Three long, straight, parallel current-carrying conductors are shown…

The correct answer is Option (4) → \( 3\times10^{-4}\,\mathrm{N} \) towards the right

Two long parallel conductors a distance \( d \) apart exert a force \( F=\mu_0 I_aI_bL/2\pi d \) on a length \( L \) of each other. Parallel currents attract and antiparallel currents repel. The middle current (downward) is opposite to both outer currents (upward), so both outer conductors repel it, in opposite directions.

From the figure, the 30 A conductor is 3 cm to the left of the middle conductor. The 20 A conductor is 5 cm to its right. In SI units the data are:

    \( L=25\,\mathrm{cm}=0.25\,\mathrm{m} \)

    \( d_1=3\,\mathrm{cm}=0.03\,\mathrm{m} \)

    \( d_2=5\,\mathrm{cm}=0.05\,\mathrm{m} \)

    \( \mu_0/2\pi=2\times10^{-7}\,\mathrm{T\,m/A} \)

The force due to the 30 A conductor, pushing the middle conductor to the right, is:

    \( \begin{aligned} F_1 &= \frac{\mu_0 I_1 I L}{2\pi d_1} \\ &= \frac{2\times10^{-7}\times30\times10\times0.25}{0.03}\,\mathrm{N} \\ &= 5\times10^{-4}\,\mathrm{N} \end{aligned} \)

The force due to the 20 A conductor, pushing it to the left, is:

    \( \begin{aligned} F_2 &= \frac{\mu_0 I_2 I L}{2\pi d_2} \\ &= \frac{2\times10^{-7}\times20\times10\times0.25}{0.05}\,\mathrm{N} \\ &= 2\times10^{-4}\,\mathrm{N} \end{aligned} \)

The net force is:

    \( \begin{aligned} F &= F_1-F_2 \\ &= 5\times10^{-4}\,\mathrm{N}-2\times10^{-4}\,\mathrm{N} \\ &= 3\times10^{-4}\,\mathrm{N} \end{aligned} \)

Therefore, the push from the nearer 30 A conductor is larger. The net force on the 25 cm length is \( 3\times10^{-4}\,\mathrm{N} \) towards the right.

NCERT (2026-27):
Force Between Two Parallel Currents, the Ampere – “We have seen from above that currents flowing in the same direction attract each other. One can show that oppositely directed currents repel each other.”