Question

A proton moving at 6.60 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.60 10-13 N. What is the angle between the proton's velocity and the field

Answer 1

Hi there!

We can use the following equation for a point charge in a magnetic field:

`\large\boxed{F_B = qv * B}`

`F_B` = Force due to magnetic field (7.6 × 10⁻¹³N)

`q` = Charge of particle (1.6 × 10⁻¹⁹ C)

`v` = velocity of particle (6.6 × 10⁶ m/s)

`B` = Magnetic field strength (1.8 T)

Or, without the cross product:

`F_B = qvBsin\theta`

θ = angle between particle's velocity and field

We can rearrange to solve for theta:

`(F_B)/(qvB) = sin\theta\\\\\theta = sin^(-1) ((F_B)/(qvB))`

Solve for theta:

`\theta = sin^(-1) ((7.6*10^(-13))/((1.6*10^(-19))(6.6*10^6)(1.80))) = \boxed{23.57^o}`