A straight wire that is $0.60 \mathrm {~m}$ and carrying a current of $2.0 \mathrm {~A}$ is placed at an angle with respect to the magnetic field of strength $0.30 \mathrm {~T}$. If the wire experiences a force of magnitude $0.18 \mathrm {~N}$, what angle does the wire make with respect to the magnetic field?
A) $30 ^ { \circ }$
B) $35 ^ { \circ }$
C) $20 ^ { \circ }$
D) $25 ^ { \circ }$
E) $60 ^ { \circ }$

Question

Quizplus · Accepted Answer

The force on a current-carrying wire in a magnetic field is given by the equation:$$F = ILB\sin	heta$$where:* F is the force on the wire (in newtons)* I is the current in the wire (in amperes)* L is the length of the wire (in meters)* B is the strength of the magnetic field (in teslas)* θ is the angle between the wire and the magnetic field (in degrees)In this problem, we are given the following information:* F = 0.18 N* I = 2.0 A* L = 0.60 m* B = 0.30 TWe can use the equation above to solve for θ:$$0.18 N = (2.0 A)(0.60 m)(0.30 T)\sin	heta$$$$\sin	heta = 0.18 N / (2.0 A)(0.60 m)(0.30 T) = 0.30$$$$	heta = \sin^{-1}(0.30) = 30^\circ$$Therefore, the angle between the wire and the magnetic field is 30 degrees.

A Straight Wire That Is 0.60 m0.60 \mathrm {~m}0.60 m And Carrying a Current Of

A Straight Wire That Is $0.60 \mathrm {~m}$ And Carrying a Current Of