A wire carries 5.00 mA of current in a region where the uniform magnetic field strength is 1.50 mT. If the net force per unit length on the wire is 4.5 $\mu$ N/m, the angle between the current and the magnetic field is
A) zero.
B) 31°.
C) 37°.
D) 53°.

Question

Quizplus · Accepted Answer

The magnetic force per unit length on a wire is given by $ F/L = I \cdot B \cdot \sin(\theta) $. Solving for $\theta$ using the given values, $\sin(\theta) = \frac{4.5 \times 10^{-6} \, \text{N/m}}{5.00 \times 10^{-3} \, \text{A} \times 1.50 \times 10^{-3} \, \text{T}}$, we find $\theta \approx 37^\circ$.

A Wire Carries 5 μ\muμ N/m, the Angle Between the Current and the Magnetic Field

A Wire Carries 5 $\mu$ N/m, the Angle Between the Current and the Magnetic Field