Question

Three long, straight and parallel wires C, D and G carrying currents are arranged as shown in Fig. The force experienced by a 25 cm length of wire C is

• A. 0.4 N
• B. 0.04 N
• C. $4\times {10}^{-3}N$
• D. $4\times {10}^{-4}N$

Answer 1

The correct option is D

$4\times {10}^{-4}N$

The magnetic field due to wire D at wire C is
$B_D=\left(\frac{{\mu }_0}{4\pi }\right)\frac{2I}{r}=\frac{{10}^{-7}\times 2\times 30}{0.03}=2\times {10}^{-4}T$
which is directed into the page.
Similarly, the field due to wire G at C is
$B_G=\frac{{10}^{-7}\times 2\times 20}{0.1}=0.4\times {10}^{-4}T$
which is directed out of the page.
Therefore, the field at the position of the wire C is
$B=B_D-B_G=2\times {10}^{-4}-0.4\times {10}^{-4}=1.6\times {10}^{-4}T$
and is directed into the page.
The force on 25 cm of wire C is
F = BIl ${90}^{\circ }=1.6\times {10}^{-4}\times 10\times 0.25$
$=4\times {10}^{-4}N$