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What is the angle between a wire carrying an 8.00-A current and the 1.20-T field it is in if 50.0 cm of the wire experiences a magnetic force of 2.40 N?

a) 30°
b) 45°
c) 60°
d) 75°

User Manh Ha
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1 Answer

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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

User Meire
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