Two long straight wires, P and Q, each carrying a current of 5 A in opposite directions, are placed parallel to each other with separation of 4 cm between them. The force on an 8 cm long section of wire Q is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.0\times10^{-5} \) N
The force between two parallel currents on a length L of either wire is \( F = \mu_0 I_1 I_2 L/2\pi d \). For currents in opposite directions the force is repulsive; the magnitude is the same either way.
\( I_1 = I_2 = 5\,\mathrm{A} \)
\( d = 4\,\mathrm{cm} = 0.04\,\mathrm{m},\ L = 8\,\mathrm{cm} = 0.08\,\mathrm{m} \)
\( \displaystyle F = \frac{\mu_0 I_1 I_2 L}{2\pi d} = \frac{2\times10^{-7} \times 5 \times 5 \times 0.08}{0.04}\,\mathrm{N} \)
\( \displaystyle F = 1.0\times10^{-5}\,\mathrm{N} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
