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.
To find the value of E such that the electron is undeflected, we can use the equation for the force experienced by a charged particle moving in both electric and magnetic fields.
The force on a charged particle moving in an electric field is given by F_electric = qE, where q is the charge of the particle and E is the electric field.
The force on a charged particle moving in a magnetic field is given by F_magnetic = qvB, where v is the velocity of the particle and B is the magnetic field.
Since we are given that the electron is moving along the negative x-axis, its velocity can be written as v = -v_0 i-hat, where v_0 is the magnitude of the velocity.
In this case, the electron is undeflected, which means that the magnetic force experienced by the electron must be equal and opposite to the electric force. Therefore, we can set F_electric = -F_magnetic.
Using the given values and equations, we can write:
qE = -qvB
Multiplying both sides by the vector i-hat (unit vector in the x-direction) to get the component in the x-direction:
qE_x = -qvB_x
Since the electron is undeflected, the net force in the x-direction must be zero, which means qE_x = 0.
Substituting the values q = -e (charge of an electron) and B_x = 0 (since the magnetic field is in the y-direction), we have:
-eE_x = 0
Simplifying, we find:
E_x = 0
Therefore, the value of E such that the electron is undeflected is E = 0.