A current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $ BC$ (radius $= b$) and $DA $ (radius $= a$) of the loop are joined by two straight wires $AB $ and $CD$. A steady current $I$ is flowing in the loop. Angle made by $AB$ and $CD$ at the origin $O$ is $30^o $. Another straight thin wire with steady current $I_1$ flowing out of the plane of the paper is kept at the origin.
The magnitude of the magnetic field $(B)$ due to the loop $ABCD$ at the origin $(O)$ is :
AIEEE 2009, Diffcult
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The magnetic field at $O$ due to current in $D A$ is
$B_{1}=\frac{\mu_{o}}{4 \pi} \frac{I}{a} \times \frac{\pi}{6} \quad$ (directed verticallyupwards)
The magnetic field at $O$ due to current in $B C$ is
$B_{2}=\frac{\mu_{o}}{4 \pi} \frac{I}{b} \times \frac{\pi}{6} \quad$ (directed vertically downwards)
The magnetic field due to current $A B$ and $C D$ at $O$ is zero.
Therefore the net magnetic field is
$B=B_{1}-B_{2} \quad$ (directed vertically upwards)
$=\frac{\mu_{o}}{4 \pi} \frac{I}{a} \frac{\pi}{6}-\frac{\mu_{o}}{4 \pi} \frac{I}{b} \times \frac{\pi}{6}=\frac{\mu_{o} I}{24}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{\mu_{o} I}{24 a b}(b-a)$
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