The magnetic field due to current carrying a circular loop of radius $5 \mathrm{~cm}$ at a point on the axis…
The magnetic field due to current carrying a circular loop of radius $5 \mathrm{~cm}$ at a point on the axis at a distance of $12 \mathrm{~cm}$ from the centre is $250 \mu \mathrm{T}$. The magnetic field at the centre of the loop is
$2529 \mu \mathrm{T}$
$4394 \mu \mathrm{T}$
$1759 \mu \mathrm{T}$
$2908 \mu \mathrm{T}$
Solution
Given, $R=5 \mathrm{~cm}=0.05 \mathrm{~m}$
$
\begin{aligned}
r & =12 \mathrm{~cm}=0.12 \mathrm{~m} \\
B & =250 \mu \mathrm{T} \\
& =250 \times 10^{-6} \mathrm{~T}
\end{aligned}
$
The magnetic field due to a current carrying circular loop at its axial line is
$
B=\frac{1}{2} \frac{\mu_0 n I R^2}{\left(R^2+r^2\right)^{3 / 2}}
$
Here, $n=1$,
$
\Rightarrow \quad B=\frac{1}{2} \frac{\mu_0 I R^2}{\left(R^2+r^2\right)^{3 / 2}}
$
The magnetic field at the centre of loop is
$
B_0=\frac{1}{2} \frac{\mu_0 I}{R}
$
Dividing Eq. (ii) by Eq. (i), we get
$
\begin{aligned}
\frac{B_0}{B} & =\frac{\mu_0 I}{2 R} \times \frac{2\left(R^2+r^2\right)^{3 / 2}}{\mu_0 I R^2} \\
& =\frac{\left(R^2+r^2\right)^{3 / 2}}{R^3} \\
\Rightarrow \quad B_0 & =B \times \frac{\left(R^2+r^2\right)^{3 / 2}}{R^3}
\end{aligned}
$
$
\begin{aligned}
& =250 \times 10^{-6} \times \frac{\left[(0.05)^2+(0.12)^2\right]^{3 / 2}}{(0.05)^3} \\
& =250 \times 10^{-6} \times \frac{(0.0169)^{3 / 2}}{(0.05)^3} \\
& =4394 \mu \mathrm{T}
\end{aligned}
$