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Sauron [17]
3 years ago
5

If the magnitude of the magnetic force on a proton is F when it is moving at 18.0 ∘ with respect to the field, what is the magni

tude of the force (in terms of F) when this charge is moving at 33.0 ∘ with respect to the field?
Physics
1 answer:
Lelu [443]3 years ago
5 0

Answer:

The force when θ = 33° is 1.7625 times of the force when θ = 18°

Explanation:

The force on a moving charge through a magnetic field is given by

F = qvB sin θ

q = charge of the moving particle

v = Velocity of the moving charge

B = Magnetic field strength

θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge

Because qvB are all constant, we can call the expression K.

F = K sinθ

when θ = 18°,

F = K sin 18° = 0.309K

when θ = 33°, let the force be F₁

F₁ = K sin 33° = 0.5446K

(F₁/F) = (0.5446K/0.309K) = 1.7625

F₁ = 1.7625 F

Hope this Helps!!!

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Explanation:

<u>Constant Acceleration Motion</u>

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