Question

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

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

Answer 1

The correct option is B

$6:5$

Step 1. Given Data:

The radius of the first wire $r_1=2cm$

The radius of the second wire $r_2=4cm$

Current through the first wire $i_1=2A$

Current through the second wire $i_2=3A$

Angle for the first wire ${\theta }_1=\frac{3\pi }{2}$

Angle for the second wire ${\theta }_2=\frac{5\pi }{3}$

Step 2. Calculate the magnetic fields:

If $B$is the magnetic field and ${\mu }_0$is the permeability constant then,

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

For the first wire $B_1=\frac{{\mu }_{0}2{\displaystyle \frac{3\pi }{2}}}{4\pi 2}$

$=\frac{3}{8}{\mu }_0$

For the second wire $B_2=\frac{{\mu }_{0}3{\displaystyle \frac{5\pi }{3}}}{4\pi 4}$

$=\frac{5}{16}{\mu }_0$

Step 3. Find the ratio:

$\frac{B_1}{B_2}=\frac{6}{5}$

Hence option (B) is correct.