An electron accelerates from rest to the right, in a horizontally directed electric field. The electron then leaves the electric field at a speed of 4.0 × 106, entering a magnetic field of magnitude 0.20 T (Tesla) directed into the screen. Calculate the magnitude and direction of the magnetic force on the electron.
x to right
y into screen
z up
F = q ( V cross B)
q = -1.6 * 10^-19 C scalar
V = 4 * 10^6 m/s in x direction
B = 0.20 T in y direction
so if q were POSITIVE force would be in UP (z direction)
however q is NEGATIVE, so force is in -z direction, down
since V and B are at 90 degrees, just multiply, sin of 90 = 1
F = 1.6*10^-19 * 4*10^6 * 0.20 Newtons downward
Well, well, well, our little electron friend is in for a ride! Let's calculate the magnitude and direction of the magnetic force on our speedy electron.
First, we need to find the velocity of the electron after it leaves the electric field. You mentioned it's 4.0 × 10^6. I assume you mean 4.0 × 10^6 m/s. Good choice for speed, Mr. Electron!
Now, we can use an equation called the Lorentz force equation, which states that the magnetic force (F) on a charged particle is equal to the charge (q) of that particle times its velocity (v) crossed with the magnetic field (B).
F = q * v * B
Since the electron carries a charge of -1.6 × 10^-19 Coulombs, and the magnetic field is 0.20 T, we can plug in those values:
F = (-1.6 × 10^-19 C) * (4.0 × 10^6 m/s) * (0.20 T)
Now, let's crunch some numbers (and please don't ask me about my crush on a pocket calculator).
F = -1.6 × 10^-19 * 4.0 × 10^6 * 0.20
Okay, *checks math glasses* let's calculate.
F = -1.28 × 10^-12 N
So, the magnitude of the magnetic force on the electron is 1.28 × 10^-12 N. That's a teeny tiny force!
But what about the direction, you may ask? Well, the magnetic force is negative, indicating it acts in the opposite direction to the electron's velocity. In other words, it's trying to slow our speedy electron down!
To be more precise, the magnetic force is perpendicular to both the velocity of the electron and the magnetic field. Since the velocity is to the right and the magnetic field is into the screen, the force will be straight down.
So, the magnitude of the magnetic force is 1.28 × 10^-12 N, and the direction is straight down. That's quite the magnetic party our electron is having!
To calculate the magnitude and direction of the magnetic force on the electron, we can use the equation for the magnetic force on a moving charge in a magnetic field:
F = q * (v x B)
where F is the magnetic force, q is the charge of the electron, v is its velocity, and B is the magnetic field.
First, let's calculate the magnitude of the magnetic force on the electron.
Given:
- velocity of the electron, v = 4.0 × 10^6 m/s
- magnitude of the magnetic field, B = 0.20 T (Tesla)
- charge of the electron, q = -1.6 × 10^-19 C (Coulombs)
Using the equation F = q * (v x B), we can calculate the magnitude of the magnetic force:
F = (-1.6 × 10^-19 C) * (4.0 × 10^6 m/s) * (0.20 T)
F = -1.6 × 10^-19 C * 4.0 × 10^6 m/s * 0.20 T
F = -1.28 × 10^-12 N
The magnitude of the magnetic force on the electron is 1.28 × 10^-12 N.
Now, let's determine the direction of the magnetic force.
The direction of the magnetic force can be obtained using the right-hand rule for the cross product. Place your right-hand fingers in the direction of the velocity of the electron (to the right) and then curl them towards the direction of the magnetic field (into the screen). Your thumb will point in the direction of the magnetic force.
In this case, using the right-hand rule, the magnetic force will be directed out of the screen or towards you.
Therefore, the magnitude of the magnetic force is 1.28 × 10^-12 N and its direction is out of the screen or towards you.
To calculate the magnitude and direction of the magnetic force on the electron, we can use the formula for the magnetic force on a moving charged particle:
F = q * v * B * sin(theta)
Where:
- F is the magnitude of the magnetic force
- q is the charge of the electron (1.6 x 10^-19 C)
- v is the velocity of the electron (4.0 x 10^6 m/s)
- B is the magnitude of the magnetic field (0.20 T)
- theta is the angle between the velocity vector and the magnetic field vector
In this case, the electron is moving horizontally to the right, which means the velocity is perpendicular to the magnetic field. So the angle theta is 90 degrees.
Now we can plug in the values:
F = (1.6 x 10^-19 C) * (4.0 x 10^6 m/s) * (0.20 T) * sin(90°)
Since sin(90°) = 1, the equation simplifies to:
F = (1.6 x 10^-19 C) * (4.0 x 10^6 m/s) * (0.20 T) * 1
Calculating this equation will give us the magnitude of the magnetic force on the electron.