Assume the proton is at rest initially.
The charge on the proton is 1.602×10−19 C
and its mass is 1.67 × 10−27 kg.
If the magnetic field is 2.85 T and the potential difference is 2.29 MV, find the magnitude
of the force on the proton after the proton
passes through the potential difference.
Answer in units of N
012 (part 3 of 3) 10.0 points
The proton will subsequently move in a circle.
How many revolutions will the proton perform per second?
Answer in units of Hz
To find the magnitude of the force on the proton, we can use the equation:
F = q * V * B
where F is the magnitude of the force, q is the charge on the proton, V is the potential difference, and B is the magnetic field.
Substituting the given values:
F = (1.602 × 10^(-19) C) * (2.29 × 10^6 V) * (2.85 T)
Simplifying the expression:
F ≈ 1.602 × 10^(-13) N
So, the magnitude of the force on the proton is approximately 1.602 × 10^(-13) N.
Now, to determine the number of revolutions the proton performs per second, we can use the formula:
f = 1 / T
where f is the frequency in Hz and T is the time period.
The time period of one revolution can be found using the formula:
T = (2 * π * m) / (q * B)
where m is the mass of the proton, q is the charge, and B is the magnetic field.
Substituting the given values:
T = (2 * π * (1.67 × 10^(-27) kg)) / ((1.602 × 10^(-19) C) * (2.85 T))
Simplifying the expression:
T ≈ 2.217 × 10^(-8) s
Now, calculating the frequency:
f = 1 / T ≈ 1 / (2.217 × 10^(-8) s)
f ≈ 4.511 × 10^7 Hz
Therefore, the proton will perform approximately 4.511 × 10^7 revolutions per second.