An alpha particle (a He nucleus, containing two protons and two neutrons and hav

Question

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64×1027 kg ) traveling horizontally at 35.8 km/s enters a uniform, vertical, 1.20 T magnetic field.

Part A

What is the diameter of the path followed by this alpha particle?

Express your answer in millimeters to three significant figures.

Part B

What effect does the magnetic field have on the speed of the particle?

Part C

What is the magnitude of the acceleration of the alpha particle while it is in the magnetic field?

Express your answer in meters per second to three significant figures.

Part D

What is the direction of the acceleration of the alpha particle while it is in the magnetic field?

Part E

Explain why the speed of the particle does not change even though an unbalanced external force acts on it.

Explanation / Answer

A. for a circular motion in magnetic field

Fc = Fm

mv^2/r = qvB

r = mv/qB

charge on alpha particle = 2e = 2*1.6*10^-19

velocity = 35.8 km/sec = 35.8*10^3 m/sec

r = 6.64*10^-27*35.8*10^3/(2*1.6*10^-19*1.2)

r = 6.19*10^-4 m

diameter = 2*r = 1.238*10^-3 m = 1.238 mm

B. For a very short time interval the displacement of the particle is in the direction of the velocity.
The magnetic force is always perpendicular to this direction so it does no work. The work-energy theorem
therefore says that the kinetic energy of the particle, and hence its speed, is constant.

W = dKE

W = F*d

W = 0, So dKE = 0

so velocity will be constant.

C. acceleration

a = Fb/m

a = qvB/m = 3.2*10^-19*35.8*10^3*1.2/(6.64*10^-27)

a = 2.07*10^12 m/sec^2

or

a = v^2/r = (35.8*10^3)^2/(6.19*10^-4) = 2.07*10^12 m/sec^2

D.The acceleration is perpendicular to v and B and so is horizontal, toward the center of curvature of the
particle’s path.

E. The unbalanced force F is perpendicular to v, so it changes the direction of v but not its magnitude, which is the speed.

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