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.