A long straight wire carries a current of 12 A. Another straight wire…

The correct answer is Option (2) → \( 4.8 \times 10^{-6}\,\mathrm{N} \)

The long wire produces a magnetic field \( B = \mu_0 I_1/2\pi d \) at the short wire, and the short wire, lying parallel to it, feels the force \( F = I_2 L B \).

    \( \displaystyle F = \frac{\mu_0 I_1 I_2 L}{2\pi d} \)

    \( I_1 = 12\,\mathrm{A} \)
    \( I_2 = 2\,\mathrm{A} \)
    \( L = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)
    \( d = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)

    \( \displaystyle F = \frac{2\times10^{-7} \times 12 \times 2 \times 0.05}{0.05}\,\mathrm{N} \)

    \( \displaystyle F = 2\times10^{-7} \times 24\,\mathrm{N} = 4.8\times10^{-6}\,\mathrm{N} \)

Here \( \mu_0/2\pi = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \), and \( L \) and \( d \) are equal, so they cancel.