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# Motional EMF - PowerPoint PPT Presentation

F E = -eE. E. -. v. F B = -evB. Motional EMF. B. -. x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x. -. -. -. -. -. -. eE = evB E = vB V ind = LE = LvB. v. V d t. x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x. L. d.

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## PowerPoint Slideshow about ' Motional EMF' - newton

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Presentation Transcript

FE = -eE

E

-

v

FB = -evB

Motional EMF

B

-

x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

-

-

-

-

-

-

eE = evB

E = vB

Vind = LE = LvB

V dt

x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

L

d

Fi = LdB Ff = L(d+vdt)B

|Vind |= dF/dt

= LvdtB/dt

= LvB

• As the negative charges accumulate at the base, a net positive charge exists at the upper end of the conductor

• As a result of this charge separation, an electric field is produced in the conductor

• Charges build up at the ends of the conductor until the downward magnetic force is balanced by the upward electric force

• There is a potential difference between the upper and lower ends of the conductor

• The potential difference between the ends of the conductor can be found by

• V = B v L

• The upper end is at a higher potential than the lower end

• A potential difference is maintained across the conductor as long as there is motion through the field

• If the motion is reversed, the polarity of the potential difference is also reversed

• Assume the moving bar has zero resistance

• As the bar is pulled to the right with velocity v under the influence of an applied force, F, the free charges experience a magnetic force along the length of the bar

• This force sets up an induced current because the charges are free to move in the closed path

• The changing magnetic flux through the loop and the corresponding induced emf in the bar result from the change in area of the loop

• The induced, motional emf, acts like a battery in the circuit

• As the bar moves to the right, the magnetic flux through the circuit increases with time because the area of the loop increases

• The induced current must in a direction such that it opposes the change in the external magnetic flux

• The flux due to the external field in increasing into the page

• The flux due to the induced current must be out of the page

• Therefore the current must be counterclockwise when the bar moves to the right

• The bar is moving toward the left

• The magnetic flux through the loop is decreasing with time

• The induced current must be clockwise to to produce its own flux into the page

• A bar magnet is moved to the right toward a stationary loop of wire (a)

• As the magnet moves, the magnetic flux increases with time

• The induced current produces a flux to the left, so the current is in the direction shown (b)

• When applying Lenz’ Law, there are two magnetic fields to consider

• The external changing magnetic field that induces the current in the loop

• The magnetic field produced by the current in the loop

• Alternating Current (AC) generator

• Converts mechanical energy to electrical energy

• Consists of a wire loop rotated by some external means

• There are a variety of sources that can supply the energy to rotate the loop

• These may include falling water, heat by burning coal to produce steam

• Basic operation of the generator

• As the loop rotates, the magnetic flux through it changes with time

• This induces an emf and a current in the external circuit

• The ends of the loop are connected to slip rings that rotate with the loop

• Connections to the external circuit are made by stationary brushed in contact with the slip rings

• The emf generated by the rotating loop can be found by

V =2 B ℓ v=2 Bℓ v sin θ

• If the loop rotates with a constant angular speed, ω, and N turns

V = N B A ω cos ω t

• V = Vmax when loop is parallel to the field

• V = 0 when when the loop is perpendicular to the field

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motorac.html#c2

Amount of angle (radian) turned in a second

wt = total angle of turn during time t

2pf = w period T = 1/f = 2p/w

t = 0:  = 0

t = t1:  = A B sin(wt1)

wt1

Vind = - d/dt

= - ABsin(wt1)/t1

Vind = - /t

= - ABwcos(wt)

AC generation

Hot

GND

http://www.howstuffworks.com/

-170 V

When we say 120 V, it means rms value!!

P  v2 or i2

vrms = 0.707va

City of Gainesville has 120,000 population. On average

approximately 200 W/person of electric power is required.

Let’s assume that GRU transmit power with 120 V. How much

current Should be carried in power line?

Pt = 120,000 x 200 W = 24,000,000 W = 24 MW

P = IV, I = P/V = 24,000,000/120 = 200,000 A

However, if we deliver power with 500,000 V,

I = 24,000,000/500,000 = 48 A

Now Joule heating due to wire resistance (R) is reduced

By (48/200,000)2 = 5.8 x 10-8

Iron Core

V

AC

dF/dt

Vp = -NpdF/dt

Vs = -NsdF/dt

Np

Ns

Vp/Vs = Np/Ns