An electron is fired horizontally at 2.5 x 10 m/s between two horizontal parallel plates 7.5 cm long, as shown in the diagram. T
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Answer:
1)
When a charge is in motion in a magnetic field, the charge experiences a force of magnitude
where here:
For the proton in this problem:
is the charge of the proton
v = 300 m/s is the speed of the proton
B = 19 T is the magnetic field
is the angle between the directions of v and B
So the force is
2)
The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.
The density of the field lines at any point tells how strong is the magnetic field at that point.
If we observe the field lines around a magnet, we observe that:
- The density of field lines is higher near the Poles
- The density of field lines is lower far from the Poles
Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.
3)
The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.
It can be applied as follows:
- Direction of index finger = direction of motion of the charge
- Direction of middle finger = direction of magnetic field
- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)
In this problem:
- Direction of motion = to the right (index finger)
- Direction of field = downward (middle finger)
- Direction of force = into the screen (thumb)
4)
The radius of a particle moving in a magnetic field is given by:
where here we have:
is the mass of the alpha particle
is the speed of the alpha particle
is the charge of the alpha particle
B = 12.2 T is the strength of the magnetic field
Substituting, we find:
5)
The cyclotron frequency of a charged particle in circular motion in a magnetic field is:
where here:
is the charge of the electron
B = 0.0045 T is the strength of the magnetic field
is the mass of the electron
Substituting, we find:
6)
When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:
1- A uniform motion in a certain direction
2- A circular motion in the direction perpendicular to the magnetic field
The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.
Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.
7)
The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:
where
N is the number of turns in the loop
is the change in magnetic flux through the loop
is the time elapsed
From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.
8)
The flux is calculated as
where
B = 5.5 T is the strength of the magnetic field
A is the area of the coil
is the angle between the direction of the field and the plane of the loop
Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is
So the flux is
See the last 7 answers in the attached document.
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Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.
Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.
We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.
In order to draw the free body diagram, first let's calculate the friction force acting on the crate:
Since the friction force is greater than the force applied, the crate will not move, and the friction force will be equal to the force applied.
The weight force is equal to 40 * 9.8 = 392 N.
So, drawing the diagram, we have:
