Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on 10 cm length of wire B.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2 \times 10^{-5}\,\mathrm{N} \), attractive
The field of wire A at wire B is \( B = \mu_0 I_A/(2\pi d) \), and the force on a length \( L \) of wire B is \( F = I_B L B \). Together, the force between parallel currents is
\( \displaystyle F = \frac{\mu_0\,I_A I_B\,L}{2\pi d} \)
\( \mu_0/2\pi = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( L = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \)
\( d = 4.0\,\mathrm{cm} = 0.04\,\mathrm{m} \)
\( \displaystyle F = \frac{2\times10^{-7} \times 8.0 \times 5.0 \times 0.1}{0.04} = 2\times10^{-5}\,\mathrm{N} \)
Currents in the same direction attract, so the force on B is attractive (towards A).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
