Two straight and parallel wires, X and Y, of lengths 16 m and 10 m, placed 3.0 cm apart, are carrying currents of 3.0 A and 4.0 A, respectively, in opposite directions. The force on 10 cm length of wire Y will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8 x 10−6 N
The wires are metres long and only 3.0 cm apart, so each can be treated as an infinitely long wire. The field of X at Y is \( \mu_0 I_X/2\pi d \), and the force on a length l of Y is
\( \displaystyle F = \frac{\mu_0 I_X I_Y\, l}{2\pi d} \)
Substituting \( \mu_0/2\pi = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \), \( d = 3.0\,\mathrm{cm} = 0.03\,\mathrm{m} \), \( l = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \):
\( \displaystyle F = \frac{2\times10^{-7} \times 3.0 \times 4.0 \times 0.1}{0.03}\,\mathrm{N} \)
\( \displaystyle F = 8\times10^{-6}\,\mathrm{N} \)
Since the currents are in opposite directions, the force is repulsive. The total lengths 16 m and 10 m are not needed.
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 – …
