This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: PHY2049 Spring 2010 Profs. P. Avery, A. Rinzler, S. Hershfield Exam 3 Solution 1. An AC power supply drives a 270 mH inductor. The power supply EMF is E = E m sin d t , where E m = 12 V and d = 120 rad/s. When the instantaneous current in the circuit is a maximum, the instantaneous emf of the power supply is Answer: 0 V Solution: The current through and voltage across an inductor are 90 out of phase. Thus, when the current is a maximum, the voltage is zero. 2. Oxygen: (A) has an intrinsic magnetic dipole moment, (B) is paramagnetic, (C) is attracted to the high field point of a non-uniform magnetic field, (D) is diamagnetic. Of these statements only the following are true, Answer: A, B & C Solution: As shown in the demonstration performed in class, oxygen is attracted to the high field point because it has an intrinsic dipole moment and is paramagnetic. 3. In the figure the instantaneous current charging the capacitor is i = 1 . 2 A. The circular capacitor plates have a radius of 20 cm. The magnitude of the magnetic field at the point labeled p that lies between the plates, 15 cm from the centerline between the plates is Answer: 9 . 10 7 T Solution: For this problem use contintegraltext vector B dvectors = 2 rB = o i d , where i d is the displacement current through a circle of radius r = 0 . 15 m . This is not the total displacement current between the disks, 1 . 2 A, because the circle of radius r = 0 . 15 m does not enclose all of the electric flux. It encloses a fraction of the electric flux: ( . 15 2 ) / ( . 2 2 ) = (0 . 75) 2 . Solve for B using 2 (0 . 15 m ) B = o (0 . 75) 2 (1 . 2 A ). 4. The magnetic flux entering the circular bottom face of a right circular cylinder is 23 . 4 T m 2 and the magnetic flux exiting the sidewall of the cylinder is 63 . 9 T m 2 . The magnetic flux on the top face must be Answer: 40 . 5 T m 2 , entering Solution: The net magnetic flux through the cylinder must be zero. The flux from two of the surfaces is 63 . 9 23 . 4 = 40 . 5 T m 2 exiting the cylinder. Thus, the remaining surface must have 40 . 5 T m 2 entering the cylinder. 5. The circular region of radius r = 50 cm shown in the figure contains a uniform electric field pointing into the page that is increasing according to E ( t ) = 2 . 4 10 5 t (in units of V/m), where t is in seconds. For the path indicated by the dotted line what is the magnitude of contintegraltext vector B dvectors , in units of T m?...
View Full Document