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W15D1: Poynting Vector and Energy FlowPowerPoint Presentation

W15D1: Poynting Vector and Energy Flow

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W15D1: Poynting Vector and Energy Flow

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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 2nd floor

Announcements

Poynting Vector and Energy Flow

Examples

- Conservation of charge:
- E and B fields exert forces on (moving) electric charges:
- Energy stored in electric and magnetic fields

Energy Flow

Power per unit area:

Poynting vector

Power through a surface

a

I

Consider the above cylindrical resistor with resistance R and current Iflowing as shown.

- What are the electric and magnetic fields on the surface of the resistor?
- Calculate the flux of the Poynting vector through the surface of the resistor (power) in terms of the electric and magnetic fields.
- Express your answer to part b) in terms of the current I and resistance R. Does your answer make sense?

L

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.

Concept Questions:Poynting Vector

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:

- Increasing in time
- Constant in time.
- Decreasing in time.
- Not enough information given to determine how Q is changing.

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:

- Increasing in
- Constant in time.
- Decreasing in time.

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

- Find the electric field E(t) at A (mag. & dir.)
- Find the magnetic field B(t) at A (mag. & dir.)
- Find the Poynting vector S(t) at A (mag. & dir.)
- What is the flux of the Poynting vector into/out of the capacitor?
- How does this compare to the time derivative of the energy stored in the electric field?
- Does this make sense?

Another look at Inductance

L

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 current I is

- increasing in time.
- constant in time.
- decreasing in time.

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

- Increasing in time
- Constant in time.
- Decreasing in time.

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.

- Find the magnetic field B(t) at D (dir. and mag.)
- Find the electric field E(t) at D (dir. and mag.)
- Find the Poynting vector S(t) at D (dir. and mag.)
- What is the flux of the Poynting vector into/out of the inductor?
- How does this compare to the time derivative of the energy stored in the magnetic field?
- Does this make sense to you?

Direction on surface of inductor with increasing current is INWARD

Direction on surface of inductor with decreasing current is OUTWARD

Dissipates

Power

Store

Energy

POWER

When

(dis)charging

Poynting Vector and EM Waves

Energy densities:

Consider cylinder:

What is the energy flow per unit area?

Direction of energy flow = direction of wave propagation

units: Joules per square meter per sec

Intensity I:

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:

Group Problem: Radiation

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.