Question

The magnetic field due to current carrying circular loop of radius 6 cm at a point on the axis at a distance of 8 cm from centre is 27 $\mu$T. The magnetic field at the centre of the current carrying loop is.

• A. 75 $\mu$T
• B. 125 $\mu$T
• C. 150 $\mu$T
• D. 250 $\mu$T

Hint:

Magnetic field on axis of circular loop: $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Magnetic field at center of circular loop: $B_{center} = \frac{\mu_0 I}{2R}$.
The ratio is $\frac{B_{center}}{B_{axis}} = \frac{(R^2+x^2)^{3/2}}{R^3} = (1 + (x/R)^2)^{3/2}$.
For calculations involving ratios of lengths (like $x/R$), units can be kept consistent (e.g., all cm) as they will cancel.
Recognize Pythagorean triples if they appear (e.g., $R=6, x=8 \implies \sqrt{R^2+x^2} = \sqrt{36+64} = \sqrt{100} = 10$).

Answer 1

The Correct Option is B

The magnetic field ($B_{axis}$) at a point on the axis of a circular current-carrying loop of radius $R$ at a distance $x$ from its center is given by: $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$ where $I$ is the current in the loop and $\mu_0$ is the permeability of free space. The magnetic field ($B_{center}$) at the center of the circular loop (where $x=0$) is: $B_{center} = \frac{\mu_0 I R^2}{2(R^2 + 0^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2)^{3/2}} = \frac{\mu_0 I R^2}{2R^3} = \frac{\mu_0 I}{2R}$. We are given: Radius of the loop $R = 6 \text{ cm} = 0.06 \text{ m}$. Distance on the axis $x = 8 \text{ cm} = 0.08 \text{ m}$. Magnetic field at this axial point $B_{axis} = 27 \text{ } \mu\text{T}$. We can find the ratio $\frac{B_{center}}{B_{axis}}$: $\frac{B_{center}}{B_{axis}} = \frac{\mu_0 I / (2R)}{\mu_0 I R^2 / (2(R^2 + x^2)^{3/2})} = \frac{1}{R} \times \frac{(R^2 + x^2)^{3/2}}{R^2} = \frac{(R^2 + x^2)^{3/2}}{R^3}$. $\frac{B_{center}}{B_{axis}} = \left(\frac{R^2 + x^2}{R^2}\right)^{3/2} = \left(1 + \frac{x^2}{R^2}\right)^{3/2}$. Substitute the given values for $R$ and $x$: $R^2 = (6 \text{ cm})^2 = 36 \text{ cm}^2$. $x^2 = (8 \text{ cm})^2 = 64 \text{ cm}^2$. $R^2 + x^2 = 36 + 64 = 100 \text{ cm}^2$. (Note: units cm will cancel out in the ratio $x^2/R^2$, so no need to convert to meters here for the ratio). $\frac{B_{center}}{B_{axis}} = \left(\frac{100 \text{ cm}^2}{36 \text{ cm}^2}\right)^{3/2} = \left(\frac{100}{36}\right)^{3/2}$. $\left(\frac{100}{36}\right)^{3/2} = \left(\left(\frac{10}{6}\right)^2\right)^{3/2} = \left(\frac{10}{6}\right)^{2 \times (3/2)} = \left(\frac{10}{6}\right)^3$. $\left(\frac{10}{6}\right)^3 = \left(\frac{5}{3}\right)^3 = \frac{5^3}{3^3} = \frac{125}{27}$. So, $\frac{B_{center}}{B_{axis}} = \frac{125}{27}$. $B_{center} = \frac{125}{27} \times B_{axis}$. Given $B_{axis} = 27 \text{ } \mu\text{T}$. $B_{center} = \frac{125}{27} \times 27 \text{ } \mu\text{T} = 125 \text{ } \mu\text{T}$. \[ \boxed{125 \text{ } \mu\text{T}} \]