Two long and parallel straight wires P and Q carrying currents of \( 8.0 \) A and \( 5.0 \) A in the same direction are separated by a distance of \( 8.0 \) cm. The force on a \( 20 \) cm section of wire P is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2\times10^{-5} \) N, attractive
Two long parallel current-carrying wires exert a magnetic force on each other, inversely proportional to their separation. Parallel currents attract and antiparallel currents repel.
\( I_{1} = 8.0\,\mathrm{A},\ I_{2} = 5.0\,\mathrm{A} \)
\( d = 8.0\,\mathrm{cm} = 0.08\,\mathrm{m},\ L = 20\,\mathrm{cm} = 0.20\,\mathrm{m} \)
\( \displaystyle F = \frac{\mu_{0}}{2\pi}\frac{I_{1}I_{2}L}{d} = \frac{(2\times10^{-7}\,\mathrm{T\,m\,A^{-1}})(8.0\,\mathrm{A})(5.0\,\mathrm{A})(0.20\,\mathrm{m})}{0.08\,\mathrm{m}} \)
\( \displaystyle F = 2\times10^{-5}\,\mathrm{N} \)
The currents are in the same direction, so the force is attractive.
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 – …
