Question

$A$ circular loop of radius $3cm$ is having a current of $12.5A$. The magnitude of magnetic field at a distance of $4cm$ on its axis is

• A. $5.65\times 10^{-5}T$
• B. $5.27\times 10^{-5}T$
• C. $6.54\times 10^{-5}T$
• D. $9.20\times 10^{-5}T$

Answer 1

Answer: (A)

$B=\frac{{\mu }_{0}I{R}^2}{2{\left(R^2+x^2\right)}^{3/2}}$
Here, $I=12.5A,R=3cm=3\times 10^{-2}m$
$x=4cm=4\times 10^{-2}m$
$\therefore B=\frac{4\pi \times 10^{-7}\times 12.5\times {\left(3\times 10^{-2}\right)}^2}{2{\left[{\left(3\times 10^{-2}\right)}^2+{\left(4\times 10^{-2}\right)}^2\right]}^{3/2}}$
$=5.65\times 10^{-5}T$