Let be the magnitude of magnetic field at center of a circular coil of radius R carrying current . Let be the magnitude of magnetic field at an axial distance 'x' from the centre. For is: [2025]

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Solution

Q.
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 centre. For $x : R = 3 : 4, \frac{B_{2}}{B_{1}}$ is: [2025]

1 4 : 5

2 16 : 25

3 64 : 125

4 25 : 16

Ans.
(3)

The magnitude of magnetic field at centre of a circular coil, $B_{1} = \frac{μ_{0} i}{2 R}$

The magnitude of magnetic field at an axial distance $x = \frac{3 R}{4}$ from the centre.

$B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}} = \frac{μ_{0} i R^{2}}{2 {(\frac{5 R}{4})}^{3}} = \frac{64}{125} (\frac{μ_{0} i}{2 R})$

$\Rightarrow \frac{B_{2}}{B_{1}} = \frac{64}{125}$

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