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:

$\large\boxed{F_B = qv \times B}$

$F_B$ = 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:
$\frac{F_B}{qvB} = sin\theta\\\\\theta = sin^{-1} (\frac{F_B}{qvB})$

Solve for theta:
$\theta = sin^{-1} (\frac{7.6*10^{-13}}{(1.6*10^{-19})(6.6*10^6)(1.80)}) = \boxed{23.57^o}$