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27 July, 21:41

A proton travels at a velocity of 3.0x10^6 m/s in a direction perpendicular to a uniform magnetic field.

If the proton experiences a magnetic force of 1.15x10-13 N, what is the magnitude of the magnetic field?

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  1. 27 July, 22:49
    0
    0.239 T

    Explanation:

    Applying,

    F = Bvqsin∅ ... Equation 1

    Where F = magnetic force, B = magnetic Field, q = charge of a proton, v = velocity of proton, ∅ = angle between the velocity and the magnetic field.

    make B the subject of the equation

    B = F / (vqsin∅) ... Equation 2

    From the question,

    Given: F = 1.15*10⁻¹³ N, v = 3.0*10⁶ m/s, ∅ = 90° (perpendicular)

    Constant: q = 1.602 x 10⁻¹⁹ C

    Substitute into equation 2

    B = 1.15*10⁻¹³ / (3.0*10⁶*1.602 x 10⁻¹⁹*sin90°)

    B = 1.15*10⁻¹³ / (4.806*10⁻¹³)

    B = 0.239 T.

    Hence the magnetic field = 0.239 T
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