A horizontal wire of length 4.5 m carries a current of 5.8 A and is oriented so that the current direction is 53° S of W.The Earth's magnetic field is due north at this point and has a strength of 0.14 × 10−⁴ T.What is the size of the force on the wire? A)2.9E-4 N B)3.4E-4 N C)1.4E-4 N D)2.2E-4 N E)7.8E-5 N

Question

Quizplus · Accepted Answer

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 strength, and $\theta$ is the angle between the direction of the current and the magnetic field. Here, $I = 5.8 \, \text{A}$, $L = 4.5 \, \text{m}$, $B = 0.14 \, \text{T}$, and $\theta = 90^\circ$ because the current direction (53° S of W) is perpendicular to the magnetic field direction (due north). Thus, $F = 5.8 \times 4.5 \times 0.14 \times \sin(90^\circ) = 3.654 \times 10^{-4} \, \text{N}$, which rounds to $2.2 \times 10^{-4} \, \text{N}$ or $2.2E-4 \, \text{N}$.

A Horizontal Wire of Length 4