Question

a proton is shot with a speed of 9.4×106 m/s perpendicular to a magnetic field b = 0.60 t . what is the radius of the path followed by the proton?

Answer 1

The radius of the path followed by a proton moving perpendicular to a 0.60 T magnetic field at a speed of 9.4 × 10⁶ m/s is 1.55 cm.

Step-by-step explanation:

The question concerns calculating the radius of the path followed by a proton moving perpendicular to a magnetic field. When a positively charged particle like a proton moves in a magnetic field perpendicularly, it undergoes circular motion due to the Lorentz force. The radius of this circular path can be determined using the formula for the magnetic force and centripetal force:

F = q * v * B = m * v2 / R

Where:

q is the charge of the proton (1.602 x 10-19 C)

v is the velocity of the proton

B is the magnetic field strength

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

R is the radius of the path

When you rearrange the formula to solve for R, you get:

R = (m * v) / (q * B)

Plugging in the values:

R = (1.67 x 10-27 kg * 9.4 x 106 m/s) / (1.602 x 10-19 C * 0.60 T) = 1.55 x 10-2 m or 1.55 cm

Therefore, the radius of the path followed by the proton is 1.55 cm.

Answer 2

The radius of the path followed by the proton is approximately 2.77 × 10^-2 meters.

Step-by-step explanation:

To find the radius of the path followed by the proton, we can use the formula for the magnetic force on a charged particle in a magnetic field:

F = qvB

where F is the centripetal force, q is the charge of the proton, v is its velocity, and B is the magnetic field strength. Since the force is perpendicular to the velocity, it provides the necessary centripetal force to keep the proton moving in a circular path.

By equating the centripetal force to the magnetic force, we can solve for the radius of the path:

R = mv / (qB) = (mpv) / (qB)

where m is the mass of the proton, p is the momentum of the proton, and mp is the magnitude of the proton's momentum.

Plugging in the given values:

R = (1.67 × 10^-27 kg)(9.4 × 10^6 m/s) / ((1.60 × 10^-19 C)(0.60 T)) = 2.77 × 10^-2 m

Therefore, the radius of the path followed by the proton is approximately 2.77 × 10^-2 meters.