Question

What are the two angles between the direction of the current and the direction of a uniform 0.0400 T magnetic field for which the magnetic force on the wire has magnitude 0.0250 N?

Answer 1

Answer:

The two values of θ are 41.03° and 138.97°.

Explanation:

The force on a current carrying wire is given by the following equation:

$\vec{F} = I\vec{L}\times \vec{B}$

The cross-product can be written with a sine term:

$F = ILB\sin(\theta)\\0.025 = IL(0.04)\sin(\theta)\\\sin(\theta) = \frac{0.025}{0.04IL}\\\theta = \arcsin(\frac{0.025}{0.04IL})$

If we assume that the wire is 0.28 m long and the current is 3.40 A, then sin(θ) becomes 0.6565.

Finally, the two values of θ are 41.03° and 138.97°.