Question

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

• A. $0.2$
• B. $6$
• C. $1$
• D. $5$

Answer 1

The correct option is C $1$

Magnetic field due to part ABC,
$B_1=\frac{{\mu }_0{I}_1}{2r}\times \frac{300}{360}$
$B_1=\frac{5\times {\mu }_0{I}_1}{6\times 2r}$

Magnetic field due to part ADC,
$B_2=\frac{{\mu }_0{I}_2}{2r}\times \frac{60}{360}$
$B_2=\frac{1\times {\mu }_0{I}_2}{6\times 2r}$

$\frac{B_2}{B_2}=\frac{5i_1}{i_2}$

Now, both the circular elements can be imagined to be in a circuit as shown in the diagram.

Since voltage is same across both the conductors.
$\therefore \frac{i_1}{i_2}=\frac{R_2}{R_1}=\frac{l_2}{l_1}$
and $\frac{l_2}{l_1}=\frac{{\theta }_2}{{\theta }_1}$
$\therefore \frac{i_1}{i_2}=\frac{{\theta }_2}{{\theta }_1}=\frac{60}{300}=\frac{1}{5}$
$\frac{B_1}{B_2}=\frac{5\times 1}{5}=1$