Question

A (1.28×10-5 C) charge is moving at (5.63×107 m/s) perpendicular to a magnetic field

of (8.91×10-4 T). What is the magnitude of the magnetic force on the charge?

A. 6.42×10 -1 N
B. 5.02×10 4 N
C. 7.21×10 2 N
D. 1.14×10 -8 N

Answer 1

Hello!

We can use the following equation for magnetic force on a moving particle:

`F_B = qv * B`

`F_B` = Magnetic Force (? N)

q = Charge of particle (1.28 × 10⁻⁵C)
v =velocity of particle (5.63× 10⁷m/s)

B = Strength of magnetic field (8.91 × 10⁻⁴T)

This is a cross-product, so the magnetic force depends on the SINE of the angle between the particle's velocity vector and the magnetic field vector.

Since the charge is moving PERPENDICULAR to the field, the angle between the velocity and magnetic field is 90°. The sine of 90° = 1, so we can simplify the equation to:

`F_B = qvB(1) = qvB\\`

Plug in the values and solve.

`F_B = (1.28*10^(-5))(5.63*10^7)(8.91*10^(-4)) = \boxed{\text{A. } 0.6421 N}`