Question

$PQRS$ is a square loop of a conducting wire. A current $I$ enters the loop at $P$ and leaves the loop at $S .$ The magnitude of magnetic field at the centre $O$ of the square is

• A. $\frac{\mu_{0}}{4 \pi} \frac{2 \sqrt{2} I}{a}$
• B. $\frac{\mu_{0}}{4 \pi} \frac{4 \sqrt{2} I}{a}$
• C. $\frac{\mu_{0}}{4} \frac{2 \sqrt{2} I}{a}$
• D. zero

Answer 1

Answer: (D)

The resistance of arm $PQRS$ is 3 times the resistances of arm $PS$. If resistance of arm P $S=R$ then resistance of arm $PQRS=3R$.
Potential difference across $PQRS$ and $PS$ will be equal, as they are connected in parallel. $ I_{1} R=I_{2} \times 3 R$
or $I_{1}=3 I_{2}\,\,\,\,\,\dots(i)$
Magnetic field at $O$ due to current through arm PS is,
$B_{1}=\frac{\mu_{0}}{4 \pi} \frac{I_{1}}{(a / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$
acting perpendicular to the loop upwards
$=\frac{\mu_{0}}{4 \pi} \frac{3 I_{2}}{(a / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ (Using (i))
Magnetic field at $O$ due to current through arm $P Q R S$ is
$B_{2}=\frac{\mu_{0}}{4 \pi} \frac{I_{2}}{(a / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right] \times 3$
acting perpendicular to the loop downwards
$\therefore $ Net magnetic field at $O$ is $B=B_{1}-B_{2}=0 $