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A proton is projected with a velocity of 7.0 km/s into field of 0.60 T perpendicular to the motion of the proton. What is the magnitude of the magnetic force that acts on the proton? (e=1.60*10^-19 C)

A. 3.4 * 10 ^ -16 N
B. 0 N
C. 6.7 * 10 ^ -16 N
D. 13 * 10 ^ -16 N
E. 4.2 * 10 ^ -16 N

1 Answer

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Final answer:

The magnetic force acting on the proton is approximately 6.7 × 10^-16 N.

Step-by-step explanation:

To calculate the magnetic force on the proton, we can use the equation:

F = qvBsinθ

where F is the force, q is the charge of the proton, v is the velocity of the proton, B is the magnetic field strength, and θ is the angle between the proton's velocity and the magnetic field.

Given that the proton has a charge of 1.60 × 10-19 C, a velocity of 7.0 km/s (which can be converted to 7000 m/s), and a magnetic field strength of 0.60 T, and since the proton's velocity is perpendicular to the magnetic field (θ = 90°), we can plug these values into the equation:

F = (1.60 × 10-19 C)(7000 m/s)(0.60 T)(sin 90°)

Calculating this gives us:

F ≈ 6.7 × 10-16 N.

Therefore, the magnitude of the magnetic force that acts on the proton is approximately 6.7 × 10-16 N.

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