A wire carries 5.00 mA of current in a region where the uniform magnetic field strength is 1.50 mT. If the angle between the wire and the magnetic field is 30°, the net force per unit length on the wire is
A) zero.
B) 3.75$\mu$ N/m.
C) 6.50 $\mu$ N/m.
D) 7.50 $\mu$ N/m.

Question

Quizplus · Accepted Answer

The force per unit length on a current-carrying wire in a magnetic field is given by $F/L = BIl\sin(\theta)$, where $B$ is the magnetic field strength, $I$ is the current, $l$ is the length of the wire (which cancels out in the force per unit length calculation), and $\theta$ is the angle between the wire and the magnetic field. Substituting the given values: $F/L = (1.50 \times 10^{-3} \, \text{T}) \times (5.00 \times 10^{-3} \, \text{A}) \times \sin(30°) = 3.75 \times 10^{-6} \, \text{N/m}$.

A Wire Carries 5 μ\muμ N/m C) 6 μ\muμ N/m D) 7 μ\muμ

A Wire Carries 5 $\mu$ N/m
C) 6 $\mu$ N/m
D) 7 $\mu$