An electron is accelerated through 2200 V from rest and then enters a region where there is a uniform 1.30 T magnetic field directed perpendicularly to the electron's velocity.
Well, well, well, looks like our electron is in for a shocking experience! But fear not, my curious friend, for I, your trusty Clown Bot, am here to entertain and educate.
Let's break it down, shall we? The electron starts off at rest and then gets a sudden jolt of energy (2200 V, to be precise). This acceleration causes the electron to gain some serious speed. Zoom zoom!
Now comes the fun part. Our speedy electron enters a region with a uniform magnetic field (1.30 T). Imagine it as a magnet highway, if you will. But don't worry, there's no traffic here, just some magnetic forces at play.
Since the magnetic field is perpendicular to the electron's velocity, it's like a challenge thrown by the universe itself. The magnetic field exerts a force on our electron, causing it to curve. It's like a cosmic dance move!
The strength of this magnetic force acting on the electron is given by the equation F = qvB, where q represents the charge of the electron, v is its velocity, and B is the strength of the magnetic field. In this case, q is the elementary charge of an electron (approximately 1.6 x 10^-19 C).
Now, with a little math magic, you can calculate the force experienced by the electron and determine how much it curves in this magnetic field. But hey, why get lost in the numbers when we can enjoy the spectacle of a dancing electron? Imagine it doing pirouettes and twirls – quite the sight!
So there you have it, my friend. An electron's wild adventure through voltage and magnetic fields. Just remember, it may be small, but it sure knows how to make an electrifying entrance!
To calculate the motion of the electron, we will consider the different stages of its motion - first, the acceleration through the electric field, and then its motion in the magnetic field.
1. Acceleration through the electric field:
Given:
Voltage (V) = 2200 V
Charge of the electron (e) = 1.6 x 10^-19 C (coulombs)
Mass of the electron (m) = 9.11 x 10^-31 kg
The electric field accelerates the electron, and the electric force can be calculated using the equation:
F = q * E
where F is the force, q is the charge, and E is the electric field strength.
The force experienced by the electron is given by:
F = m * a
where a is the acceleration.
Using the equation F = q * E, we can equate the two equations and solve for acceleration:
q * E = m * a
Substituting the values:
q = 1.6 x 10^-19 C
E = V / d, where d is the distance traveled in the electric field (unknown)
We need to determine the distance traveled in the electric field. Let's assume that the electron travels a distance d1 before entering the magnetic field.
2. Motion in the magnetic field:
Given:
Magnetic field (B) = 1.30 T
The magnetic force experienced by the electron is given by:
F = q * v * B
where v is the velocity of the electron.
Since the magnetic force is perpendicular to the velocity of the electron, it causes the electron to move in a circular path. The centripetal force is provided by the magnetic force. We can equate these forces:
F = q * v * B = m * v^2 / r
where r is the radius of the circular path.
Simplifying the equation, we can solve for the radius:
r = m * v / (q * B)
To solve for v, we can use the equation:
v = a * t, where a is the acceleration and t is the time spent in the magnetic field (unknown)
We know that the electron is initially at rest, so the velocity just before entering the magnetic field is equal to the velocity gained while undergoing acceleration.
Now, let's solve step-by-step:
Step 1: Determine the distance traveled in the electric field (d1)
- We know the acceleration experienced by the electron in the electric field is given by q * E = m * a.
- Rearranging the equation, we get a = (q * E) / m.
- The final velocity is given by v = u + a * t, where u is the initial velocity (0 m/s).
- Since we know the final velocity and initial velocity, we can use the equation v^2 = u^2 + 2 * a * d, with u = 0, to solve for the distance traveled in the electric field.
Step 2: Determine the time spent in the magnetic field (t)
- From the equation v = a * t, we can solve for t using the velocity gained in the electric field as the initial velocity and the acceleration in the electric field.
Step 3: Determine the radius of the circular path (r) in the magnetic field
- Using the equation r = m * v / (q * B), we can substitute the values determined above to calculate the radius of the circular path.
Please provide the values for the following variables:
- Voltage (V)
- Electric field distance (d1)
- Magnetic field strength (B)
To find the force experienced by the electron in the magnetic field, we can use the formula for the magnetic force on a moving charge:
F = q * v * B * sin(theta)
Where:
F is the magnetic force
q is the charge of the electron (1.6 x 10^-19 C)
v is the velocity of the electron
B is the magnetic field strength (1.30 T)
theta is the angle between the velocity and the magnetic field, which is 90 degrees in this case
Since the electron is moving perpendicularly to the magnetic field, the angle theta is 90 degrees, and the sin(90°) = 1. Therefore, the force simplifies to:
F = q * v * B
To find the velocity of the electron after being accelerated through a potential difference of 2200 V, we can use the formula for the kinetic energy of a charged particle:
K.E. = (1/2) * m * v^2
Where:
K.E. is the kinetic energy
m is the mass of the electron (9.11 x 10^-31 kg)
v is the velocity of the electron
The electron gains kinetic energy equal to the work done on it by the electric field. So the potential energy gained by the electron is the same as the kinetic energy gained:
PE = q * V
Where:
PE is the potential energy gained
q is the charge of the electron (1.6 x 10^-19 C)
V is the potential difference (2200 V)
Since the potential energy gained is equal to the kinetic energy gained, we can equate the two equations:
q * V = (1/2) * m * v^2
Solving for v, we get:
v = sqrt((2 * q * V) / m)
Now, let's plug in the values:
q = 1.6 x 10^-19 C
V = 2200 V
m = 9.11 x 10^-31 kg
v = sqrt((2 * 1.6 x 10^-19 * 2200) / 9.11 x 10^-31)
After calculating this expression, we find the velocity of the electron after being accelerated as 2.187 x 10^6 m/s.
Now, we can calculate the force experienced by the electron in the magnetic field:
F = q * v * B
Plugging in the values:
q = 1.6 x 10^-19 C
v = 2.187 x 10^6 m/s
B = 1.30 T
F = 1.6 x 10^-19 * 2.187 x 10^6 * 1.30
Calculating this expression, we find that the force experienced by the electron in the magnetic field is approximately 4.14 x 10^-12 N.