Question

Magnetic field at the centre of a coil in the form of a square of side $2cm$ carrying a current of $1.414A$ is

• A. $8\times 10^{-5}T$
• B. $7\times 10^{-5}T$
• C. $1.5\times 10^{-5}T$
• D. $6\times 10^{-5}T$

Answer 1

Answer: (A)

${\beta }_{\text{centre }}=\frac{4\times {\mu }_0}{4\pi }\times \frac{I}{(a/2)}\left(\sin 45^{\circ }+\sin 45^{\circ }\right)$
$=4\times \frac{{\mu }_0}{4\pi }\times \frac{2I}{a}\times \frac{2}{\sqrt{2}}$
$=\frac{{\mu }_{0}I\times 2\sqrt{2}}{\pi at}$
$=\frac{4\pi \times 10^{-7}\times 1.414\times 2\times \sqrt{2}}{\pi \times 2\times 10^{-2}}$
$=8\times 10^{-5}T$