One side of a square loop (dimensions of 15 cm * 15 cm) runs parallel to a current-carrying wire I_w = 5.0 A, and 3.0 A runs through the loop. The net force on the loop is 2.7 * 10^-6 N. The distance between the side of the loop and the wire is A) 5.0 cm. B) 10 cm. C) 15 cm. D) 20 cm.

Question

Quizplus · Accepted Answer

The net force on the loop is due to the magnetic field created by the current-carrying wire. The force on a current-carrying wire in a magnetic field is given by $F = ILB\sin(\theta)$, where $I$ is the current, $L$ is the length of the wire, $B$ is the magnetic field, and $\theta$ is the angle between the wire and the magnetic field direction. In this case, $\theta = 90^\circ$, so $\sin(\theta) = 1$. The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$, where $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$ is the magnetic constant, and $I$ is the current through the wire. Given the net force and the currents, we can solve for $r$, but the exact calculation depends on understanding the interaction between the loop and the wire's magnetic field. The correct distance can be determined by applying these principles and matching the given net force, leading to the conclusion that the distance is 10 cm.

One Side of a Square Loop (Dimensions of 15 Cm