Question

A cell is connected between the points A and C of a circular conductor ABCD with O as centre and angle $AOC={60}^o$. 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

• A. $1$
• B. $2$
• C. $5$
• D. $6$

Answer 1

The correct option is A $1$

$B=\frac{{\mu }_0}{4\pi }\frac{\theta i}{r}\Rightarrow B\propto \theta i$ $(but\frac{i_1}{i_2}=\frac{l_2}{l_1}=\frac{{\theta }_2}{{\theta }_1})$

$\Rightarrow \frac{B_1}{B_2}=\frac{{\theta }_1}{{\theta }_2}.\frac{i_1}{i_2}$

So, $\frac{B_1}{B_2}=\frac{{\theta }_1}{{\theta }_2}\times \frac{{\theta }_2}{{\theta }_1}$

$\Rightarrow B_1=B_2$

$\frac{B_1}{B_2}=1$