Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current $I$. Let $B_2$ be the magnitude of magnetic field at an axial distance ' $x$ ' from the center. For $x: R=3: 4, \frac{B_2}{B_1}$ is :

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{B}_1=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} \qquad \mathrm{B}_2=\mathrm{B}_1 \sin ^3 \theta \\ & \therefore \frac{\mathrm{~B}_2}{\mathrm{~B}_1}=\sin ^3 \theta=\left(\frac{4}{5}\right)^3=\frac{64}{125}\end{aligned}$

Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current $I$…

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