Two long straight wires each carrying an electric current of 6.0 A are placed parallel to each other at a separation of 3.0 cm. The magnitude of the force experienced by 12 cm length of a wire will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.88 x 10−5 N
Each wire lies in the magnetic field of the other, so a length L of either wire feels the force between two parallel currents.
\( \displaystyle F = \frac{\mu_0 I_1 I_2 L}{2\pi d} \)
\( I_1 = I_2 = 6.0\,\mathrm{A} \)
\( d = 3.0\,\mathrm{cm} = 3.0\times10^{-2}\,\mathrm{m} \)
\( L = 12\,\mathrm{cm} = 0.12\,\mathrm{m} \)
\( \displaystyle F = \frac{(4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}})(6.0\,\mathrm{A})(6.0\,\mathrm{A})(0.12\,\mathrm{m})}{2\pi\,(3.0\times10^{-2}\,\mathrm{m})} \)
\( \displaystyle F = \frac{2\times10^{-7}\times36\times0.12}{3.0\times10^{-2}}\,\mathrm{N} \)
\( \displaystyle F = 2.88\times10^{-5}\,\mathrm{N} \)
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 – …
