Question

A proton travels with a speed of 4.95 x 106 m/s at an angle of 57° with the direction of a magnetic field of magnitude 0.190 T in the positive x-direction. (a) What is the magnitude of the magnetic force on the proton? N (b) What is the proton's acceleration? (Give the magnitude.) m/s2 Need Help? Read It

Answer 1

(a) The magnitude of the magnetic force on the proton is approximately 1.76 x 10^-14 N.

(b) The magnitude of the proton's acceleration is approximately 2.87 x 10^12 m/s².

Step-by-step explanation:

Part (a): Magnetic Force

Identify relevant information:

Proton speed (v) = 4.95 x 10^6 m/s

Angle between velocity and magnetic field (θ) = 57°

Magnetic field strength (B) = 0.190 T

Formula for magnetic force:

F = q * v * B * sin(θ)

where:

F is the magnetic force

q is the charge of the proton (1.6 x 10^-19 C)

v is the proton's velocity

B is the magnetic field strength

θ is the angle between the velocity and the magnetic field

Calculate the magnetic force:

F = (1.6 x 10^-19 C) * (4.95 x 10^6 m/s) * (0.190 T) * sin(57°)

F ≈ 1.76 x 10^-14 N

Part (b): Proton's Acceleration

Formula for magnetic force:

F = q * v * B * sin(θ)

Formula for acceleration:

a = F / m

where:

a is the acceleration

m is the mass of the proton (1.67 x 10^-27 kg)

Calculate the acceleration:

a = (1.76 x 10^-14 N) / (1.67 x 10^-27 kg)

a ≈ 2.87 x 10^12 m/s²