A straight wire carries a current of 40 A in a uniform magnetic field (magnitude = 80 mT). If the force per unit length on this wire is 2.0 N/m, determine the angle between the wire and the magnetic field.
A) either 39° or 141°
B) either 25° or 155°
C) either 70° or 110°
D) either 42° or 138°
E) either 65° or 115°

Question

Quizplus · Accepted Answer

The force per unit length on a current-carrying wire in a magnetic field is given by $F/L = IB\sin(\theta)$, where $I$ is the current, $B$ is the magnetic field strength, and $\theta$ is the angle between the wire and the magnetic field. Rearranging for $\theta$, we get $\sin(\theta) = F/(IB)$. Substituting the given values, $F/L = 2.0 \, \text{N/m}$, $I = 40 \, \text{A}$, and $B = 80 \, \text{mT} = 0.08 \, \text{T}$, we find $\sin(\theta) = 2.0 / (40 \times 0.08) = 0.625$. Solving for $\theta$, we get $\theta = \arcsin(0.625)$, which gives $\theta \approx 39^\circ$ or $180^\circ - 39^\circ = 141^\circ$ due to the sine function's symmetry.

A Straight Wire Carries a Current of 40 a in a Uniform