Two long and parallel straight wires A and B carrying currents of 10.0 A and 5.0 A in the same direction are separated by a distance of 5.0 cm. Estimate the force on a 10 cm section of wire A.

$2 × 10^{-5} N$

The correct answer is Option (1) → $2 × 10^{-5} N$

Given:

$I_1 = 10\,A$

$I_2 = 5\,A$

$r = 5.0\,cm = 0.05\,m$

$l = 10\,cm = 0.1\,m$

Magnetic field due to wire B at distance r:

$B = \frac{\mu_0 I_2}{2 \pi r}$

$B = \frac{4\pi \times 10^{-7} \times 5}{2 \pi \times 0.05}$

$B = \frac{2 \times 10^{-7} \times 5}{0.05}$

$B = 2 \times 10^{-5}\,T$

Force on wire A:

$F = I_1 l B$

$F = 10 \times 0.1 \times 2 \times 10^{-5}$

$F = 2 \times 10^{-5}\,N$

Since currents are in the same direction, force is attractive

Final Answer: Force on 10 cm section of wire A = $2 × 10^{-5} N$ (attractive).