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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 magnitude of the force felt by a charged particle moving in a magnetic field is given by the following formula: F = qvBsinθ 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 - θ is the angle between the velocity of the particle and the magnetic field Given: - q = 1.60218 × 10^-19 C (charge of a proton) - v = 1.53 × 10^6 m/s (speed of the proton) - B = 0.338 T (magnetic field strength) - θ = 33.6 degrees First, convert the angle from degrees to radians: θ (in radians) = 33.6 * (π / 180) Next, plug these values into the formula: F = (1.60218 × 10^-19 C) * (1.53 × 10^6 m/s) * (0.338 T) * sin(33.6 degrees in radians) Finally, calculate the result. Remember that the magnetic force is usually very small since the charge of the particle is small.

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