JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 6)
Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current I. Let $\mathrm{B}_2$ be the magnitude of magnetic field at an axial distance ' $x$ ' from the center. For $x: \mathrm{R}=3: 4, \frac{\mathrm{~B}_2}{\mathrm{~B}_1}$ is :
$64: 125$
$25: 16$
$4: 5$
$16: 25$
Explanation
_2nd_April_Morning_Shift_en_6_1.png)
$$\begin{aligned} & B_1=\frac{\mu_0 i}{2 R} \quad \quad B_2=B_1 \sin ^3 \theta \\ & \therefore \frac{B_2}{B_1}=\sin ^3 \theta=\left(\frac{4}{5}\right)^3=\frac{64}{125} \end{aligned}$$
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