Question

1. A bubble chamber is a device used to study the trajectories of elementary particles such as electrons and protons. The image below shows the trajectories of some typical particles, one of which is indicated by a white arrow. Suppose this particle is moving in the direction indicated by the black arrow; the particle follows a curved path due to the presence of a magnetic field perpendicular to the plane of the photo.

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1. For this question, we assume that the magnetic field is directed into the page. Does this particle have a positive charge or a negative charge? Explain your answer.
2. Describe how the trajectory changes when the charge on the particle is increased by a factor of 3.
3. Describe how the trajectory changes when the mass is decreased by a factor of 10.

Answer 1

a)

Here the charged particle in the perpendicular magnetic field directing into the plane of the paper is executing the circular motion trajectory in the Anti clockwise direction as shown in the figure given as marked by the black arrow.

Here,By the Flemmings left hand rule,

In which the middle finger indicates the direction of the velocity and the fore finger indicates the direction of the magnetic field,So,the thumb indicates the direction of the force as well for a positive charge.Also the direction of the force will be opposite for the case of the negetive charge as well.

Here we have for the point A considered where it marked with black arrow,

We have the velocity directing towards +Y axis,and the magnetic field is directing into the plane(-Z axis),Also the force is directing to the centre of the circle(Centripetal force)ie -X axis.

So,According to Flemming's left hand rule,We got the force is directing to the centre of the circle in the given direction itself(-X axis).So,since there is no change in the direction of the force or same to the actual one,So

The charge taken is positive charge(+ve).

b)

Here the centripetal force for the circular motion is provided by the magnetic force on the charged particle.

Let velocity of charged particle of charge be . And mass of particle,.

Let radius of the circular path,

magnitude of the magnetic field,

So,Centripetal force,$F_c=\frac{mv^2}{r}$

Also,the magnetic force,

FB = quBsin 90º = quB

ie,$\frac{mv^2}{r}=qvB$

So,Radius of the circular trajectory,$r=\frac{mv}{qB}$

Here given that the charge of the particle increased by a factor of 3.

ie,Final charge,$q'=3q$

Here since all others m,v and B remains same,we have,radius of trajectory,$r\alpha \frac{1}{q}$

So,$\frac{r}{r'}= \frac{q'}{q}=\frac{3q}{q}=3$

So,The final radius of the circular trajectory,$r'=\frac{r}{3}$

So,final radius of the trajectory of the circular path of the charged particle will be reduced to 1/3th of the initial radius.

Or the final radius of the trajectory is reduced by a factor of 3.

c)

we have found the radius of the trajectory,$r=\frac{mv}{qB}$

If the mass of the particle is decreased by a factor of 10.

ie,the final mass of the particle,$m'=\frac{m}{10}$

We have when v,q and B remans constant,mass,$m\alpha r$

So,$\frac{r}{r'}=\frac{m}{m'}=\frac{m}{m/10}=10$

So,$r'=\frac{r}{10}$

So,The final radius of the trajectory of the parcticle becomes 1/10th of initial radius.

ie,Final radius of the trajectory of particle is decreased by a factor of 10.

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