Three long, straight, parallel wires A, B and C lie in one plane, in this order. The distance between A and B is 5 cm, and the distance between B and C is also 5 cm. Each wire carries a current of 10 A in the same direction. The magnitude of the net magnetic force on each metre of wire A is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(6\times10^{-4}\ \mathrm{N}\)
Parallel currents attract. So wire A is pulled towards B and also towards C. Both pulls act in the same direction, so they add. Each force per unit length is \( f=\mu_0I_aI_b/2\pi d \), where \( d \) is the distance from A.
Here each current is \( 10\ \mathrm{A} \), \( d_{AB}=5\ \mathrm{cm}=0.05\ \mathrm{m} \) and \( d_{AC}=10\ \mathrm{cm}=0.10\ \mathrm{m} \). Also \( \mu_0/2\pi=2\times10^{-7}\ \mathrm{T\,m\,A^{-1}} \).
\( \begin{aligned} f_{AB} &= \frac{\mu_0 I_a I_b}{2\pi d_{AB}} \\ &= \frac{2\times10^{-7}\times10\times10}{0.05} \\ &= 4\times10^{-4}\ \mathrm{N\,m^{-1}} \end{aligned} \)
\( \begin{aligned} f_{AC} &= \frac{2\times10^{-7}\times10\times10}{0.10} \\ &= 2\times10^{-4}\ \mathrm{N\,m^{-1}} \end{aligned} \)
\( \begin{aligned} f &= f_{AB}+f_{AC} \\ &= (4+2)\times10^{-4} \\ &= 6\times10^{-4}\ \mathrm{N\,m^{-1}} \end{aligned} \)
NCERT (2026-27):
Force Between Two Parallel Currents, the Ampere – “Parallel currents attract, and antiparallel currents repel.”
Force Between Two Parallel Currents, the Ampere – “Let fba represent the magnitude of the force Fba per unit length. Then, …”
\( f_{ba}=\frac{\mu_0 I_a I_b}{2\pi d} \)\( \qquad (4.19) \)