phys1002_sln04_sm2_2007

# phys1002_sln04_sm2_2007 - TUTORIAL 4 SOLUTIONS Semester 2...

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Unformatted text preview: TUTORIAL 4 SOLUTIONS Semester 2, 2007 1. MOTION OF POINT CHARGES IN A MAGNETIC FIELD A particle with mass m and charge q will move in a circular orbit with radius R = mv/qB . If we consider two particles, call them 1 and 2, we find the ratios of their radii to be R 1 R 2 = m 1 v 1 q 2 m 2 v 2 q 1 . (1) If the particles have the same kinetic energy E = mv 2 / 2 we have m 1 v 2 1 = m 2 v 2 2 (2) which implies v 1 v 2 = r m 2 m 1 . (3) This gives R 1 R 2 = r m 1 m 2 q 2 q 1 . (4) Hence R d R p = r m d m p q p q d = √ 2 · 1 = √ 2 , (5) R α R p = r m α m p q p q α = √ 4 · (1 / 2) = 1 . (6) 2. THE CYCLOTRON. (a) The cyclotron frequency is given as f = 1 T = qB 2 πm so that for this problem we have: f = 1 . 6 × 10- 19 × 1 . 4 2 π × 1 . 67 × 10- 27 s- 1 = 2 . 13 × 10 7 s- 1 = 21 . 3 MHz . (b) The maximum energy of the protons when they emerge is their total kinetic energy which is given as K = 1 2 mv 2 = q 2 B 2 r 2 2 m , so that we have: K = (1 . 6 × 10- 19 ) 2 × 1 . 4 2 × . 7 2 2 × 1 . 67 × 10- 27 J = 7 . 36 × 10- 12 J = 46 MeV . (c) From the above equations we see that both the frequency and the Kinetic energy are inversely proportional to the mass, so if deuterons are used, both quantities will be reduced by a factor of 2. 3. TORQUE ON A CURRENT LOOP. (a) The magnetic moment of the coil is given by ~μ = NIA ˆ n where N is the number of turns, I is the current, and A = πr 2 is the area enclosed by the circular coil. The magnitude of the magnetic moment is therefore μ = NIA = 50 · 2 A ·...
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phys1002_sln04_sm2_2007 - TUTORIAL 4 SOLUTIONS Semester 2...

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