Question

A cell is connected between the points A and C of a circular conductor $ABCDA$ of centre $O\angle AOC=60^{\circ }$ If $B_1$ and $B_2$ are the magnitudes of the magnetic fields at $O$ due to currents in $ABC$ and $ADC$ respectively, the ratio $B_1/B_2$, is

• A. 6
• B. 5
• C. 1
• D. 1/5

Answer 1

Answer: (C)

Let $l_1,l_2$ be the lengths of the two parts $ABC$ and $ADC$ of the conductor and $\rho$ be resistance per unit length of the conductor.
The resistance of the part $ABC$ will be
$R_1=\rho l_1/A$
The resistance of the part $ADC$ will be
$R_2=\rho l_2/A$
Potential difference across $AC$ must be equal $=I_1{R}_1=I_2{R}_2$
or $I_1\rho l_1/A=I_2\rho l_2/A$ or $I_1{l}_1=I_2{l}_2\dots (i)$
Magnetic field at $O$ due to a current $I_1$ flowing in $ABC$ is given by
$B_1=\frac{{\mu }_0{I}_1{\theta }_1}{4\pi r}=\frac{{\mu }_0{I}_1{l}_1}{4\pi r^2}$ (As $l_1={\theta }_{1}r)$
Magnetic field at $O$ due to current $I_2$ flowing in $ADC$ is given by
$B_2=\frac{{\mu }_0{I}_2{\theta }_2}{4\pi r}=\frac{{\mu }_0}{4\pi }\frac{I_2{l}_2}{r^2}$
(As $l_2={\theta }_{2}r)$
$\therefore \frac{B_1}{B_2}=1$ [Using (i)]