Two parallel straight wires are $7.0 \mathrm {~cm}$ apart and $50 \mathrm {~m}$ long. Each one carries a $18 - \mathrm { A }$ current in the same direction. One wire is securely anchored, and the other is attached in the center to a movable cart. If the force needed to move the wire when it is not attached to the cart is negligible, with what magnitude force does the wire pull on the cart? $\left( \mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } \right)$
A) $93 \mathrm { mN }$
B) $66 \mathrm { mN }$
C) $37 \mathrm { mN }$
D) $46 \mathrm { mN }$

Question

Quizplus · Accepted Answer

The force per unit length between two parallel currents is given by the formula $F/L = \mu_0 I_1 I_2 / (2\pi d)$, where $F$ is the force, $L$ is the length of the wires, $\mu_0$ is the permeability of free space, $I_1$ and $I_2$ are the currents in the wires, and $d$ is the distance between the wires. Substituting the given values, $F/L = (4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}) \times (18 \, \text{A})^2 / (2\pi \times 0.07 \, \text{m})$, and then multiplying by the length of the wire ($50 \, \text{m}$) gives the total force. The calculation yields a force of approximately $46 \, \text{mN}$, making option D correct.

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Two Parallel Straight Wires Are $7.0 \mathrm {~cm}$ Apart And $50 \mathrm {~m}$