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
Show Hint
Instead of solving the complete calculation with the length multiplication, look at the ratio $\frac{I}{r}$ for the outer wires affecting the middle wire.
For wire $D$: $\frac{30\text{ A}}{3\text{ cm}} = 10$.
For wire $G$: $\frac{20\text{ A}}{2\text{ cm}} = 10$.
Since both produce identical repulsion ratios from opposite sides, the forces balance perfectly to Zero automatically!
Concept:
The magnetic force per unit length between two long, straight parallel current-carrying conductors separated by a distance $r$ is given by Ampere's force law:
\[
\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}
\]
The nature of the magnetic force is governed by the relative directions of the currents:
• Parallel currents (flowing in the same direction) attract each other.
• Anti-parallel currents (flowing in opposite directions) repel each other.
By using the principle of superposition, the net force experienced by a segment of wire $C$ is the vector sum of the forces exerted individually on it by wire $D$ and wire $G$.
Step 1: Convert all given values into standard SI units.
From the provided diagram, let us write down the parameters:
• Current in wire $D$ ($I_D$) = $30\text{ A}$ (pointing upwards)
• Current in wire $C$ ($I_C$) = $10\text{ A}$ (pointing downwards)
• Current in wire $G$ ($I_G$) = $20\text{ A}$ (pointing upwards)
• Distance between wire $D$ and wire $C$ ($r_1$) = $3\text{ cm} = 3 \times 10^{-2}\text{ m}$
• Distance between wire $C$ and wire $G$ ($r_2$) = $2\text{ cm} = 2 \times 10^{-2}\text{ m}$
• Target length of wire $C$ ($L$) = $25\text{ cm} = 0.25\text{ m}$
Step 2: Calculate the force acting on wire $C$ due to wire $D$ ($\vec{F}_{CD}$).
Since wire $D$ has an upward current and wire $C$ has a downward current, they are anti-parallel. Therefore, wire $D$ will repel wire $C$. This repulsive force pushes wire $C$ away from wire $D$, directed toward the right (+x direction).
Calculating the magnitude of this force ($F_{CD}$):
\[
F_{CD} = \frac{\mu_0 I_D I_C}{2\pi r_1} \times L
\]
Using the value $\frac{\mu_0}{2\pi} = 2 \times 10^{-7}\text{ T}\cdot\text{m/A}$:
\[
F_{CD} = \frac{(2 \times 10^{-7}) \times 30 \times 10}{3 \times 10^{-2}} \times 0.25
\]
Simplifying the fractions:
\[
F_{CD} = \frac{6 \times 10^{-5}}{3 \times 10^{-2}} \times 0.25 = (2 \times 10^{-3}) \times 0.25 = 0.5 \times 10^{-3}\text{ N} \quad (\text{directed to the Right})
\]
Step 3: Calculate the force acting on wire $C$ due to wire $G$ ($\vec{F}_{CG}$).
Similarly, wire $G$ has an upward current and wire $C$ has a downward current. Being anti-parallel, wire $G$ will also repel wire $C$. This repulsive force pushes wire $C$ away from wire $G$, directed toward the left (-x direction).
Calculating the magnitude of this force ($F_{CG}$):
\[
F_{CG} = \frac{\mu_0 I_G I_C}{2\pi r_2} \times L
\]
\[
F_{CG} = \frac{(2 \times 10^{-7}) \times 20 \times 10}{2 \times 10^{-2}} \times 0.25
\]
Simplifying the fractions:
\[
F_{CG} = \frac{4 \times 10^{-5}}{2 \times 10^{-2}} \times 0.25 = (2 \times 10^{-3}) \times 0.25 = 0.5 \times 10^{-3}\text{ N} \quad (\text{directed to the Left})
\]
Step 4: Compute the net force acting on the 25 cm section of wire $C$.
Taking the rightward direction as positive, let us sum the two force vectors:
\[
F_{\text{net}} = F_{CD} - F_{CG}
\]
\[
F_{\text{net}} = (0.5 \times 10^{-3}\text{ N}) - (0.5 \times 10^{-3}\text{ N}) = 0\text{ N}
\]
Since the two repulsive forces are exactly equal in magnitude but point in opposite directions, they cancel each other out completely, leaving the net force as zero.
