Question

Two thin long parallel wires separated by a distance \(\mathrm{y}\) are carrying a current I Amp each. The magnitude of the force per unit length exerted by one wire on other is (a) \(\left[\left(\mu_{0} I^{2}\right) / y^{2}\right]\) (b) \(\left[\left(\mu_{o} I^{2}\right) /(2 \pi \mathrm{y})\right]\) (c) \(\left[\left(\mu_{0}\right) /(2 \pi)\right](1 / y)\) (d) $\left[\left(\mu_{0}\right) /(2 \pi)\right]\left(1 / \mathrm{y}^{2}\right)$

Short answer

The correct answer is (b) \(\left[\left(\mu_{0} I^{2}\right) /(2 \pi \mathrm{y})\right]\).

Step-by-step solution

Write down the formula for the force between two current-carrying wires

The formula for the force F per unit length (F/L) between two current-carrying wires is given by: \(\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}\) where - \(\mu_0\) is the magnetic constant (permeability of free space), - \(I_1\) and \(I_2\) are the currents in the two wires, - \(d\) is the distance between the wires. Since both wires have the same current I, we replace \(I_1\) and \(I_2\) with I in the formula: \(\frac{F}{L}=\frac{\mu_0 I^2}{2\pi d}\)

Find the correct substitution for the distance

Given that the distance between the wires is y, we need to substitute the variable d with y to get the formula: \(\frac{F}{L}=\frac{\mu_0 I^2}{2\pi y}\)

Compare the result with the given options

The formula we found is \(\frac{F}{L}=\frac{\mu_0 I^2}{2\pi y}\), which matches the option (b). Therefore, the correct answer is (b) \(\left[\left(\mu_{0} I^{2}\right) /(2 \pi \mathrm{y})\right]\).