A copper wire of length 1 m and diameter 1 mm is oriented such that it is perpendicular to a magnetic field of strength 5 T.The wire carries a current of 3 A across its length.What is the magnitude of the force on this wire?

Question

Quizplus · Accepted Answer

The force on a current-carrying conductor 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 conductor, and $\theta$ is the angle between the direction of the current and the direction of the magnetic field. Since the wire is perpendicular to the magnetic field, $\theta = 90^\circ$, and $\sin(90^\circ) = 1$. Thus, $F = 5 \, \text{T} \times 3 \, \text{A} \times 1 \, \text{m} = 15 \, \text{N}$.

A Copper Wire of Length 1 M and Diameter 1