Question 4: Magnetism and Matter

Question

The magnetic field at the centre of a current carrying circular loop of radius $R$ is $16 \mu \mathrm{~T}$. The magnetic field at a distance $x=\sqrt{3} R$ on its axis from the centre is
$\_\_\_\_$ $\mu \mathrm{T}$.

Accepted Answer

The magnetic field at the centre of a circular loop of radius R carrying current i is given by:
$$ \mathrm{B}_{\mathrm{c}}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} $$
We are given that $\mathrm{B}_{\mathrm{c}}=16 \mu \mathrm{~T}$.
The magnetic field at a point on the axis at a distance x from the centre is given by:
$$ \mathrm{B}_{\mathrm{a}}=\frac{\mu_0 \mathrm{i} \mathrm{R}^2}{2\left(\mathrm{R}^2+\mathrm{x}^2\right)^{\frac{3}{2}}} $$
The ratio of magnetic field at the centre to the axis,
$\frac{\mathrm{B}_{\mathrm{C}}}{\mathrm{B}_{\mathrm{a}}}=\frac{\left(\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}}\right)}{\left(\frac{\mu_0 \mathrm{i} \mathrm{R}^2}{2\left(\mathrm{R}^2+\mathrm{x}^2\right)^{\frac{3}{2}}}\right)}$
$\Rightarrow \mathrm{B}_{\mathrm{a}}=\mathrm{B}_{\mathrm{c}}\left[\frac{\mathrm{R}^3}{\left(\mathrm{R}^2+\mathrm{x}^2\right)^{3 / 2}}\right]$
Given the distance $x=\sqrt{3} R$,
$$ \Rightarrow B_a=B_c\left[\frac{R^3}{\left(R^2+(\sqrt{3} R)^2\right)^{3 / 2}}\right]=\frac{B_c}{8} $$
$$\Rightarrow $$ $$ \mathrm{B}_{\mathrm{a}}=\frac{16 \mu \mathrm{~T}}{8}=2 \mu \mathrm{~T} $$
Hence, the correct option is (B).

Answer

8

Answer

2

Answer

4

Answer

$2 \sqrt{2}$

Magnetism and Matter

The magnetic field at the centre of a current carrying circular loop of radius R is 16 \mu \mathrm{~T}. The magnetic field at a distance x=\sqrt{3} R on its axis from the centre is \_\_\_\_ \mu \mathrm{T}.

Choose the correct answer: