Concept explainers
(a)
The proof that
(a)
Answer to Problem 20P
It is proved that
Explanation of Solution
Formula used:
The expression for magnetic field is given by,
The expression for angle is given by,
Calculation:
The magnetic field is calculated as,
Conclusion:
Therefore, it is proved that
(b)
The proof that
(b)
Answer to Problem 20P
It is proved that
Explanation of Solution
Formula used:
The expression for electric field is given by,
The expression for flux is given by,
Calculation:
The expression for flux is calculated as,
Conclusion:
Therefore, it is proved that
(c)
The proof that
(c)
Answer to Problem 20P
It is proved that
Explanation of Solution
Formula used:
The expression for flux from (b) is given by,
Calculation:
Integrate equation (1) from
Conclusion:
Therefore, it is proved that
(d)
The proof that
(d)
Answer to Problem 20P
It is proved that
Explanation of Solution
Formula used:
The expression for displacement current is given by,
Calculation:
The summation of current and displacement current is calculated as,
Conclusion:
Therefore, it is proved that
(e)
The proof that result from generalized ampere’s law is similar to part (a).
(e)
Answer to Problem 20P
It is proved that result from generalized ampere’s law is similar to part (a).
Explanation of Solution
Formula used:
The expression for generalized form of ampere’s law is given by,
Calculation:
The magnetic field is calculated as,
Conclusion:
Therefore, it is proved that result from generalized ampere’s law is similar to part (a).
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Chapter 30 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- A circular loop of wire with a radius of 4.0 cm is in a uniform magnetic field of magnitude 0.060 T. The plane of the loop is perpendicular to the direction of the magnetic field. In a time interval of 0.50 s, the magnetic field changes to the opposite direction with a magnitude of 0.040 T. What is the magnitude of the average emf induced in the loop? (a) 0.20 V (b) 0.025 V (c) 5.0 mV (d) 1.0 mV (e) 0.20 mVarrow_forwardTwo long coaxial copper tubes, each of length L, are connected to a battery of voltage V. The inner tube has inner radius o and outer radius b, and the outer tube has inner radius c and outer radius d. The tubes are then disconnected from the battery and rotated in the same direction at angular speed of radians per second about their common axis. Find the magnetic field (a) at a point inside the space enclosed by the inner tube r d. (Hint: Hunk of copper tubes as a capacitor and find the charge density based on the voltage applied, Q=VC, C=20LIn(c/b) .)arrow_forward. A solenoid has radius 5.80 mm, length 11.0 cm, 5000 turns, and is placed with its axis of symmetry along the x-axis, through the origin. A vector normal to the opening of the solenoid points to the right. We measure the resistance of the solenoid to be 14.0 Ω. The solenoid is in a region where the temperature is 49.0°C and initially, there is an external magnetic field of 0.30 T in the +x direction. Then the magnetic field is turned off and drops to 0 T over 50.0 milliseconds a. What is the magnitude of the average induced emf during the 50.0 milliseconds while the magnetic field magnitude decreases to 0? b. What is the direction of the induced current, as viewed from the right? Answer clockwise, counterclockwise, or zero and show work or explain in words. c. What is the magnitude of the induced current? d. What is the magnitude and direction of the induced magnetic field?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning