Exam2_solutions_Fall11

# Exam2_solutions_Fall11 - PHY2049 Fall11 Exam 2 ...

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Unformatted text preview: PHY2049 Fall11 Exam 2 Exam 2 Solutions 1. A horizontal power line carries a current of 2900A from south to north. Earth's magnetic field, with a magnitude of 60μT, is directed toward the north with a dip angle 60° downward into the Earth relative to the horizontal. Find the magnitude and direction (use compass directions) of the magnetic force acting on a 100 m length of power line. (1) 15 N, West (2) 15 N, East (3) 17.5 N, West (4) 8.7 N, East (5) 17.5 N, East The magnetic field and the power line both point north, but the magnetic field points into the Earth at an angle of 60°. i S N 60° B The force on a current carrying wire is given by: F = iL × B So the force points west with magnitude: F = iLB sin θ = ( 2900 ) (100 ) 60 × 10−6 ( sin 60 ) = 15 N ( Answer: 15 N ) y − − − − − − E x e- + + + + + + + PHY2049 Fall11 Exam 2 2. A beam of electrons (“cathode rays”) is sent between two parallel electric plates with an electric field between them of 2x10^4 N/C j^ s . If the electron beam travels perpendicular to the electric field with a velocity of 4.2x 10^7 m/s in the +i direction, what magnetic field is necessary (direction and magnitude) so that the electrons continue traveling in a straight line without deflection by the electric field? Balance forces: F = q(E + v × B ) = 0 ⇒E+ v×B = 0 Eˆ ⇒ B = k so that ˆ × k = − ˆ jˆ i v E = 2 × 104 N/C v = 4.2 × 107 m/s ⇒B= 2 × 104 N/C = 0.48 mT 4.2 × 107 3. The magnetic field of a solenoid (long compared to its radius) is used to keep a proton in a perfectly circular orbit. The solenoid has 1000 windings per meter of length and has a radius of 1 m. If the proton has a velocity magnitude of v = 1.5 ×106 m/s , what is the minimum current needed to keep the proton orbiting within the confines of the solenoid in a plane perpendicular to the solenoid axis? The proton mass is mp = 1.67 ×10−27 kg and its charge is q = +1.6 × 10−19 C . r p (1) 12.5 A (2) 78,000 A (3) 25,000 A (4) 12,500 A (5) 1.6x10- 7 A B = µ0 ni mv = qBr = qr µ0 ni ( )( )( ) 1.67 × 10−27 kg 1.5 × 106 m/s mv ⇒i= = = 12.5 A qr µ0 n 1.6 × 10-19 C (1m ) 4π 10−7 (1000 ) ( ) PHY2049 Fall11 Exam 2 4. A series RLC circuit is driven by sinusoidally- varying EMF source with a maximum amplitude of 125V. The resistance R=100 Ohm, the inductance L = 2x10^- 3 H, and the capacitance C = 0.1 uF. At what frequency (cycles/sec) will the amplitude of the current be a maximum? (1) 11 kHz (2) 11 Hz (3) 7.1 kHz (4) 2x10- 5 Hz (5) 1x10- 5 Hz εm im = 2 R 2 + ( X L − XC ) The amplitude of the current is a maximum when X L = XC ⇒ ω = 2π f = f= 1 2π LC 1 LC = 11.3 kHz 5. An alternating EMF source drives a series RLC circuit with a maximum amplitude 9.0V. The phase angle of the current is +45º. When the potential difference across the capacitor reaches its maximum positive value of +6V, what is the potential difference across the inductor (including sign)? (1) - 12: 4 V (2) - 8 V (3) 6.4 V (4) - 0.4 V (5) 0 V ε = ε d sin ω t i = im sin (ω t − φ ) im sin (ω t − φ − 90 ) ωC This is a maximum when ω t − φ − 90 = 90 , ω t = φ + 180 So the potential across the resistor is: i i ΔVR = m sin (ω t − φ ) = m sin (180 ) = 0 R R Now apply Kirchoff’s loop rule to solve for the potential across the inductor: ε − ΔVC − ΔVL − ΔVR = 0 ΔVC = ΔVL = ε ( t ) − ΔVC = ε d sin (180 + φ ) − ΔVC = 9 ( −0.707 ) − 6 = −12.4V PHY2049 Fall11 Exam 2 6. A square metal loop of side length l=0.5m is pulled out of a uniform magnetic field with a velocity of v=2 m/s. The magnetic field has a strength of B=0.25 T and is directed perpendicular to the surface of the loop. One side of the square is aligned with the edge of the field region when the pulling first starts. What is the magnitude of the induced EMF in the loop as it is pulled? B v (1) 0.25 V (2) 5 V (3) 0.125 V (4) 0.0625 V (5) 1 V The induced EMF is given by Faraday’s Law: dΦ B d d ε= = ( BA ) = ⎡ BL ( L − vt ) ⎤ ⎦ dt dt dt ⎣ ε = BLv = 0.25V 7. An inductor of inductance L=10 H and resistance R = 2 Ohm is plugged into a DC source of EMF at t=0. How long does it take for the current through the inductor to reach 80% of its maximum? (1) 8.0 s (2) 5.0 s (3) 1.1 s (4) 10 s (5) 2.0 s The solution to Kirchoff’s loop rule for LR circuits gives the current i(t): ε L i ( t ) = 1 − e− t /τ L τ L = = 5s R R εε 0.8 = 1 − e− t /τ L RR − t /τ L e = 1 − 0.8 ( ) ( t = − ln 0.2 τL t = −τ L ln 0.2 = 8 s ) PHY2049 Fall11 Exam 2 8. The current through an inductor with inductance L=0.1 H is shown by the graph, with the direction from left to right through the inductor as shown. What is the EMF across the inductor (VL= Vright - Vleft), including sign, at t = 1 ms? (1) –350 V (2) 0.35 V (3) –7 V (4) 3500 V (5) –35 V di Δi VL = − L = − L dt Δt 7A = − ( 0.1 H ) = −350V 2 × 10 −3 s 9. A parallel plate capacitor has circular plates with a radius R=2cm and a time- dependent electric field between them of (3 x 106 V/m- s) t . What is the magnitude ...
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