A rectangular wire coil with dimensions 21.0 cm by 45.0 cm and containing one turn is placed in a uniform magnetic field of 7.80 T. The coil carries a current of 18.0 A. Calculate the magnitude of the magnetic force exerted on the right side of the coil.
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The force on a segment of wire carrying current in a magnetic field is given by F = I * L * B * sin(θ), where: - F is the force, - I is the current passing through the wire, - L is the length of the wire segment, - B is the magnetic field strength, and - θ is the angle between the wire segment and the direction of the magnetic field.
For the rectangular coil of wire, we are interested in the right side of the loop, which has a length 'a'. Given the data: - a (length of the wire) = 21.0 cm = 0.21 m, - The magnetic field (B) = 7.80 T, - The current (I) is 18.0 A, and - θ is 90 degrees because the wire segment on the right side is perpendicular to the direction of the magnetic field in this scenario.
We know that sin(90°) = 1, therefore:
F = I * L * B * sin(θ) F = 18.0 A * 0.21 m * 7.80 T * 1 F = 18.0 A * 0.21 m * 7.80 T F = 29.484 N
The magnitude of the force on the right side of the loop is 29.484 newtons, directed outwards from the loop along the axis of the side, according to the right-hand rule for the direction of forces in a magnetic field.