Question

Three long, straight parallel wires, carrying current, are arranged as shown in figure. The force experienced by a 25 cm length of wire C is

• A. $10^{-3}N$
• B. $2.5\times 10^{-3}N$
• C. zero
• D. $1.5\times 10^{-3}N$

Answer 1

Answer: (C)

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.
The magnetic field due to wire $G$ at $C$ is
$B_G=\frac{10^{-7}\times 2\times 20}{0.02}=2\times 10^{-4}T$
which is directed out of the page.
Therefore, the net magnetic field at the position of the wire $C$ is
$B=B_G-B_D=2\times 10^{-4}-2\times 10^{-4}$ = zero.
The force on 25 cm of wire $C$ is
$F=BIl\sin \theta$ = zero.