A proton moving at 4.80 106 m/s through a magnetic field of magnitude 1.74 T experiences a magnetic force of magnitude 7.00 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest. Enter only positive values between 0 and 360.)

31.55° and 148.45°

Explanation:

Formula for calculating the force experiences by the proton placed in a magnetic field is as expressed below;

F = qvBsinθ where;

F is the magnetic force experienced by the proton

q is the charge on the proton

v is the velocity of the proton

B is the magnetic field

θ is the angle between the proton's velocity and the field (Required)

Given parameters

F =  7.00 * 10⁻¹³N

q = 1.602*10⁻¹⁹C

v = 4.80 * 10⁶ m/s

B = 1.74 T

θ  =?

From the formula F = qvBsinθ;

sinθ = F/qvB

sinθ = 7.00 * 10⁻¹³/1.602*10⁻¹⁹* 4.80 * 10⁶*1.74

sinθ =  7.00 * 10⁻¹³/13.38*10⁻¹³

sinθ = 0.5231689 * 10⁰

sinθ = 0.5231689

θ = sin⁻¹0.5231689

θ = 31.55°

The following are the positive values of the angle between 0° and 360°

Sin is positive in the first and second quadrant. In the second quadrant the angle is equal to 180°-31.55° = 148.45°.

Hence the possible values of the angle from smallest to largest are 31.55° and 148.45°

d

explanation:

i think the answer is d. but im not sure!