# A charged particle of mass m = 0.38 kg moves along a horizontal frictionlessplane through a 0.8-m long velocity selector as shown in the figure below. Refer to the Image After leaving the velocity...

A charged particle of mass m = 0.38 kg moves along a horizontal frictionless

plane through a 0.8-m long velocity selector as shown in the figure below.

Refer to the Image

After leaving the velocity selector, the particle moves along a semicircular

path until it strikes a phosphorescent screen. If the entire motion (from start

to end) takes 2.0-seconds, determine the charge of the particle. Assume that

B = 0.3 T and E = 5.0 V/m.

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### 1 Answer

We are looking to find the charge of a particle which can escape a velocity selector from the centre of its aperture.

The equation that relates charge, magnetic field strength, velocity and force is the Lorenz equation:

`F=q(E+vxxB)`

where F is the force, E is the electric field, v is the velocity and B is the strength of the magnetic field.

For the particle to escape the velocity selector, the force on the particle from the Lorenz force must be 0. This is because inside the velocity selector, the electric field will be applying a force downwards whilst the magnetic field will be applying an opposing force in upwards. We can tell it is this way around as once the particle leaves the electric. As the particle is attracted downwards in a magnetic field pointing upwards, we can also tell that the particle must be negatively charged.

As the velocity of the particle affects the magnitude of the force exerted onto it by the magnetic field, but not the electric field, the electric field and magnetic field can be set such that only particles with a certain velocity will travel in a straight line within the selector.

All of this means that we can plug this into the Lorenz equation and get:

`0=q(E+vxxB)`

which can be rearranged to be:

`qE=-q(vxxB)`

As the charge in this equation is a scalar quantity, it can be divided out to show that:

`E=-vxxB`

As we already know that the force in this scenario affects the particle in only one dimension, we can drop the vector quantities of these variables and simply use their magnitude, which comes out so that:

`|E|=|v||B|`

which can be rearranged such that:

`|v|=|E|/|B|`

We can now plug in the information we know to get the speed of the particle:

`|v|=5/0.3~~16.67`

Now as the charge cancelled out of the equation, it cannot be calculated by simply plugging this information into the Lorenz equation which describes the force inside the selector. We do however have some further information in that the particle is moving in circular motion in the magnetic field once it leaves the selector.

As we know the speed of the particle and the time it takes to hit the screen, we can calculate the distance it has travelled on the semicircular path which will give us the radius of the circle. From here we can work out the acceleration and hence the force exerted by the magnetic field and the charge.

It says in the question that the particle moves for 2.0 seconds. The velocity of the particle is related to the distance it travelled by:

`vt=s`

where s is the distance travelled and t is the time it took to travel that distance. Knowing the time and velocity, we can see that the total distance travelled by the particle is:

`s_t=5/0.3*2m`

and from this we can take off the length of the selector to find that the distance travelled by the particle in the semicircle (and thus the circumference of the semicircle) is:

`s_(c/2)=s_t-0.8=(5xx2)/0.3-0.8m`

The circumference of a circle is given by:

`C=2pir`

where r is the radius of a circle. Thus the radius of the semicircle is:

`r=S_(c/2)/pi`

In circular motion, the acceleration is of a fixed intensity perpendicular to the velocity. The equation we need is one which links the magnitude of the acceleration with the magnitude of the velocity of the particle movement and the length of the radius with which it moves around in the circle. This equation for constant circular motion is:

`a=v^2/r`

as explained in the second reference link below. Newton's second law of motion states that:

`F=ma`

These equations can be put together with the Lorenz equation for a particle in a magnetic field to give:

`|F|=|q|vB=(mpiv^2)/(s_(c/2))`

and rearranged to give this in terms of the magnitude of the charge:

`|q|=|(mpiv^2)/(s_(c/2)vB)|=|(mpiv)/(s_(c/2)B)|`

We now know all of those values! Lets plug them in. I'll be ignoring the units for to make the equation look better, but will add them in at the end.

`|q|=|(mpiv)/(s_(c/2)B)|=(0.38xxpixx(5/0.3))/((5xx2)/(0.3)-0.8xx0.3)~~0.60123C`

As we know the particle must be negatively charged, the final result is that the particle's charge is:

`q~~-0.60123C`

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