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?

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