Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Electrostatics wikipedia , lookup
Work (physics) wikipedia , lookup
Neutron magnetic moment wikipedia , lookup
Electrical resistance and conductance wikipedia , lookup
Field (physics) wikipedia , lookup
Magnetic field wikipedia , lookup
Magnetic monopole wikipedia , lookup
Electromagnetism wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Superconductivity wikipedia , lookup
Version A Physics 122 Midterm Examination #2 March 22, 2006 Name: Recitation Section: Lab Section: Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Total 1. A conducting bar slides on rails separated by a distance l=1.5 m; the rails are electrically connected with a resistance R=0.1 . There is a homogeneous magnetic field B=1 T pointing into the ground (vertical into this page). The bar moves with constant velocity v = 2 m/s to the left. × × × × × × × × × × × Fapp × × × × × × × × × × a) Calculate the change of the magnet flux per unit time for the loop connecting resistance R and the sliding bar. b) Find the induced potential difference (also referred to as emf) across the bar. c) What is the direction (clockwise or counter clockwise) and magnitude of the current induced? d) Find the magnitude of the constant external force pulling the bar Fapp. Solution: a) the change of magnetic flux with time is proportional to change of area (negative) B m Tm 2 Bvl 1T 1.5m 2 3 and magnetic field, t s s b) According to Faraday’s law the induced emf is equal to to minus change of flux emf 3V c) Lenz’s law requires that induction opposes change, since change is due to decreasing flux the induced magnetic field must point into plane and consequently the current flows clockwise. d) Force on a rode of length l caring a current I in magnetic field emf 3V is: F I lB lB 1.5m 1T 45 N R 0.1 2. Consider the following AC circuit which is powered by a household outlet with RMS voltage Vrms=120 V and frequency f=60 Hz. The circuit includes a resistor with resistance R=200, an inductor with self inductance L=200mH, and a capacitor with capacity C=20F. a) Draw a phasor diagram for the circuit. b) Calculate the impedance of the circuit. c) Calculate the phase angle between current and voltage in the circuit. d) What is the power delivered by the AC source to this circuit? e) What is the RMS voltage across the capacitor? Solution: a) need to calculate impedanze for Inductor and Capacitor: X L L 2 fL 2 60 Hz 0.2 H 75.4 XC 1 1 1 132.6 C 2 fC 2 60 Hz 2 105 F b) total impedance given by: Z R2 ( X L X C )2 208 X XC c) phase angle: arctan L R o 17.7 d) need to consider phase angle in AC circuit: V 2 Pav I rmsVrms cos rms cos 66W Z e) VCrms X C I rms 76.5V f) Two parallel wires carry current of 1 A and 2 A, respectively. The currents are flowing in opposite direction. If the wires are separated by 10 cm find: 1.00 A 2.00 A d=10cm a) The location in horizontal direction where the magnetic 2.00 A field is zero. b) The direction of the magnetic field at P1 and P2. c) The magnitude and direction of the force on a 10 cm long segment of the left wire due to the magnetic field created by the right wire. d) The magnitude and direction of the force on a 10 cm long segment of the right wire due to the field created by the left wire. Solution: a) at x0 magnetic field will be zero if: B1 ( x0 ) B2 ( x0 ) 0 . Here B1 is the field at a distance x0 from wire with I1=2A and B2 is the field at a distance x0+d from wire with I2=-1A. 0 I1 0 I 2 I2 I 0 1 I 2 x0 I 1 x0 I 1 d 2 x0 2 ( x0 d ) x0 d x0 x0 I1 2A d 10cm 20cm I 2 I1 ( 1 A) 2 A B is zero 20 cm left of wire with 2A. b) To get field at P1 and P2 contribution of both wires must be added (use P1: 0 2 107 2 N A2 ): B( x 10cm ) 0 I 1 I N 2A 1 A 2 2107 2 2 x x d A 0.1m 0.2m B( x 10cm ) 30 107 T Right hand rule gives direction: out of plane P2: B( x 30cm ) 0 I1 I N 2A 1 A 2 2107 2 2 x x d A 0.3m 0.2m B( x 30cm ) 3.3 107 T Right hand rule give direction in to plane c) Calculate force on 10 cm segments of wire with I2=1A due to magnetic field around wire with I1=2A. First calculate field: B(d ) 0 I1 2 d force on wire segment now is: F I 2 lB(d ) 0 l N 10cm I1 I 2 2 107 2 (1 A)2 A 4 107 N 2 d A 10cm according to right hand rule force points to left. d) Opposite direction and equal magnitude. 2. A positive charge of 1 C and a mass of 1 g is accelerated through a potential difference V = 1000V and shot into a magnetic field B=1T pointing vertically into the plane. 1km V Solution c) a) What is the velocity of the charge after it leaves the plates but before it reaches the magnetic field? b) Calculate the radius of curvature of the path the charge will take in the magnetic field. c) Draw the path approximately to scale. Solution: a) Charge accelerated through electrical potential V, convert potential to kinetic energy: PE qV 106 C 1000V 103 J 1 PE KE mv 2 2 v 2 PE m 2 103 Nm m 1.41 3 10 kg s b) Force due to magnetic field provides centripetal force: mv 2 mv FB qvB R 1410 m R qB The primary side of a transformer is powered by an AC source. The coil on the primary side has N1=4 loops while the one on the secondary side has N2=8 loops. AC power source Vrms = 20 V f = 50 Hz R = 100 a) What is the maximum (or peak) voltage and frequency on the secondary side? b) Calculate the power delivered to the resistor on the secondary side. c) Find the current flowing through the loop on the primary side. Solution: a) Frequency unchanged, i.e. 50 Hz. Peak voltage on secondary N side: V2max 2V2 rms 2 2 V1rms 56.6V N1 b) Pav V2 rms 2 16W R c) power on primary and secondary side the same: I1rms Pav 0.8 A V1rms