JEE Main 2025 Apr 2 Shift 1, Physics Q14: Magnetic Field due to Electric Current

JEE Main2025Apr 2, Shift 1Physics

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
D
25 : 16

Solution

Magnetic field of a circular coil:

At the centre: $B_{1} = \frac{μ _{0} I}{2 R}$ .

At axial distance $x$ : $B_{2} = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}$ .

Taking the ratio:

$\frac{B _{2}}{B _{1}} = \frac{R ^{3}}{( R ^{2} + x ^{2} ) ^{3/2}} = (\frac{R}{R ^{2} + x ^{2}})^{3} = sin^{3} θ$

where $sin θ = \frac{R}{R ^{2} + x ^{2}}$ .

With $x : R = 3 : 4$ (so $x = 3 k$ , $R = 4 k$ ):

$R^{2} + x^{2} = 16 k^{2} + 9 k^{2} = 5 k$

$sin θ = \frac{4}{5}$

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

$∴$ Option (3) is correct.

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JEE Main 2025 Apr 2 Shift 1, Physics Q14: Magnetic Field due to Electric Current

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