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. $I_{1}R=I_2\times 3R$
or $I_1=3I_2\dots (i)$
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
$=\frac{{\mu }_0}{4\pi }\frac{3I_2}{(a/2)}\left[\sin 45^{\circ }+\sin 45^{\circ }\right]$ (Using (i))
Magnetic field at $O$ due to current through arm $PQRS$ 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 $B=B_1-B_2=0$