W15D1: Poynting Vector and Energy Flow. Today ’ s Readings: Course Notes: Sections 13.6, 13.12.3-13.12.4 . Final Math Review Week 15 Tues from 9-11 pm in 32-082 Final Exam Monday Morning May 20 from 9 am-12 noon Johnson Athletic Center Track 2 nd floor . Announcements. Outline.
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Course Notes: Sections 13.6, 13.12.3-13.12.4
Poynting Vector and Energy Flow
Power per unit area:
Power through a surface
Consider the above cylindrical resistor with resistance R and current Iflowing as shown.
On surface of resistor direction is INWARD
What is the magnetic field?
So we had to modify Ampere’s Law:
Integration direction counter clockwise for line integral requires that unit normal points out page for surface integral.
Current positive out of page. Negative into page.
Electric flux positive out of page, negative into page.
B field clockwise, line integral negative.
B field counterclockwise, line integral positive.
The figures above show a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the charge Q is:
Answer: 3. The charge Q is decreasing in time
Use the Ampere-Maxwell Law. Choose positive unit normal out of plane. Because the magnetic field points clockwise line integral is negative hence positive electric flux (out of the plane of the figure on the right) must be decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q.
The figures above show a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the energy stored in the electric field is:
Answer: 1. The the energy stored in the electric field is increasing in time
The direction of the Poynting Flux S (= E x B) inside the capacitor is inward. Therefore electromagnetic energy is flowing inward, and the energy in the electric field inside is increasing.
A circular capacitor of spacing d and radius R is in a circuit carrying the steady current i shown. Ignore edge effects. At time t = 0 it is uncharged
The figures above show a side and top view of a solenoid carrying current I with electric and magnetic fields E and B at time t. The current I is
Answer: 3. The current I is decreasing in time
Use Faraday’s law. Choose positive unit normal out of plane. Because the electric field points counterclockwise line integral is positive, therefore the positive magnetic flux must be decreasing (out of the plane of the figure on the right). Hence B is decreasing. Thus I must be decreasing, since B is proportional to I.
The figures above show a side and top view of a solenoid carrying current I with electric and magnetic fields E and B at time t. The energy stored in the magnetic field is
Answer: 3. The energy stored in the magnetic field is decreasing in time.
The Poynting Flux S (= E x B) inside the solenoid is directed outward from the center of the solenoid. Therefore EM energy is flowing outward, and the energy stored in the magnetic field inside is decreasing.
A solenoid of radius a and length h has an increasing current I(t) as pictured.
Consider a point D inside the solenoid at radius r (r < a). Ignore edge effects.
Direction on surface of inductor with increasing current is INWARD
Direction on surface of inductor with decreasing current is OUTWARD
What is the energy flow per unit area?
Direction of energy flow = direction of wave propagation
units: Joules per square meter per sec
EM waves transport energy:
They also transport momentum:
They exert a pressure:
This is only for hitting an absorbing surface. For hitting a perfectly reflecting surface the values are doubled:
A light bulb puts out 100 W of electromagnetic radiation. What is the time-average intensity of radiation from this light bulb at a distance of one meter from the bulb? What are the maximum values of electric and magnetic fields, E and B, at this same distance from the bulb? What is the pressure this radiation will exert on a very small perfectly conducting plate at 1 meter. For simplicity, you may assume the radiation is a plane wave of wavelength λ.