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A proton moving at 4.80 10⁶ m/s through a magnetic field of 1.52 T experiences a magnetic force of magnitude 7.95 10⁻¹³ N. What is the angle between the proton's velocity and the field?

Respuesta :

Angle between the proton's velocity and the field is 42.86°

Explanation:

We know that

          Force = Charge x velocity x Magnetic field x sin of angle between velocity and field.

          F =qvB sinθ

Here we have

            q = 1.602 x 10⁻¹⁹ C

             v = 4.80 x 10⁶ m/s

            B = 1.52 T

            F = 7.95 x 10⁻¹³ N

Substituting

           F =qvB sinθ

           7.95 x 10⁻¹³ =  1.602 x 10⁻¹⁹ x 4.80 x 10⁶ x 1.52  sinθ

            θ = 42.86°

Angle between the proton's velocity and the field is 42.86°

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