An electron travels at a speed of 1.0×104 through a uniform magnetic field whose magnitude is 1.2×10−3 . What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field are perpendicular?

F = MA

F = qv*B
ma = qvb
a = qbv/m
m = mass of electron
q = charge of electron

To find the magnitude of the magnetic force on the electron, you can use the formula:

F = q * v * B * sin(θ)

Where:
- F is the magnitude of the magnetic force
- q is the charge of the electron (1.6 × 10^-19 C)
- v is the velocity of the electron (1.0 × 10^4 m/s)
- B is the magnitude of the magnetic field (1.2 × 10^-3 T)
- θ is the angle between the velocity and the magnetic field (90°, since they are perpendicular)

Let's substitute the given values into the formula:

F = (1.6 × 10^-19 C) * (1.0 × 10^4 m/s) * (1.2 × 10^-3 T) * sin(90°)

Since sin(90°) = 1, we can simplify the equation:

F = (1.6 × 10^-19 C) * (1.0 × 10^4 m/s) * (1.2 × 10^-3 T) * 1

Calculating the values:

F = (1.6 × 1.0 × 1.2) × (10^-19 × 10^4 × 10^-3) C m/s T

F = 1.92 × 10^-18 C m/s T

The magnitude of the magnetic force on the electron is 1.92 × 10^-18 Newtons.