Calculate the magnitude of the force due to a magnetic field of 0.40T on an electron moving perpendicular to…

Analyze the question: An electron moves perpendicular to a magnetic field, so we need the magnetic force on a moving charge.

Concept: The force on a charge q moving with speed v in a magnetic field B is

F = Bqv sin θ

For perpendicular motion, θ = 90°, so sin θ = 1 and F = Bqv.

Given:

• B = 0.40 T
• v = 4.00 × 10⁶ m/s
• q = e = 1.60 × 10⁻¹⁹ C

Solution:

F = (0.40)(1.60 × 10⁻¹⁹)(4.00 × 10⁶)

Multiply the numbers first: 0.40 × 1.60 × 4.00 = 2.56

Add the powers of 10: 10⁻¹⁹ × 10⁶ = 10⁻¹³

F = 2.56 × 10⁻¹³ N

Check against the options:

• A. 6.40 × -10 (garbled, not valid)
• B. 6.40 × 10⁻¹³ N (this would result if you used 1.00 instead of 0.40 for B, or made an arithmetic slip)
• C. 2.56 × 10⁻¹³ N ✔
• D. 1.60 × 10⁻¹² N
• E. 2.30 × 10⁻¹¹ N

Tip: For perpendicular motion, just multiply B × q × v. Do the decimal numbers first (0.40 × 1.60 × 4.00 = 2.56), then handle the powers of 10 separately. If the velocity is parallel to the field, the force is zero because sin 0° = 0.