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The minimum radius of a field bend on trade size 1 "1/4 RMC is ____.

a. 8
b. 8AWG
c. 30
d. 15

User DennyHiu
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1 Answer

5 votes

Final answer:

The angle between a wire and the magnetic field can be found using the inverse sine of the quotient of the force and the product of current, length of wire, and field strength. When the wire is rotated to make a 90° angle with the field, the force is maximized and is calculated simply as the product of current, length of wire, and field strength without any trigonometric functions.

Step-by-step explanation:

The student's question refers to the magnetic forces experienced by a wire carrying current in a magnetic field. To find the angle between the wire and the magnetic field given the force, current, length of wire, and intensity of magnetic field, we use the formula for magnetic force on a current-carrying conductor: F = I L B sin(θ), where F is the force, I is the current, L is the length of the wire in the magnetic field, B is the magnetic field, and θ is the angle. Rearranging for θ, we get θ = sin-1(F / (I L B)). Plugging in the values, we find the angle is:

θ = sin-1(2.40 N / (8.00 A * 0.50 m * 1.20 T))

To answer part (b), when the angle is 90 degrees, we have maximum force since sin(90°) = 1. Therefore, the force is simply the product of the current, length of the wire, and the magnetic field strength without any angle consideration. Thus, the force on the wire at 90° is:

F = 8.00 A * 0.50 m * 1.20 T

F = 4.80 N

User Jimmy Fagerholm
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