A velocity selector consists of electric and magnetic fields described by the expressions vector E = E k hat bold and vector B = B j hat bold, with B = 0.0140 T. Find the value of E such that a 670 -eV electron moving along the negative x axis is undeflected.
kV/m
To find the value of E that will cause the electron to be undeflected, we can use the principles of velocity selection in a velocity selector.
In a velocity selector, the force experienced by a charged particle in an electric field is balanced by the force experienced in a magnetic field, resulting in no net force and therefore no deflection.
The force experienced by a charged particle in an electric field can be calculated using the equation:
F = qE
where F is the force, q is the charge of the particle, and E is the electric field strength.
The force experienced by a charged particle in a magnetic field can be calculated using the equation:
F = qvB
where F is the force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.
In this case, the charged particle is an electron with charge -e (where e is the elementary charge) and a velocity of v = -670 m/s (since it is moving along the negative x-axis).
Setting the forces equal to each other, we have:
qE = qvB
Since the charge of the electron is -e, we can rewrite the equation as:
(-e)E = (-e)(-670)(0.0140)
Simplifying, we have:
E = (670)(0.0140) V/m
Calculating the value, we get:
E ≈ 9.38 V/m
Therefore, the value of E needed to cause the electron to be undeflected is approximately 9.38 V/m.