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An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64×10−27 kg ) traveling horizontally at 35.0 km/s enters a uniform, vertical, 1.16 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.

Respuesta :

Answer:

A) Diameter = 12.522 x 10^(-4) m

B) It makes the speed to remain constant.

C) acceleration = 2.6988 x 10^(12) m/s²

D) Direction of acceleration is horizontally and radially inward the curvature particles path.

E) It's because the force is perpendicular to the velocity and as such it changes the magnitude but not the direction.

Explanation:

A) From Lorentz force equation;

F = qvB sinθ

Now the angle between the magnetic field and the velocity is 90° and sin 90 = 1,thus the magnitude of the force is now;

F = qvB

Now, this magnitude must be equal to the force given by Newton's second law. Thus;

F = ma

But, in this case, acceleration is directed radially inwards and thus;

a = v²/r

Thus, F = (mv²)/r

Now since the magnitude from lorentz force equation must be equal to that of Newton's second law, thus;

qvB = (mv²)/r

So, r is radius and since there are two protons and the charge of one proton is 1.6 x 10^(-19),

q = 2 x 1.6 x 10^(-19) = 3.2 x 10^(-19)C

And v = 35km/s or 35000m/s

So qvB = (mv²)/r

Making r the subject, we obtain;

r = mv/qB

r = (6.64×10^(−27) x 35000)/(3.2 x 10^(-19) x 1.16) = 6.261 x 10^(-4) m

We know that Diameter = 2 x Radius

Thus, D = 2 x 6.261 x 10^(-4) = 12.522 x 10^(-4) m

B) The force acting on the alpha particle is at 90° angle and thus perpendicular to the direction of motion and the integral of the work done will be zero. Thus, we can say that the force produces a radially inward acceleration and the velocity is perpendicular to this acceleration. Hence we can conclude that the acceleration makes the velocity constant because it only changes the direction and not the magnitude.

C) From initially, we saw that F from Lorentz force equation is equal to that from Newton's second law.

Thus ; qvBsinθ = ma

Thus, a = (qvBsinθ)/m

We saw that the angle is 90 and as such sin 90 = 1.

So we can calculate a as follows;

a = [3.2 x 10^(-19) x 35000 x 1.16]/(6.64×10^(−27) = 2.6988 x 10^(12) m/s²

D) The direction of the acceleration is horizontally and radially inward the curvature particles path because the acceleration is perpendicular to the magnetic field and the velocity.

E) The speed of the particle does not change even though an unbalanced external force acts on it because the force is perpendicular to the velocity and as such it changes the magnitude but not the direction.