Final answer:
The angle between a wire carrying an 8.00-A current and the magnetic field is 30°, found by using the formula for magnetic force on a current-carrying wire in a magnetic field. When the wire is rotated to be perpendicular (90°) to the field, the force experienced by the wire is 4.80 N. Therefore, the correct answer is a) 30°.
Step-by-step explanation:
The question asks to find the angle between a current-carrying wire and a magnetic field, given certain parameters such as current, magnetic field strength, and the resulting magnetic force. The angle can be determined using the formula F = I*L*B*sin(θ), rearranging it to solve for θ. Here, F is the magnetic force, I is the current, L is the length of the wire, and B is the magnetic field strength.
Substituting the given values:
2.40 N = 8.00 A * 0.50 m * 1.20 T * sin(θ)
Solving for sin(θ), we get:
sin(θ) = (2.40 N)/(8.00 A * 0.50 m * 1.20 T)
sin(θ) = (2.40)/(4.80)
sin(θ) = 0.5
Therefore, the angle θ for which sin(θ) = 0.5 is 30°, which corresponds to option (a).
Additionally, the force on the wire if rotated to 90° with the field can be calculated using the same formula:
F = I*L*B*sin(90°)
Since sin(90°) = 1, the force would be:
F = 8.00 A * 0.50 m * 1.20 T
F = 4.80 N