JEE Physics 2023 Question with Solution

Given: Two parallel straight wires carry equal currents. Initially, each wire carries $10 \, \text{A}$ and the separation is $5 \, \text{cm}$. In the second case, the distance is halved and the currents are doubled.

Find: The relation between $F_2$ and $F_1$ for the same wire length.

For two long parallel wires, force per unit length is

$$F = \frac{\mu_0 i_1 i_2}{2\pi d}$$

So the force is directly proportional to

$$F \propto \frac{i_1 i_2}{d}$$

Initially,

$$F_1 \propto \frac{(10)(10)}{5}$$

After doubling both currents and halving the distance,

$$F_2 \propto \frac{(20)(20)}{5/2}$$

Therefore,

$$\frac{F_2}{F_1} = \frac{\frac{(20)(20)}{5/2}}{\frac{(10)(10)}{5}}$$ $$= \frac{400}{5/2} \times \frac{5}{100}$$ $$= 400 \times \frac{2}{5} \times \frac{5}{100}$$ $$= 8$$

Hence,

$$F_2 = 8F_1$$

Therefore, the correct option is A.

The solution labels the correct option as B, but its working clearly gives $F_2 = 8F_1$, which matches option A.