A straight $1.0 - \mathrm { m }$ long wire is carrying a current. The wire is placed perpendicular to a magnetic field of strength $0.20 \mathrm {~T}$. If the wire experiences a force of $0.60 \mathrm {~N}$, what is the current in the wire?

Question

Quizplus · Accepted Answer

The force experienced by a current-carrying wire in a magnetic field is given by the equation $F = BIL$, where $F$ is the force, $B$ is the magnetic field strength, $I$ is the current, and $L$ is the length of the wire. Rearranging this equation to solve for $I$, we get $I = \frac{F}{BL}$. Substituting the given values, $I = \frac{0.60}{0.20 \times 1.0} = 3.0 \mathrm {~A}$.

A Straight 1.0−m1.0 - \mathrm { m }1.0−m Long Wire Is Carrying a Current

A Straight $1.0 - \mathrm { m }$ Long Wire Is Carrying a Current