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 magnitude of the

Question

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 magnitude of the force (in terms of F) when this charge is moving at 33.0 ∘ with respect to the field?

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Cherry 4 years 2021-08-19T09:27:38+00:00 1 Answers 14 views 0

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  1. minhkhoi
    0
    2021-08-19T09:28:50+00:00
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    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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