Final answer:
The magnitude of the force on a 65.5 cm long wire carrying 15.0 A at right angles to a 47.5 mT magnetic field is approximately 0.4656 N.
Step-by-step explanation:
To find the magnitude of the force on a current-carrying wire in a magnetic field, we can use the formula derived from Lorentz force law which is F = ILB sin(θ), where F is the force in Newtons, I is the current in Amperes, L is the length of the wire in meters, B is the magnetic field strength in Teslas, and θ is the angle between the direction of the current and the magnetic field (in radians or degrees).
For our given problem, we have a wire of length 65.5 cm (which is 0.655 m after conversion), carrying a current of 15.0 A, placed perpendicular (at a 90° angle) to a magnetic field of 47.5 mT (which is 0.0475 T after conversion). Using the formula: F = 15.0 A * 0.655 m * 0.0475 T * sin(90°). Since the sinusoid of 90 degrees is 1, the formula simplifies to F = 15.0 A * 0.655 m * 0.0475 T. Calculating this we get F ≈ 0.4656 N, which is the magnitude of the force on the wire.