The magnetic field due to a current carrying loop of radius $3 \mathrm{~cm}$ at a point on its axis at a…
The magnetic field due to a current carrying loop of radius $3 \mathrm{~cm}$ at a point on its axis at a distance of $4 \mathrm{~cm}$ from its centre of $54 \mu \mathrm{T}$. Then, the value of the magnetic field at the centre of the loop is
$250 \mu \top$
$150 \mu \top$
$75 \mu \top$
$125 \mu \top$
Solution
$B_{\text {axis }}=54 \times 10^{-6}=\frac{\mu N I r^2}{2\left(r^2+x^2\right)^{3 / 2}}$
Here, $\quad r=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$
and $x=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$.
From above relation, we get
$
\mu N I=\frac{54 \times 10^{-2} \times 2 \times\left(25 \times 10^{-4}\right)^{3 / 2}}{\left(3 \times 10^{-2}\right)^2}
$
So, $B_{\text {centre }}=\frac{\mu N I}{2 r}=250 \mu \mathrm{T}$