Question

A cell is connected between the points A and C of a circular conductor ABCD with 0 as centre and angle $\text{AOC}\text{= 6}{\text{0}}^{\text{o}}\text{.}$ If ${\text{B}}_{\text{1}}$ and ${\text{B}}_2$ 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

Answer: (A)

$\begin{array}{l}B=\frac{{\mu }_0}{4\pi }\frac{\theta i}{r}\Rightarrow B\propto \theta i\left(\text{ but }\frac{i_1}{i_2}=\frac{l_2}{l_1}=\frac{{\theta }_2}{{\theta }_1}\right)\\ \Rightarrow \frac{B_1}{B_2}=\frac{{\theta }_1}{{\theta }_2}\cdot \frac{i_1}{i_2}\\ \text{ So, }\frac{B_1}{B_2}=\frac{{\theta }_1}{{\theta }_2}\times \frac{{\theta }_2}{{\theta }_1}\\ \Rightarrow B_1=B_2\\ \frac{B_1}{B_2}=1\end{array}$