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 $\mathrm{x}: \mathrm{R}=3: 4, \frac{\mathrm{~B}_{2}}{\mathrm{~B}_{1}}$ is:

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Magnetic Effects of Current

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:

(A) $4 : 5$
(B) $16 : 25$
(C) $64 : 125$ ✓ Correct
(D) $25 : 16$

Answer: (C)

Solution

On the axis of a current loop the field falls as

B_{2} = B_{1} (\frac{R ^{2}}{R ^{2} + x ^{2}})^{3/2}

. With

x : R = 3 : 4

\frac{R ^{2}}{R ^{2} + x ^{2}} = \frac{16}{16 + 9} = \frac{16}{25}

, so

\frac{B _{2}}{B _{1}} = (\frac{16}{25})^{3/2} = (\frac{4}{5})^{3} = \frac{64}{125}

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option

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