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Question: A proton travels with a speed of 1.53×106 m/s at an angle of 33.6 ◦ with a magnetic field of 0.338 T pointed in the y direction. The charge of proton is 1.60218 × 10−19 C. What is the magnitude of the magnetic force on the proton?

The force experienced by a charged particle moving in a magnetic field is given by the equation: F = q * v * B * sin(θ) where: - F is the magnetic force, - q is the charge of the particle, - v is the speed of the particle, - B is the magnetic field strength, and - θ is the angle between the velocity of the particle and the magnetic field direction. In this case: - The charge of proton (q) = 1.60218 × 10^-19 C, - The speed of proton (v) = 1.53×10^6 m/s, - The magnetic field strength (B) = 0.338 T, and - The angle between the velocity of the particle and the magnetic field direction (θ) = 33.6 degrees. Substituting these values into the equation gives: F = (1.60218 × 10^-19 C) * (1.53×10^6 m/s) * (0.338 T) * sin(33.6) Please note the sin function takes angles in radians, so we must convert θ from degrees to radians before doing the calculation. There are (pi/180) radians per degree, so the angle in radians is: θ_radians = (33.6 degrees) * (pi/180) Substituting this into the equation gives: F = (1.60218 × 10^-19 C) * (1.53×10^6 m/s) * (0.338 T) * sin(33.6 * pi/180) Therefore, you can calculate the force by using this equation.

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