Question

Three long, straight parallel wires, carrying current, are arranged as shown in figure. The force experienced by $25 cm$ 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
$\vec{B}_{D}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I}{r}=\frac{10^{-7} \times 2 \times 30}{0.03}$
$=2 \times 10^{-4} T$
which is directed into the page.
The magnetic field due to wire $G$ at $C$ is
$\vec{B}_{G}=\frac{10^{-7} \times 2 \times 20}{0.02}=2 \times 10^{-4} T$
which is directed out of the page.
Therefore, the 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