In the current-carrying conductor (AOCDEFG) a | vedclass

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(B) Using the Biot-Savart law,the magnetic field at a point on the axis of a straight current-carrying conductor is zero. Therefore,the magnetic field

Vedclass · Accepted Answer

$\frac{\mu_0 I}{8}\left(\frac{R_1+R_2}{R_1 R_2}\right)$

Vedclass · Answer

(B) Using the Biot-Savart law,the magnetic field at a point on the axis of a straight current-carrying conductor is zero. Therefore,the magnetic fields due to the straight segments $AO$,$OC$,$DE$,and $FG$ are zero at point $O$.The magnetic field at the center of a full circular loop is $B = \frac{\mu_0 I}{2R}$.For a circular arc subtending an angle $	heta$ at the center,the magnetic field is $B = \frac{\mu_0 I 	heta}{4\pi R}$.Here,both arcs $CD$ and $EF$ subtend an angle of $90^\circ$ or $\frac{\pi}{2}$ radians at point $O$.For arc $CD$ with radius $R_1$:$B_{CD} = \frac{\mu_0 I (\pi/2)}{4\pi R_1} = \frac{\mu_0 I}{8 R_1}$ (directed into the plane of the paper).For arc $EF$ with radius $R_2$:$B_{EF} = \frac{\mu_0 I (\pi/2)}{4\pi R_2} = \frac{\mu_0 I}{8 R_2}$ (directed into the plane of the paper).Since both fields are in the same direction,the total magnetic field at $O$ is:$B = B_{CD} + B_{EF} = \frac{\mu_0 I}{8 R_1} + \frac{\mu_0 I}{8 R_2} = \frac{\mu_0 I}{8} \left( \frac{1}{R_1} + \frac{1}{R_2} ight) = \frac{\mu_0 I}{8} \left( \frac{R_1 + R_2}{R_1 R_2} ight)$.

In the current-carrying conductor $(AOCDEFG)$ as shown,the magnetic induction at the point $O$ is ($R_1$ and $R_2$ are radii of arcs $CD$ and $EF$ respectively,$I$ = current in the loop,$\mu_0$ = permeability of free space).

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