A proton moving at 6.40 106 m/s through a magnetic field of magnitude 1.72 T experiences a magnetic force of magnitude 7.20 10-13 N. What is the angle between the proton's velocity and the field

24.08°

Explanation:

From Electromagnetism,

F = qBvsin∅ Equation 1

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

make ∅ the subject of the equation

∅ = sin⁻¹(F/qBv) Equation 2

Given: F = 7.20×10⁻¹³ N, q = 1.602 x 10⁻¹⁹ C, B = 1.72 T, v = 6.4×10⁶ m/s

Substitute into equation 2

∅ = sin⁻¹[7.20×10⁻¹³/(1.602 x 10⁻¹⁹× 1.72× 6.4×10⁶)]

∅ = sin⁻¹[0.408]

∅ = 24.08°.

Hence the angle between the velocity and the Field = 24.08°

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank