A proton moves perpendicularly to a mag-

netic field that has a magnitude of 3.09×10^−2 T.
The charge on a proton is 1.60×10−19 C.
What is the speed of the particle if the
magnitude of the magnetic force on it is 3.50*10^−14N?
Answer in units of m/s.

To find the speed of the particle, we can use the equation for the magnetic force on a charged particle moving through a magnetic field.

The formula for the magnetic force (F) on a charged particle is given by:

F = q * v * B * sin(θ)

Where:
F is the magnetic force.
q is the charge on the particle.
v is the velocity of the particle.
B is the magnetic field strength.
θ is the angle between the velocity vector and the magnetic field vector.

In this case, the velocity of the proton is perpendicular to the magnetic field, so the angle between the velocity and the magnetic field is 90 degrees. Therefore, sin(θ) = 1.

So the equation becomes:

F = q * v * B

We can rearrange the equation to solve for the velocity (v):

v = F / (q * B)

Now we can plug in the known values:

F = 3.50 * 10^(-14) N (the magnitude of the magnetic force on the proton)
q = 1.60 * 10^(-19) C (the charge on the proton)
B = 3.09 * 10^(-2) T (the magnetic field strength)

v = (3.50 * 10^(-14) N) / (1.60 * 10^(-19) C * 3.09 * 10^(-2) T)

Now, we can calculate the value of v:

v ≈ 7.22 * 10^5 m/s

Therefore, the speed of the particle is approximately 7.22 * 10^5 m/s.