Physics 30 Dec 2000 Diploma Exam

# The thumb indicates the direction of the protons

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Unformatted text preview: ing Lenz’s Law, the induced magnetic field will oppose the deflecting magnetic field. Therefore, the deflecting magnetic field between the magnets is upward. • Calculate the radius of the path of a proton travelling at 2.50 × 106 m/s Fm = Fc qvB⊥ = r= mv r T 2 mv qB⊥ r= T 2π r 2π m = qB v mv r = qB (1.67 × 10 −27 kg)(2.50 × 106 m/s) (1.60 × 10− 19 C)(0.863 T) r = 0.0302 m = • Calculate the final speed of a proton after it passes once between the D’s, if its initial velocity entering the space between the D’s is 2.50 × 106 m/s. Method I: Conservation of Energy ∆E = ∆Ek = Vq = vf = = vf 1 2 mvf 2 − 1 2 mvi 2 2(Vq + 1 mvi 2 ) 2 m 2[(20 000 V)(1.60 × 10− 19 C) + 1 2 (1.67 × 10− 27 kg)(2.50 × 106 m/s) 2 ] 1.67 × 10− 27 kg = 3.18 × 106 m/s Note: The incorrect solution Vq = 1 2 mvf 2 gives a speed of 1.96 × 106 m/s Method II: Dynamics and Kinematics & E = Fe = Vq d Fe V = q d & E q = ma = ma a= a= Vq dm (20 000 V)(1.60 × 10−19 C) (5.00 × 10− 2 m)(1.67 × 10−27 kg) a = 3.832 × 1013 m/s2 vf 2 = vi 2 + 2 ad (2.50 × 106 m/s) 2 + 2(3.832 × 1013 m/s 2 )(5.00 × 10− 2 m) vf = vf = 3.18 × 106 m/s • The speed of a particle moving with circular motion and the time it takes the particle to complete one circular orbit are given by the formulas v= 2π R 2π m and T = T qB⊥ Beginning with force equations from the tearout Physics Data Sheet, derive the formula for the period, 2π m T= qB⊥ Method I Fm = qvB⊥ = qB⊥ = T = Method II Fc Fm mv2 r 2π R 1 m T R 2π m qB⊥ Unit of T = = = qvB⊥ = 2π R q B⊥ T T 2π m are equivalent to seconds. qB⊥ • Show that the units of T= = 2π m qB⊥ kg (C)(T ) kg kg ⋅m/s2 (C) (C/s)( m) 1 1 s Unit of T = s = = Fc 4π 2 mR T2 4π 2 mR T2 2π m qB⊥...
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