Question #153213

2-1. Compute rL for the following cases if v0 is negligible: 

(a) A 10-keV electron in the earth's magnetic field of 5 x 10-5 T. 

(b)· A solar wind proton with streaming velocity 300 km/sec, B = 5 x 10-9 T. 

(c) A 1-keV He+ ion in the solar atmosphere near a sunspot, where B = 

5 X 10-2 T. 

(d) A 3.5-MeV He++ ash particle in an 8-T DT fusion reactor.


1
Expert's answer
2020-12-31T08:06:44-0500

a)


rL=9.11031(1.61019)510510103(1.61019)2(9.11031)=3.37 mr_L=\frac{9.1\cdot10^{-31}}{(1.6\cdot10^{-19})5\cdot10^{-5}}\sqrt{\frac{10\cdot10^3(1.6\cdot10^{-19})}{2(9.1\cdot10^{-31})}}=3.37\ m

b)


rL=1.671027(300000)(1.61019)5109=626250 mr_L=\frac{1.67\cdot10^{-27}(300000)}{(1.6\cdot10^{-19})5\cdot10^{-9}}=626250\ m

c)


rL=(4)(1.671027)(1.61019)51021103(1.61019)2(4)(1.671027)=0.091 mr_L=\frac{(4)(1.67\cdot10^{-27})}{(1.6\cdot10^{-19})5\cdot10^{-2}}\sqrt{\frac{1\cdot10^3(1.6\cdot10^{-19})}{2(4)(1.67\cdot10^{-27})}}=0.091\ m

d)


rL=(4)(1.671027)2(1.61019)83.5106(1.61019)2(4)(1.671027)=0.017 mr_L=\frac{(4)(1.67\cdot10^{-27})}{2(1.6\cdot10^{-19})8}\sqrt{\frac{3.5\cdot10^6(1.6\cdot10^{-19})}{2(4)(1.67\cdot10^{-27})}}=0.017\ m


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS