Question

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

$8\times 10^{-5}T$

Answer 2

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