Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
I= Problem 2: A 5.0-mm-diameter proton beam carries a total current of = 1.5 mA. The current density in the proton beam, which increases with distance from the center, is given by J = Jedge (r/R), where R is the radius of the beam and Jedge is the current density at the edge. Determine the value of Jedge- a) Fig. 3 shows the cross section of the beam. Compute the current dI flowing through the ring of radius r and width dr shown in the figure. Notice that for small dr the area of the ring can be approximated by the area of a rectangle that you can get by "unrolling" the ring.

I= Problem 2: A 5.0-mm-diameter proton beam carries a total current of = 1.5 mA. The current density in the proton beam, which increases with distance from the center, is given by J = Jedge (r/R), where R is the radius of the beam and Jedge is the current density at the edge. Determine the value of Jedge- a) Fig. 3 shows the cross section of the beam. Compute the current dI flowing through the ring of radius r and width dr shown in the figure. Notice that for small dr the area of the ring can be approximated by the area of a rectangle that you can get by "unrolling" the ring.

icon
Related questions
Question

Hello, I need help with PART A,PART AND PART C because I don't understand this problem and I really need help can. you. label which one is which 

**Problem 2:** A 5.0-mm-diameter proton beam carries a total current of \( I = 1.5 \, \text{mA} \). The current density in the proton beam, which increases with distance from the center, is given by \( J = J_{\text{edge}} (r/R) \), where \( R \) is the radius of the beam and \( J_{\text{edge}} \) is the current density at the edge. Determine the value of \( J_{\text{edge}} \). **a)** Fig. 3 shows the cross section of the beam. Compute the current \( dI \) flowing through the ring of radius \( r \) and width \( dr \) shown in the figure. Notice that for small \( dr \) the area of the ring can be approximated by the area of a rectangle that you can get by “unrolling” the ring. **Diagram Explanation:** The figure shows a circle, representing the cross-section of the proton beam. A small ring within the circle has an infinitesimal thickness \( dr \). This ring can be approximated as a rectangle with width \( dr \) and length \( 2\pi r \). **b)** Sum up the contributions from all rings by integrating \( dI \) with respect to the radial coordinate \( r \), \[ I = \int_{r=0}^{r=R} dI \] Express \( J_{\text{edge}} \) as a function of \( I \) and \( R \) and compute its value. **c)** How many protons per second are delivered by this proton beam?
Transcribed Image Text:**Problem 2:** A 5.0-mm-diameter proton beam carries a total current of \( I = 1.5 \, \text{mA} \). The current density in the proton beam, which increases with distance from the center, is given by \( J = J_{\text{edge}} (r/R) \), where \( R \) is the radius of the beam and \( J_{\text{edge}} \) is the current density at the edge. Determine the value of \( J_{\text{edge}} \). **a)** Fig. 3 shows the cross section of the beam. Compute the current \( dI \) flowing through the ring of radius \( r \) and width \( dr \) shown in the figure. Notice that for small \( dr \) the area of the ring can be approximated by the area of a rectangle that you can get by “unrolling” the ring. **Diagram Explanation:** The figure shows a circle, representing the cross-section of the proton beam. A small ring within the circle has an infinitesimal thickness \( dr \). This ring can be approximated as a rectangle with width \( dr \) and length \( 2\pi r \). **b)** Sum up the contributions from all rings by integrating \( dI \) with respect to the radial coordinate \( r \), \[ I = \int_{r=0}^{r=R} dI \] Express \( J_{\text{edge}} \) as a function of \( I \) and \( R \) and compute its value. **c)** How many protons per second are delivered by this proton beam?
Expert Solution
Step 1

Solution:

The diameter of the proton beam is, 

d=5 mm  =5×10-3 mm

The total current of the proton beam is, 

I=1.5 mA =1.5×10-3A

The current density is given by the following,

J=JedgerR                                                                                   ......1Here, Jedge-current density at the edgeR-Radius of the beam

steps

Step by step

Solved in 5 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer