The magnetic field due to a current carrying loop of radius 3 cm at a point on its axis at a distance of 4 cm from its centre is 54$\mu$T. Then the value of the magnetic field at the centre of the loop is ____

Given data:

At a point located 4 cm from the center of a current-carrying loop with a radius of 3 cm, the magnetic field strength is measured as $54\mu T$.

Formula used: $B=\frac{{\mu }_0}{4\pi }\left[2\pi \frac{a^2}{{\left(a^2+x^2\right)}^{\frac{3}{2}}}\right]$

Detailed Solution:

The magnetic field generated by a circular loop carrying current with a radius of a, at a distance of x from the center, on the axis is described as follows:

$B=\frac{{\mu }_0}{4\pi }\left[2\pi \frac{a^2}{{\left(a^2+x^2\right)}^{\frac{3}{2}}}\right]$

$$\begin{gathered} \Rightarrow 54\times {10}^{-6}=\frac{{10}^{-7}\times I\times 2\pi \times {\left(3\times {10}^{-2}\right)}^2}{{\left[{\left(3\times {10}^{-2}\right)}^2+{\left(4\times {10}^{-2}\right)}^2\right]}^{\frac{3}{2}}} \\ \Rightarrow I=11.94A \end{gathered}$$

$$\begin{gathered} \therefore \text{Magnetic field at the center = }\frac{{\mu }_{0}I}{4\pi a}2\pi \\ =\frac{{10}^{-7}\times 11.94\times 2\pi }{3\times {10}^{-2}} \\ =250\mu T \end{gathered}$$

Ans: Therefore, the correct answer is $250\mu T$. Hence, option A is correct.