A 3.0 cm section of a horizontal wire, carrying a 6.7 A current from east to west, is placed in a horizontal north-to-south 0.42 T uniform magnetic field in the lab. What is the magnitude of the force on the 3.0 cm section of the wire?
A) 0.084 N
B) 0.17 N
C) 0.25 N
D) 0.0017 N
E) 0.0025 N

Question

Quizplus · Accepted Answer

The magnitude of the force on a current-carrying wire in a magnetic field is given by $F = BIL\sin(\theta)$, where $B$ is the magnetic field strength, $I$ is the current, $L$ is the length of the wire, and $\theta$ is the angle between the direction of the current and the direction of the magnetic field. Here, $B = 0.42 \, \text{T}$, $I = 6.7 \, \text{A}$, $L = 0.03 \, \text{m}$ (since 3.0 cm = 0.03 m), and $\theta = 90^\circ$ (since the current flows east to west and the magnetic field is north to south, they are perpendicular). Thus, $\sin(90^\circ) = 1$, and the force is $F = 0.42 \times 6.7 \times 0.03 \times 1 = 0.0846 \, \text{N}$, which rounds to $0.084 \, \text{N}$.

A 30 Cm Section of a Horizontal Wire, Carrying a 6