An electron in an electron beam experiences a downward force of 2.0 x 10^-14 N while traveling in a magnetic field of 8.8 x 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

V = 1.5 * 10^6 m/s

To find the magnitude of the velocity of the electron, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field:

F = qvB

Where:
- F is the force experienced by the electron
- q is the charge of the electron (1.60×10^−19 C)
- v is the magnitude of the velocity of the electron
- B is the magnitude of the magnetic field (8.8 x 10 ^-2 T)

Rearranging the formula to solve for v:

v = F / (qB)

Substituting the given values:

v = (2.0 x 10^-14 N) / (1.60×10^−19 C * 8.8 x 10 ^-2 T)

v ≈ 1.70 x 10^5 m/s

Therefore, the magnitude of the velocity of the electron is approximately 1.70 x 10^5 m/s.

To find the magnitude of the velocity of the electron, we can use the equation for the magnetic force on a moving charged particle:

F = qvB,

where F is the force, q is the charge of the particle, v is the velocity, and B is the magnetic field strength.

Given:
Force (F) = 2.0 × 10^-14 N,
Charge (q) = -1.60 × 10^-19 C (electron's charge),
Magnetic field (B) = 8.8 × 10^-2 T.

From the equation above, we can rearrange it to solve for velocity (v):

F = qvB,

v = F / (qB).

Substituting the given values:

v = (2.0 × 10^-14 N) / (-1.60 × 10^-19 C × 8.8 × 10^-2 T).

Now, let's simplify and calculate the magnitude of the velocity:

v = (2.0 × 10^-14 N) / (-1.60 × 10^-19 C × 8.8 × 10^-2 T)
= -2.0 × 10^-14 / (1.60 × 10^-19 × 8.8 × 10^-2)
= -2.0 / (1.60 × 8.8) × 10^-14 / 10^-19 × 10^2
= -0.1136 × 10^-3 × 10^2
= -11.36 × 10^-3
= -11.36 × 10^-3 m/s.

Since velocity is a vector quantity, the negative sign indicates that the electron is traveling in the opposite direction of the magnetic field (west in this case).

Therefore, the magnitude of the velocity of the electron is approximately 11.36 × 10^-3 m/s.