Question

$A$ circular loop of radius $3 \,cm$ is having a current of $12.5\, A$. The magnitude of magnetic field at a distance of $4 \,cm$ 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)

$B=\frac{\mu_{0}IR^{2}}{2\left(R^{2}+x^{2}\right)^{3/ 2}}$
Here, $ I=12.5\,A, R=3\,cm =3\times10^{-2}\,m$
$x = 4 cm=4\times10^{-2}\,m$
$\therefore B=\frac{4\pi\times10^{-7}\times12.5\times\left(3\times10^{-2}\right)^{2}}{2\left[\left(3\times10^{-2}\right)^{2} +\left(4\times10^{-2}\right)^{2}\right]^{3 /2}}$
$=5.65\times10^{-5}\,T $