hw05 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 5...

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 5. 26.08, 26.32, 26.39, 26.43; Q27.02, Q27.07 26.08. a) J = n p ev p = (8.70 cm -3 )(10 6 cm 3 /m 3 )(1.60 ×10 -19 C)(470 ×10 3 m/s) J = 6.54 ×10 -7 A/m 2 b) i = JA , where A is the area that intercepts the protons. Since Earth looks like a disk from the Sun, A = π R 2 , where R is the radius of Earth. i = J R 2 = (6.54 ×10 -7 A/m 2 ) (6.37 ×10 6 m) 2 i = 8.34 ×10 7 A 26.32. J = ( ne ) v d , so ne J v d = . J 1 , the current density in section 1 is needed. Information is given to calculate J 2 , the current density in section 2. Then, since the current in section1 must equal the current in section 2, J 1 A 1 = J 2 A 2 , or J 1 ( R 1 2 ) = J 2 ( R 1 2 ), so 2 2 2 2 2 1 1 2 2 1 2 1 4 1 2 1 2 / J J J R R J R R J = = = = . To get J 2 use, m 10 69 . 1 m) 00 . 2 /( V) 10 0 . 10 ( / 8 6 2 × × = = = - - ρ L V E J = 296 A/m
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This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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hw05 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 5...

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