Question

A 15. 0 ampere current flows through a 4. 75 m section of wire, perpendicular to a 0. 250 t magnetic field. what is the magnitude of the force exerted by the field on the wire section?

Answer 1

Answer:

Approximately $17.8\; {\rm L}$.

Explanation:

Consider a straight wire section of length $L$ carrying a current of $I$. If that wire is placed in a magnetic field of strength $B$ at an angle of $\theta$ relative to the field, the magnitude of the magnetic force on this wire would be:

$F = B\, I\, L\, \sin(\theta)$.

In this question, $\theta = 90^{\circ}$ since the wire is perpendicular to the magnetic field. The magnitude of the magnetic force on this wire would be:

$\begin{aligned}F &= B\, I\, L\, \sin(\theta) \\ &= (0.250\; {\rm T}) \, (15.0\; {\rm A}) \, (4.75\; {\rm m})\, (\sin(90^{\circ}) \\ &\approx 17.8\; {\rm N} \end{aligned}$.