A long horizontal wire P carries a current of 50 A. A second long wire Q lies parallel to P, 1 cm directly above it. The mass per unit length of Q is \(1\times10^{-3}\ \mathrm{kg\,m^{-1}}\). Wire Q stays at rest, held up only by the magnetic force due to P. The current in Q is (take \(g=10\ \mathrm{m\,s^{-2}}\) and \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 10 A, opposite to the current in P
Wire Q lies in the magnetic field of P, so it feels a force per unit length \( f=\mu_0I_PI_Q/2\pi d \). To hold Q up, this force must push Q away from P. Antiparallel currents repel, so the current in Q must be opposite to the current in P. For balance, the force per unit length equals the weight per unit length, \( (m/l)\,g \).
Here \( d=1\ \mathrm{cm}=0.01\ \mathrm{m} \).
\( \begin{aligned} \frac{\mu_0 I_P I_Q}{2\pi d} &= \frac{m}{l}\,g \\ \frac{4\pi\times10^{-7}\times50\times I_Q}{2\pi\times0.01} &= 1\times10^{-3}\times10 \\ 1\times10^{-3}\,I_Q &= 1\times10^{-2} \\ I_Q &= 10\ \mathrm{A} \end{aligned} \)
Therefore, Q carries 10 A, opposite to the current in P.
NCERT (2026-27):
Force Between Two Parallel Currents, the Ampere – “Let fba represent the magnitude of the force Fba per unit length. Then, …”
\( f_{ba}=\frac{\mu_0 I_a I_b}{2\pi d} \)\( \qquad (4.19) \)
Force Between Two Parallel Currents, the Ampere – “Parallel currents attract, and antiparallel currents repel.”