Question

ABCD is a square loop made of a uniform conducting wire. A current enters the loop at A and leaves at D. The magnetic field is

• A. zero only at the centre of the loop
• B. maximum at the centre of the loop
• C. zero at all points outside the loop
• D. zero at all points inside the loop

Answer 1

Answer: (A)

zero only at the centre of the loop

Out of the current I entering A, $\frac{I}{4}$ goes along ABCD and

$\frac{3I}{4}$ goes along AD.

$(\overrightarrow{B}{)}_{AD}=\frac{{\mu }_0\frac{3I}{4}}{4\pi \frac{a}{2}}(\sin {45}^0+\sin {45}^0)=\frac{{\mu }_{0}3I}{4\sqrt{2}\pi a}\hat{k}$

$(\overrightarrow{B}{)}_{BC}=\frac{{\mu }_0\frac{I}{4}}{4\pi \frac{a}{2}}(\sin {45}^0+\sin {45}^0)=\frac{{\mu }_{0}I}{4\sqrt{2}\pi a}(-\hat{k})$

$(\overrightarrow{B}{)}_{AB}=\frac{{\mu }_0\frac{I}{4}}{4\pi \frac{a}{2}}(\sin {45}^0+\sin {45}^0)=\frac{{\mu }_{0}I}{4\sqrt{2}\pi a}(-\hat{k})$

$(\overrightarrow{B}{)}_{CD}=\frac{{\mu }_0\frac{I}{4}}{4\pi \frac{a}{2}}(\sin {45}^0+\sin {45}^0)=\frac{{\mu }_{0}I}{4\sqrt{2}\pi a}(-\hat{k})$

$(\overrightarrow{B{)}_{total}}=(\overrightarrow{B}{)}_{AD}+(\overrightarrow{B}{)}_{AB}+(\overrightarrow{B}{)}_{BC}+(\overrightarrow{B}{)}_{CD}=0$