Question

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?

Answer 1

0.239 T

Step-by-step 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