solutions_chapter21

# P d b d t p 5 ri na 5 1 40.0 v 21 0.150 a 2 1 200 2 p

This preview shows pages 2–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P D B D t P 5 RI NA 5 1 40.0 V 21 0.150 A 2 1 200 2 p 1 3.00 3 10 2 2 m 2 2 5 10.6 T / s. I 5 E R 5 NA D B / D t R . E 5 N P DF B D t P 5 NA P D B D t P . f 5 0°, F B 5 BA cos f . E 5 N P DF B D t P . I 5 E R 5 0.0171 V 0.600 V 5 0.0285 A. E 5 A D B D t 5 1 0.0900 m 2 21 0.190 T / s 2 5 0.0171 V. f 5 0°, F B 5 BA cos f . E 5 P DF B D t P . D 2 5 D / " 2 . D 1 5 D . D 2 2 5 1 2 D 1 2 , A 5 1 2 p D 2 A 2 5 A 1 / 2. F B 5 BA cos f . 21-2 Chapter 21 21.10. Set Up: At let the plane of the coil be perpendicular to the magnetic field, so at Then where is the angular velocity of the coil. B Since the field is uniform, A sketch of is given in Figure 21.10a. Figure 21.10 Solve: (a) The instantaneous value of the induced emf equals the rate of change of the flux and this is the slope of the curve in Figure 21.10a. From point A to point B this slope is negative and increasing in magnitude and from B to C it is negative and decreasing in magnitude. From point C to D this slope is positive and increasing and from D to E it is positive and decreasing. After point E the behavior of repeats. Therefore, the induced emf increases and decreases periodically and reverses sign. (b) The induced emf versus time is sketched in Figure 21.10b. 21.11. Set Up: The maximum emf is produced when the opposite sides of the coil are moving perpendicular to the field with speed The maximum emf induced in each of the two sides on one turn is The maximum total emf is Solve: (a) (b) The emf changes polarity as the coil rotates and the average emf is zero. Reflect: The maximum emf is proportional to the angular velocity of the coil. As increases, the flux through the coil changes more rapidly. 21.12. Set Up: The field of the induced current is directed to oppose the change in flux in the primary circuit. Solve: (a) The magnetic field in A is to the left and is increasing. The flux is increasing so the field due to the induced current in B is to the right. To produce magnetic field to the right, the induced current flows through R from right to left. (b) The magnetic field in A is to the right and is decreasing. The flux is decreasing so the field due to the induced cur- rent in B is to the right. To produce magnetic field to the right the induced current flows through R from right to left. (c) The magnetic field in A is to the right and is increasing. The flux is increasing so the field due to the induced cur- rent in B is to the left. To produce magnetic field to the left the induced current flows through R from left to right. (d) The magnetic field of A is constant so the flux through B is constant and there is no induced emf and no induced current. 21.13. Set Up: The field of the induced current is directed to oppose the change in flux....
View Full Document

{[ snackBarMessage ]}

### Page2 / 17

P D B D t P 5 RI NA 5 1 40.0 V 21 0.150 A 2 1 200 2 p 1...

This preview shows document pages 2 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online