JEE Physics 2025 Question with Solution

Given: A circular coil of radius $R$ carries current $I$. Magnetic field at the center is $B_1$ and magnetic field at axial distance $x$ is $B_2$. Also, $x : R = 3 : 4$.

Find: The ratio $\frac{B_2}{B_1}$.

For a circular coil, the magnetic field at the center is

$$B_1 = \frac{\mu_0 I}{2R}$$

and the magnetic field at a point on its axis is

$$B_2 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$

Now form the ratio:

$$\frac{B_2}{B_1} = \frac{\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}}{\frac{\mu_0 I}{2R}}$$

Cancel the common factor $\frac{\mu_0 I}{2}$:

$$\frac{B_2}{B_1} = \frac{R^3}{(R^2 + x^2)^{3/2}}$$

Given

$$\frac{x}{R} = \frac{3}{4} \implies x = \frac{3R}{4}$$

Substitute this in the ratio:

$$\frac{B_2}{B_1} = \frac{R^3}{\left(R^2 + \left(\frac{3R}{4}\right)^2\right)^{3/2}}$$ $$\frac{B_2}{B_1} = \frac{R^3}{\left(R^2 + \frac{9R^2}{16}\right)^{3/2}} = \frac{R^3}{\left(\frac{25R^2}{16}\right)^{3/2}}$$

Now evaluate the denominator:

$$\left(\frac{25R^2}{16}\right)^{3/2} = \left(\frac{5R}{4}\right)^3 = \frac{125R^3}{64}$$

Therefore,

$$\frac{B_2}{B_1} = \frac{R^3}{125R^3/64} = \frac{64}{125}$$

Therefore, the correct option is C and the ratio is $\frac{64}{125}$.

Given: Radius of coil $R$, current $I$, and axial distance ratio $x:R = 3:4$.

Find: The value of $\frac{B_2}{B_1}$.

Concept used: The magnetic field is maximum at the center of a circular coil and decreases along the axis. Use the field expressions at the center and at an axial point.

1. At the center,

$$B_1 = \frac{\mu_0 I}{2R}$$

2. On the axis at distance $x$,

$$B_2 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$

3. Divide the second expression by the first:

$$\frac{B_2}{B_1} = \frac{\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}}{\frac{\mu_0 I}{2R}} = \frac{R^3}{(R^2 + x^2)^{3/2}}$$

4. From $x:R = 3:4$,

$$x = \frac{3R}{4}$$

5. Substitute into the ratio:

$$\frac{B_2}{B_1} = \frac{R^3}{\left(R^2 + \frac{9R^2}{16}\right)^{3/2}} = \frac{R^3}{\left(\frac{25R^2}{16}\right)^{3/2}}$$

6. Simplify:

$$\left(\frac{25R^2}{16}\right)^{3/2} = \frac{125R^3}{64}$$

So,

$$\frac{B_2}{B_1} = \frac{R^3}{125R^3/64} = \frac{64}{125}$$

Therefore, the ratio is $\frac{64}{125}$.