9 Problems on Initial Kinetic Energy in Magnetism - Exam 3 | PHYS 4620, Exams of Physics

Download 9 Problems on Initial Kinetic Energy in Magnetism - Exam 3 | PHYS 4620 and more Exams Physics in PDF only on Docsity! Phys 4620, Spring 2006 Exam #3 1. In the problem set we found the (classical) rate of energy loss of the electron in a hydrogen atom. It was too damn big! In the Phys 2020 magnetic fields lab, we accelerate electrons through a potential difference of about 150 V. Then they are made to go in a circular path by a magnetic field which is perpendicular to the plane of their motion. The radius of the path is about 3.0 cm. a) What is the speed of the electrons as they move on the circular path? Do we need to worry about relativity in studying their motion? If not, why not? The speed of the electrons is the same as what they have after being accelerated through the potential; since their kinetic energy is then T = |q∆V | = e(150 V) = 150 eV = 2.4 × 10−17 J then assuming Newtonian mechanics for which T = 1 2 mv2, the speed of the electrons is v2 = 2T m = 2(2.4 × 10−17 J) (9.11 × 10−31 kg) = 5.26 × 10 13 m2 s2 =⇒ v = 7.26 × 106 m s This speed is significantly less than c (though not < 1%) so if we just want a couple decimal places of accuracy we can ignore relativity. b) What is the magnitude of the acceleration of the electrons? Recalling our happy days in Phys 2110 we calculate ac = v2 r = (7.26 × 106 m s )2 (0.030 m) = 1.76 × 1015 m s2 c) Make the low–velocity approximation made in the text and find the total power radiated by the electrons. We’ll approximate v  c; then we can use the Larmor formula, P ≈ µ0q 2a2 6πc = (4π × 10−7)(1.6 × 10−19)2(1.76 m s2 )2 6π(3.00 × 108 m s ) J s = 1.76 × 10−23 J s d) At this rate of energy loss, how long would it take the electrons to lose 10% of their kinetic energy? Is this number bigger than a 2020 lab period? 1 10% of the initial KE is 2.4 × 10−18 J. With ∆E = Pt, the time to lose this much energy is t = ∆E P = (2.4 × 10−18 J) (1.76 × 10−23 J/s)1.36 × 10 4 s = 38 hr This is much longer than a lab period! The electrons in the 2020 lab don’t lose much energy due to radiation. 2. In the lab reference frame, particle A moves in the +x direction with speed 1 2 c and particle B moves in the +x direction with speed 3 4 c. What is the speed of particle B in the reference frame of particle A? In the frame of A, the lab moves at vx = −c/2. And B moves at +34c with respect to the lab. The do a relativistic addition of these velocities and get: vBA = − c 2 + 3 4 c 1 + (−1/2)(3/4)c 2 c2 = c/4 1 − 3/8 = 8 5 c 4 = 2 5 c 3. What is the (relativistic) kinetic energy of a proton which has a speed of 0.9c ? (Use mpc 2 = 938.27 MeV.) Energy of the proton is E = mpc 2 √ 1 − u2/c2 = mpc 2 √ 1 − ( 0.9c c )2 = 2.29mpc 2 This gives E = 2.15 × 103 MeV = mpc2 + T so then T = 2153 MeV− 938.27 MeV = 1214 MeV 4. A proton with a kinetic energy of 1.0 GeV is incident on a stationary proton (in the lab frame). a) Find the momentum of the incident proton in the lab frame. (Use mpc 2 = 938.27 MeV.) In the lab frame, the energy of the moving proton is Ep = mpc 2 + 1.0 GeV = 1938 MeV then with E2p = p 2c2 + m2c4 we get p2c2 = (1938 MeV2 − (938 MeV)2 = 2.87 × 106 MeV2 so pc = 1696 MeV or p = 1696 MeV/c 2 which is impossible! 6. What does it mean to say that a mathematical object is a 4-tensor? Specifically, what property does the object tµν have to have? If tµν is a 4-tensor then the corresponding quantity in the reference frame S is given by t̄µν = ΛµσΛ ν λt σλ where tσλ is the (set of) values of t in frame S and Λµσ is the Lorentz transformation matrix relating frames S and S . 7. If we let µ = 3 in the relativistic “inhomogeneous” Maxwell equation, we have ∂F 3ν ∂xν = µ0J 3 Show how this is the same as one of the Maxwell equations written in our old vector notion. Recall that xν = (ct, x, y, z) and J3 = Jz and using the given entries for F 3ν (3rd row) we get: ∂F 3ν ∂xν = ∂(−Ez/c) ∂(ct) + ∂(By) ∂x + ∂(−Bx) ∂y + 0 = µ0Jz or ( ∂By ∂x − ∂Bx ∂y ) − 1 c2 ∂Ez ∂t = µ0Jz This is the zth component of the vector equation ∇× B = µ0J + 1 c2 ∂E ∂t , the Ampere--Maxwell equation. 8. If we let µ = 0 and ν = 1 in the definition of the µν element of the EM field tensor we get F 01 = ∂A1 ∂x0 − ∂A 0 ∂x1 Show that this agrees with the relations we previously had between the fields and the po- tentials. Use A1 = Ax A 0 = V/c x1 = x 1 = x x0 = −x0 = −ct F 01 = Ex/c then get Ex c = ∂Ax ∂(−ct) − ∂(V/c) ∂x 5 or Ex = − ∂V ∂x − ∂Ax ∂t This is the x component of E = −∇V − ∂A ∂t 6 Useful Equations ∫ b a (∇T ) · dl = T (b) − T (a) ∫ V (∇ · v)dτ = ∮ S v · da ∫ S (∇× v) · da = ∮ P v · dl Spherical: dl = dlr r̂ + dlθ θ̂ + dlφ φ̂ dτ = r 2 sin θ dr dθ dφ Gradient: ∇T = ∂T ∂r r̂ + 1 r ∂T ∂θ θ̂ + 1 r sin θ ∂T ∂φ φ̂ (1) Divergence: ∇ · v = 1 r2 ∂ ∂r (r2vr) + 1 r sin θ ∂ ∂θ (sin θvθ) + 1 r sin θ ∂vφ ∂φ (2) Curl: ∇×v = 1 r sin θ [ ∂ ∂θ (sin θ vφ) − ∂vθ ∂φ ] r̂+ 1 r [ 1 sin θ ∂vr ∂φ − ∂ ∂r (rvφ) ] θ̂+ 1 r [ ∂ ∂r (rvθ) − ∂vr ∂θ ] φ̂ (3) Laplacian: ∇2T = 1 r2 ∂ ∂r ( r2 ∂T ∂r ) + 1 r2 sin θ ∂ ∂θ ( sin θ ∂T ∂θ ) + 1 r2 sin2 θ ∂2T ∂φ2 (4) Cylindrical: dl = dls ŝ + dlφ φ̂ + dlz ẑ dτ = s ds dφ dz Gradient: ∇T = ∂T ∂s ŝ + 1 s ∂T ∂φ φ̂ + ∂T ∂z ẑ (5) Divergence: ∇ · v = 1 s ∂ ∂s (svs) + 1 s ∂vφ ∂φ + ∂vz ∂z (6) Curl: ∇× v = ( 1 s ∂vz ∂φ − ∂vφ ∂z ) ŝ + ( ∂vs ∂z − ∂vz ∂s ) φ̂ + 1 s [ ∂ ∂s (svφ) − ∂vs ∂φ ] ẑ (7) Laplacian: ∇2T = 1 s ∂ ∂s ( s ∂T ∂s ) + 1 s2 ∂2T ∂φ2 + ∂2T ∂z2 (8) 7