An electron in an electron beam experiences a downward force of 1.9×10^−14 N while traveling in a magnetic field of 6.4 × 10^−2 T west. The charge on a proton is 1.60×10−19.
a) What is the magnitude of the velocity?
b) What is its direction?
To find the magnitude of the velocity of the electron in the electron beam, we can use the formula for the magnetic force on a charged particle:
F = qvB
Where:
F is the force experienced by the particle
q is the charge of the particle
v is the velocity of the particle
B is the strength of the magnetic field
In this case, the force experienced by the electron is given as 1.9×10^−14 N, the charge of an electron is -1.60×10^−19 C (negative because it's an electron), and the strength of the magnetic field is 6.4 × 10^−2 T. We want to find the magnitude of the velocity (|v|).
a) Let's rearrange the formula to solve for v:
F = qvB
|v| = F / (qB)
Substituting the given values:
|v| = (1.9×10^−14 N) / ((-1.60×10^−19 C) * (6.4 × 10^−2 T))
Now, let's calculate |v|:
|v| = (1.9×10^−14) / ((-1.60×10^−19) * (6.4 × 10^−2))
|v| ≈ 1.86 × 10^5 m/s
Therefore, the magnitude of the velocity of the electron is approximately 1.86 × 10^5 m/s.
b) To determine the direction of the velocity, we need to consider the force experienced by the electron. Since the force is downward (in the negative y-direction), and the charge of the electron is negative, we can conclude that the electron is traveling in the opposite direction of the force.
Hence, the direction of the electron's velocity is upward or in the positive y-direction.
To solve this problem, we can use the equation for the magnetic force on a charged particle:
F = q * v * B
Where:
- F is the force experienced by the charged particle,
- q is the charge of the particle,
- v is the velocity of the particle, and
- B is the magnetic field strength.
We are given:
- F = 1.9 × 10^(-14) N
- q = 1.6 × 10^(-19) C (charge on a proton)
- B = 6.4 × 10^(-2) T (magnetic field strength)
a) To find the magnitude of the velocity v, we can rearrange the equation and solve for v:
F = q * v * B
v = F / (q * B)
Substituting the given values:
v = (1.9 × 10^(-14) N) / (1.6 × 10^(-19) C * 6.4 × 10^(-2) T)
v ≈ 1.95 × 10^4 m/s
Therefore, the magnitude of the velocity is approximately 1.95 × 10^4 m/s.
b) To find the direction of the velocity, we need to determine the direction of the force acting on the electron.
The force experienced by a charged particle moving in a magnetic field is perpendicular to both the velocity and the magnetic field. Using the right-hand rule, we can determine that the force on an electron in a magnetic field is pointing out of the page if the electron is moving downward.
Since the force experienced by the electron is downward, the velocity would be perpendicular to the force and would be directed towards the right.
Therefore, the direction of the velocity is to the right.