Q122 / 180 · COMEDK 2013 · Physics
Three long straight wires $A, B$ and $C$ are carrying currents as shown in figure, then the resultant force on $B$ is directed
- Aperpendicular to the plane of paper and outward
- Bperpendicular to the plane of paper and inward
- Ctowards $A$
- Correct answerDtowards $C$
Solution
The magnetic force between two current carrying wires for current in same direction is attractive.
Thus, force on $B$ due to the magnetic field produced due to wire $A$ is
$$
\begin{aligned}
F_{\mathrm{A}} &=B_{A} I_{B} l=\frac{\mu_{0} \times I_{A}}{2 \pi d} \times I_{B} \times l \\
&=\frac{\mu_{0} \times 1 \times 2 \times l}{2 \pi d}=\frac{2 \mu_{0} l}{2 \pi d}, \text { towards } A \quad \ldots(\text { i })
\end{aligned}
$$
Similarly, force due to field of wire $C$,
$$
\begin{aligned}
& F_{C}=B_{C} I_{B} l=\frac{\mu_{0} I_{C} I_{B} l}{2 \pi d}=\frac{\mu_{0} \times 3 \times 2 \times l}{2 \pi d} \\
\Rightarrow \quad & F_{C}=\frac{6 \mu_{0} l}{2 \pi d}, \text { towards } C
\end{aligned}
$$
From Eqs. (i) and (ii) we get
$$
F_{C}>F_{A}
$$
$\therefore$ The resultant force on $B$ is directed towards $C$.
Hence, the correct answer is option: 'D'.