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Three long straight wires $A, B$ and $C$ are carrying currents as shown in figure. Then, the resultant force on $B$ is directed
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Verified Answer
The correct answer is:
towards $C$
Magnetic field at any point due to a straight current carrying wire is given by
$$
B=\frac{\mu_{0} I}{2 \pi d}
$$
where, $d$ is the distance of the point from the wire and $I_{1}$ is current flowing in the wire.
Force due to this magnetic field on current carrying wire is given by
$$
F=B I L
$$
where, $L$ is the length of the wire.
$\therefore$ Force on wire $B$ due to wire $A, F_{A}=B I L$
$$
\Rightarrow \quad F_{A}=\frac{\mu_{0} \times 1}{2 \pi d} \times 2 \times L
$$
Force on wire $B$ due to wire $C, F_{C}=B I L$
$$
\Rightarrow \quad F_{C}=\frac{\mu_{0} \times 3}{2 \pi d} \times 2 \times L
$$
It is clear that $F_{C}>F_{A}$. Hence, net force is towards wire $C$.
$$
B=\frac{\mu_{0} I}{2 \pi d}
$$
where, $d$ is the distance of the point from the wire and $I_{1}$ is current flowing in the wire.
Force due to this magnetic field on current carrying wire is given by
$$
F=B I L
$$
where, $L$ is the length of the wire.
$\therefore$ Force on wire $B$ due to wire $A, F_{A}=B I L$
$$
\Rightarrow \quad F_{A}=\frac{\mu_{0} \times 1}{2 \pi d} \times 2 \times L
$$
Force on wire $B$ due to wire $C, F_{C}=B I L$
$$
\Rightarrow \quad F_{C}=\frac{\mu_{0} \times 3}{2 \pi d} \times 2 \times L
$$
It is clear that $F_{C}>F_{A}$. Hence, net force is towards wire $C$.
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