Question

Two thin parallel wires carry currents along the same direction. The force experienced by one due to the other is:

• A. Perpendicular to the lines and attractive
• B. Perpendicular to the lines and repulsive
• C. Zero
• D. Parallel to the lines

Hint:

Remember the simple rule for currents: - Like currents attract: Currents flowing in the same direction pull the wires together. - Unlike currents repel: Currents flowing in opposite directions push the wires apart.

Answer 1

The Correct Option is A

Concept: The force between two parallel current-carrying conductors is determined by combining Ampere's Right-Hand Grip Rule (to find the magnetic field direction produced by the first wire) and the Lorentz Force Formula (to find the force exerted on the second wire carrying current through that field). The force vector $\vec{F}$ acting on a length $L$ of a wire carrying current $I$ inside a magnetic field $\vec{B}$ is given by: \[ \vec{F} = I(\vec{L} \times \vec{B}) \]

Step 1: Find the direction of the magnetic field from Wire 1.
Let Wire 1 carry a current $I_1$ vertically upward along the $+z$ axis. According to Ampere's Right-Hand Rule, the magnetic field lines $\vec{B}_1$ encircle the wire. At the location of parallel Wire 2 (positioned to the right along the $+x$ axis), these field lines point straight into the page (along the $-\hat{a}_y$ direction).

Step 2: Apply the cross product to find the force direction on Wire 2.
Wire 2 carries current $I_2$ in the same direction (upward, along $+\hat{a}_z$). The force per unit length vector on Wire 2 is: \[ \vec{F}_{21} = I_2 (\hat{a}_z \times \vec{B}_1) = I_2 \left[ \hat{a}_z \times (-B_1 \hat{a}_y) \right] \] Using the cyclic vector identity $\hat{a}_z \times \hat{a}_y = -\hat{a}_x$: \[ \vec{F}_{21} = -B_1 I_2 (-\hat{a}_x) = +B_1 I_2 \hat{a}_x \] Wait, let's re-verify the right hand rule coordinate orientation. Let Wire 1 be at $x=0$, Wire 2 be at $x=d$. Current $\vec{I}_1$ is in $+z$ direction. At $x=d$, field $\vec{B}_1$ points in $+\hat{a}_y$ direction. Then force on Wire 2 is $\vec{I}_2 \times \vec{B}_1 = (I_2 \hat{a}_z) \times (B_1 \hat{a}_y) = -I_2 B_1 \hat{a}_x$. Since the force vector points in the $-\hat{a}_x$ direction (towards Wire 1), it pulls Wire 2 directly toward Wire 1. This means the force is attractive and directed along the line perpendicular to the wires. Therefore, the force is perpendicular to the lines and attractive. This matches Option (A).