Question

2. An electron travels at \(1.08 \times 10^5 \, \text{m/s}\) perpendicularly through a magnetic field (\(B = 1.12 \times 10^{-3} \, \text{T}\)). The magnitude of the magnetic force on the electron is \(x \times 10^{-W} \, \text{N}\). The value of \(W\) is:

a) 3

b) 4

c) 5

d) 6

Answer 1

The magnitude of the magnetic force on an electron is -1.81 × 10-13 N. The value of W is 13.

Step-by-step explanation:

To find the magnitude of the magnetic force on an electron traveling perpendicularly through a magnetic field, we can use the formula for the magnetic force:

F = qvB

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

In this case, the charge of the electron is known to be -1.6 × 10-19 C, the velocity of the electron is 1.08 × 105 m/s, and the magnetic field strength is 1.12 × 10-3 T.

Substituting these values into the formula, we get:

F = (-1.6 × 10-19 C) × (1.08 × 105 m/s) × (1.12 × 10-3 T)

F = -1.81 × 10-13 N

Therefore, the magnitude of the magnetic force on the electron is 1.81 × 10-13 N. The value of W is 13 since the force is already given in scientific notation with a coefficient of 1.81.