This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Faradays Law and Inductance Problem 1 (medium): A solenoid has length l , radius a 1 , and N 1 coil turns. A time dependent current flows through this solenoid in the counterclockwise direction, and is given by I 1 ( t ) = c + ft where c > 0 and f > 0. 1: Located within this solenoid, and concentric with it, is a circular ring of radius a 2 < a 1 and resistance R , where the plane of the ring is perpendicular to the axis of the solenoid. What are the magnitude and direction of the current induced in the ring? How does your answer change if f < 0 in the above expression for the current in the solenoid? 2: Suppose that instead of a ring, we have a second solenoid of radius a 2 < a 1 , length l , and N 2 coil turns. The second solenoid is concentric with, and entirely contained within, the first solenoid. What is the mutual inductance between the two solenoids? 3: Suppose the first (outer) solenoid has counterclockwise current I 1 , and the second (inner) solenoid has counterclockwise current I 2 . What total magnetic energy is con tained within the two solenoids? & & & & & & & & & Solution: We first write the magnetic field due to the solenoid at a distance r from its central axis: 1 B 1 ( r < a 1 ) = N 1 I 1 l B 1 ( r > a 1 ) = 0 Everywhere inside the solenoid, the magnetic field is uniform, and because I 1 is coun terclockwise, this field is directed out of the page. Everywhere outside the solenoid, the magnetic field is zero. The ring of radius a 2 < a 1 is contained within the solenoid. The magnetic flux through the ring is therefore B = B 1 ( r < a 1 ) a 2 2 = N 1 I 1 a 2 2 l The emf induced in the ring is due to the change in magnetic flux through the ring. E = d B dt = N 1 a 2 2 l dI 1 dt = N 1 a 2 2 l d dt ( c + ft ) = N 1 fa 2 2 l Let I r denote the induced current in the ring. Because the ring has resistance R , we have E = I r R . Therefore, the induced current in the ring is I r = E R = N 1 fa 2 2 Rl The  sign means that the induced current flows clockwise around the ring. This is assuming that f > 0, so that the current in the solenoid is increasing in the counterclock wise direction. If, on the other hand, we have f < 0, then the current in the solenoid is decreasing in the counterclockwise direction, or possibly increasing in the clockwise di rection. In this case, the induced current in the ring has the same magnitude, but flows counterclockwise around the ring. We now turn to part 2 . Let I 1 denote the current through the outer solenoid. The inner solenoid has area a 2 2 and N 2 turns. We wish to compute the flux 21 through the inner solenoid due to the magnetic field of the outer solenoid: 21 = B 1 ( r < a 1 ) a 2 2 = N 1 I 1 a 2 2 l 2 The mutual inductance between the inner solenoid and outer solenoid is therefore M 21 = N 2 21 I 1 = N 1 N 2 a 2 2 l Note that we can also compute the mutual inductance by assuming that a current...
View
Full
Document
This note was uploaded on 01/29/2010 for the course PHYSICS 6B 318036810 taught by Professor Waung during the Spring '09 term at UCLA.
 Spring '09
 WAUNG

Click to edit the document details