An electron in an electron beam experiences a downward force of 0.53×10^(−14)N while traveling in a magnetic field of 2.4×10^(−2)T west. The charge on a proton is 1.60×10^(−19).
a) What is the magnitude of the velocity? Answer in units of m/s.
b)What is its direction? (West, None of These, East, South, North)
To find the magnitude of the velocity of the electron, we can use the formula for the magnetic force experienced by a charged particle:
F = q * v * B
Where:
F is the force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field strength
We are given the force experienced by the electron (F = 0.53×10^(-14) N), the charge of a proton (q = 1.60×10^(-19) C), and the magnetic field strength (B = 2.4×10^(-2) T).
a) To find the magnitude of the velocity, we rearrange the formula and solve for v:
v = F / (q * B)
Plugging in the given values:
v = (0.53×10^(-14) N) / (1.60×10^(-19) C * 2.4×10^(-2) T)
Let's calculate the value of v:
v ≈ 1.39×10^5 m/s
Therefore, the magnitude of the velocity of the electron is approximately 1.39×10^5 m/s.
b) The direction of the velocity is not directly given in the question. However, we can deduce the direction based on the given magnetic field's direction, which is west.
Since the electron experiences a downward force in the magnetic field, based on the right-hand rule for the magnetic force on a moving charged particle, we can determine that the direction of the electron's velocity is north.
Therefore, the direction of the electron's velocity is North.