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 start by using the formula for the magnetic force on a charged particle moving through a magnetic field:
F = q * v * B
where:
F is the magnetic force (3.50 × 10^−14 N),
q is the charge of the particle (1.60 × 10^−19 C),
v is the speed of the particle (what we're trying to find), and
B is the magnitude of the magnetic field (3.09 × 10^−2 T).
Rearranging the formula to solve for v, we have:
v = F / (q * B)
Now we can plug in the given values to calculate the speed:
v = (3.50 × 10^−14 N) / ((1.60 × 10^−19 C) * (3.09 × 10^−2 T))
Using the given values, we can simplify the expression:
v = (3.50 × 10^−14 N) / (4.96 × 10^−21 C∙T)
Dividing 3.50 × 10^−14 N by 4.96 × 10^−21 C∙T, we get:
v ≈ 7.05 × 10^6 m/s
Therefore, the speed of the proton is approximately 7.05 × 10^6 m/s.