Question

A wire $A$, bent in the shape of an arc of a circle, carrying a current of $2\text{A}$ and having radius $2\text{cm}$ and another wire $B$, also bent in the shape of arc of a circle, carrying a current of $3\text{A}$ and having radius of $4\text{cm}$, are placed as shown in the figure. The ratio of the magnetic fields due to the wires $A$ and $B$ at the common centre $O$ is :

• A. $2:5$
• B. $6:5$
• C. $4:6$
• D. $6:4$

Answer 1

The correct option is B $6:5$

Given:
$I_A=2A$
$r_A=2cm$
$I_B=3A$
$r_B=4cm$

Arc of wire A make an angle at the centre,
${\theta }_A=2\pi -\frac{\pi }{2}=\frac{3\pi }{2}$

Arc of wire B make an angle at the centre,
${\theta }_B=2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$

As we know that Magnetic field due to an arc at the centre of loop is given by,

$B=\frac{{\mu }_{0}I\theta }{4\pi r}$

$\therefore \frac{B_A}{B_B}=\frac{\frac{{\mu }_0{I}_A{\theta }_A}{4\pi r_A}}{\frac{{\mu }_0{I}_B{\theta }_B}{4\pi r_B}}$

$=\frac{I_A}{I_B}\cdot \frac{r_B{\theta }_A}{r_A{\theta }_B}$

$=\frac{2}{3}\cdot \frac{4}{2}\cdot \frac{3\pi /2}{5\pi /3}=\frac{6}{5}$

Hence, $\left(B\right)$ is the correct answer.