1.2}{\pi} \times 30}{2 \times 1 \times 10^{-2} \times 360}$ $\begin{aligned} & =\frac{2 \times 10^{-7}}{10^{-2} \times 10}=2 \times 10^{-6} \\ & =2 \times 10^3 \times 10^{-9} \mathrm{~T}=2000 \mathrm{nT}\end{aligned}$ Net field at centre $\begin{aligned} & =B_2 \odot-B_1 \otimes=2000-1000 \\ & =1000 \mathrm{nT}=10^3 \times 10^{-9} \mathrm{~T} \\ & =10^{-6} \mathrm{~T}=1 \mu \mathrm{T}\end{aligned}$

The magnitude of magnetic field at $O$ due to a current carrying loop as shown in the figure is, where $O$…

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