Problem 3: A proton in a cyclotron gains AK = 2eAV of kinetic energy per revolution, where AV is the potential between the dees. Although the energy gain comes in small pulses, the proton makes so many revolutions that it is reasonable to model the energy as increasing at the constant rate P = dk = where T is the period of the cyclotron motion. This is power input because it is a rate of increase of energy. Find an expression for r(t), the radius of a proton's orbit in a cyclotron, in terms of m, e, B, P, and t. Assume that r = 0 at t = 0. 3 a) Using the formula for the radius of the cyclotron orbit, r = mg, work out a formula for the kinetic energy K of the proton in terms of m, e, B, and r. dk b) Show that equation d = P results in the following differential equation for r as a function of t: rd = m. Make sure to show all the steps that lead to this equation. dt dt e²B². c) This differential equation can be solved by separating the variables. For that, rewrite it as r dr = Pm e² B² dt and integrate both sides. The expression on the left side should be integrated from r = 0 (center) to r(t) (an intermediate radius at the moment of time t), while the expression on the right side should be integrated from 0 to t. From here, obtain the expression for r(t) in terms of m, e, B, P, and t.

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Hello, I truly need help with part A,part B and part C because I really don't know how to do it is there any chance that you can help and can you also label them thank you

**Problem 3:**
A proton in a cyclotron gains \(\Delta K = 2e \Delta V\) of kinetic energy per revolution, where \(\Delta V\) is the potential between the dees. Although the energy gain comes in small pulses, the proton makes so many revolutions that it is reasonable to model the energy as increasing at the constant rate \(P = \frac{dK}{dt} = \frac{\Delta K}{T}\), where \(T\) is the period of the cyclotron motion. This is power input because it is a rate of increase of energy. Find an expression for \(r(t)\), the radius of a proton’s orbit in a cyclotron, in terms of \(m\), \(e\), \(B\), \(P\), and \(t\). Assume that \(r = 0\) at \(t = 0\).

a) Using the formula for the radius of the cyclotron orbit, \(r = \frac{mv}{qB}\), work out a formula for the kinetic energy \(K\) of the proton in terms of \(m\), \(e\), \(B\), and \(r\).

b) Show that equation \(\frac{dK}{dt} = P\) results in the following differential equation for \(r\) as a function of \(t\): 
\[
r \frac{dr}{dt} = \frac{Pm}{e^2B^2}.
\]
Make sure to show all the steps that lead to this equation.

c) This differential equation can be solved by separating the variables. For that, rewrite it as 
\[
r \, dr = \frac{Pm}{e^2B^2} \, dt
\]
and integrate both sides. The expression on the left side should be integrated from \(r = 0\) (center) to \(r(t)\) (an intermediate radius at the moment of time \(t\)), while the expression on the right side should be integrated from \(0\) to \(t\). From here, obtain the expression for \(r(t)\) in terms of \(m\), \(e\), \(B\), \(P\), and \(t\).
Transcribed Image Text:**Problem 3:** A proton in a cyclotron gains \(\Delta K = 2e \Delta V\) of kinetic energy per revolution, where \(\Delta V\) is the potential between the dees. Although the energy gain comes in small pulses, the proton makes so many revolutions that it is reasonable to model the energy as increasing at the constant rate \(P = \frac{dK}{dt} = \frac{\Delta K}{T}\), where \(T\) is the period of the cyclotron motion. This is power input because it is a rate of increase of energy. Find an expression for \(r(t)\), the radius of a proton’s orbit in a cyclotron, in terms of \(m\), \(e\), \(B\), \(P\), and \(t\). Assume that \(r = 0\) at \(t = 0\). a) Using the formula for the radius of the cyclotron orbit, \(r = \frac{mv}{qB}\), work out a formula for the kinetic energy \(K\) of the proton in terms of \(m\), \(e\), \(B\), and \(r\). b) Show that equation \(\frac{dK}{dt} = P\) results in the following differential equation for \(r\) as a function of \(t\): \[ r \frac{dr}{dt} = \frac{Pm}{e^2B^2}. \] Make sure to show all the steps that lead to this equation. c) This differential equation can be solved by separating the variables. For that, rewrite it as \[ r \, dr = \frac{Pm}{e^2B^2} \, dt \] and integrate both sides. The expression on the left side should be integrated from \(r = 0\) (center) to \(r(t)\) (an intermediate radius at the moment of time \(t\)), while the expression on the right side should be integrated from \(0\) to \(t\). From here, obtain the expression for \(r(t)\) in terms of \(m\), \(e\), \(B\), \(P\), and \(t\).
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