A cell is connected between the points $A\ \ and\ C$ of a circular conductor $ABCD$ with $O$ as centre and angle $AOC = 60{^\circ}.$ If $B_{1}\ and\ B_{2}$ are the magnitudes of the magnetic fields at $O$ due to the currents in $ABC\ and\ \ ADC$ respectively, then ratio $\frac{B_{1}}{B_{2}}$ is [image:4_image32.emf]

Question

Dr. Abubacker · Accepted Answer

(c) From Biot-Savart law the magnetic field at the centre is directly proportional to the length of current carrying segment.
$	herefore\frac{B_{1}}{B_{2}} = \frac{length\ of\ ABC}{length\ of\ ADC}$
$\ \ \ \ \ \ \ \ \ \  = \frac{angle\ subtended\ by\ ABC}{angle\ subtended\ by\ ADC}$
$\ \ \ \ \ \ \ \ \ \  = \frac{(360{^\circ} - 60{^\circ})}{60{^\circ}} = \frac{300}{60} = \frac{5}{1}$

Q. A cell is connected between the points $A\ \ and\ C$ of a circular conductor $ABCD$ with $O$ as centre and angle $AOC = 60{^\circ}.$ If $B_{1}\ and\ B_{2}$ are the magnitudes of the magnetic fields at $O$ due to the currents in $ABC\ and\ \ ADC$ respectively, then ratio $\frac{B_{1}}{B_{2}}$ is

Explanation

Reviewed by Dr. Abubacker

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