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 2011 Profs. P. Avery, S. Hershfield Exam 3 Solution 1. The current i in a long wire is going up as shown in the figure, but decreasing in magnitude. What is the direction of the induced current in the left loop and the right loop. (List the direction of the induced current in the left loop first.) Answer: counterclockwise, clockwise Solution: Because the current in the wire is going up, the magnetic field is coming out of page on the left side of the wire and into the page on the right side. The magnitude of the flux is decreasing because the current is decreasing. Con- sequently, via Lenz’s law the induced current acts to produce a flux coming out of the page on the left (counterclockwise) and into the page on the right (clockwise). 2. A metal rod is forced to move with constant velocity along two parallel metal rails, which are connected with a strip of metal (see figure). A magnetic field of 0.5 T points out of the page. The separation between the rails is L = 30 cm, the rod has resistance 40Ω, and velocity 70 cm/s . What force must be applied to the rod to keep it moving at the constant velocity? Answer: 4 × 10 − 4 N Solution: The magnitude of the induced emf is d Φ B /dt = BdA/dt = BvL , and consequently the induced current is i = Bvl/R . The magnetic force on the rod is iLB . This is the force which must be applied to keep the rod moving at a constant velocity. 3. The magnetic field is 2 T into the page and 1 T out of the page in regions 1 and 2 respectively (see figure). The magnitude of both fields is decreasing at a rate of- . 005 T/s . If region 1 has area 0.3 m 2 and region 2 has area 0.6 m 2 , what is contintegraltext vector E · dvectors for the path of integration shown in the figure? Answer: 1 . 5 × 10 − 3 V Solution: From the direction of the integral indicated in the figure positive flux is coming out of the page. Thus, the flux is negative in region 1 and positive in region 2; however, the flux is decreasing in magnitude so d Φ B /dt is positive in region 1 and negative in region 2. Finally, the integral contintegraltext vector E · dvectors is equal to- d Φ B /dt so the contribution from region 1 is negative and the one from region 2 is positive.- d Φ B /dt = 0 . 005(0 . 6- . 3) Volts....
View Full Document