If a proton in a proton beam experiences a downward force of 2.0X10-14N while traveling in a magnetic field of 8.3X10-2T west, what is the velocity?
To find the velocity of a proton in a magnetic field, we can use the formula for the magnetic force on a moving charged particle:
F = q * v * B
Where:
F is the magnetic force
q is the charge of the particle (in this case, the charge of a proton is +1.6 × 10^-19 C)
v is the velocity of the particle
B is the magnetic field strength
Rearranging the formula to solve for velocity (v), we have:
v = F / (q * B)
Let's substitute the given values into the equation:
F = 2.0 × 10^-14 N
q = +1.6 × 10^-19 C
B = 8.3 × 10^-2 T
Now we can calculate the velocity:
v = (2.0 × 10^-14 N) / ((1.6 × 10^-19 C) * (8.3 × 10^-2 T))
Before performing the calculation, let's simplify the expression:
v = (2.0 × 10^-14 N) / (1.328 × 10^-20 C * T)
v = 1.504 × 10^6 m/s
Therefore, the velocity of the proton is approximately 1.504 × 10^6 m/s.