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A proton traveling at 3.70 km/s suddenly enters a uniform magnetic field

of. 710 T, traveling at an angle of 55 degrees. Find the magnitude of this force this magnetic field exerts on the proton.

User Pixeltrix
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1 Answer

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To find the magnitude of the force that the magnetic field exerts on the proton, we can use the formula:

F = qvBsinθ

where F is the force, q is the charge of the particle (in this case, a proton with a charge of +1.602 x 10^-19 C), v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.

Plugging in the values given, we get:

F = (1.602 x 10^-19 C)(3.70 x 10^3 m/s)(0.710 T)sin(55°)
F ≈ 1.51 x 10^-14 N

Therefore, the magnitude of the force that the magnetic field exerts on the proton is approximately 1.51 x 10^-14 N.
User Shibankar
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