A wire in a uniform magnetic field of 0.350 T carries a current of 3.50 A.If the magnitude of the magnetic force per unit length on the wire is 0.278 N/m,what is the angle between the magnetic field and the current?
A)13.1°
B)19.9°
C)4.56°
D)52.6°

Question

Quizplus · Accepted Answer

The magnetic force per unit length is given by:$$F_L=IB\sin	heta$$where:* $F_L$ is the magnetic force per unit length* $I$ is the current* $B$ is the magnetic field* $	heta$ is the angle between the magnetic field and the currentSolving for $	heta$, we get:$$	heta=\sin^{-1}\left(\frac{F_L}{IB}ight)$$Plugging in the given values, we get:$$	heta=\sin^{-1}\left(\frac{0.278	ext{ N/m}}{(3.50	ext{ A})(0.350	ext{ T})}ight)=13.1^\circ$$Therefore, the correct answer is A.

A Wire in a Uniform Magnetic Field of 0