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28 magnetic induction

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induced currents due to magnetic force qv x B

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Faraday’s Law

• induced current due to qvB force can be determined in terms of magnetic flux

• flux definition same as before, with B replacing E

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Faraday’s Law

• emf induced around closed loop:

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Lenz’s Law

• The change in externally applied magnetic flux is opposed by the responsive magnetic flux change of a circuit.

• indicated by (-) sign in Faraday’s Law:

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current direction determined from Lenz’s Law

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Lenz’s Law: induced magnetic force opposes the motion

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A metallic wire loop is in a uniform magnetic field.

Determine if there is a current induced in the loop when

•loop is stationary

•loop moves left or right

•loop moves upward out of the field region

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A wire loop moves from a region with no magnetic field into a region with a uniform magnetic field pointing into the page. What is the direction of induced current in loop?

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The loop is now entirely inside the B-field region

Apply Faraday’s Law to determine induced emf.

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Example F.L.

• A loop of area 0.45 sq.m. is rotated 180 deg. in 0.15 seconds in uniform B = 1.2 tesla such that maximum flux change occurs. Calculate average emf induced.

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flux through coil

• N = #turns of wire

• flux = NBAcos.

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simple motor

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Ex. The loop was inserted in 0.05s. The emf induced is

= NB(A)/t

= (80)(0.6)(ax/t)

= (80)(0.6)(0.2)(x/t)

= (80)(0.6)(0.2)(x/t)

= (80)(0.6)(0.2)(0.15/0.05)

= 28.8 volts

B = 0.6T

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motional emf

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motional emf: Eddy Currents produce resistive force.

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The property of an electric circuit whereby an electromotive force is produced in the circuit by a change of current in the circuit itself. (McGraw-Hill Science Dict.)

L = (flux due to current)/current

self-inductance, L

dt

dIL

dt

LId

dt

d

)(

SI Unit: Henry, 1 H = 1 Wb/A

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n= #turns/m

Example: n = 10,000 turns per meter, length 1.0m, and Area = 1.0m2.

L = (12.6x10-7)(10,000)2(1)(1) = 126H

L, solenoid

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Magnetic Energydt

dIL

dt

dILIIP

dtdt

dIILPdt

“-” indicates power absorbed by inductor from battery. inductor gets a “+” of this.

Example: energy stored in a 126 henry coil with 1,000A

Um = ½ (126)(1,000)2 = 6.3x107 J = 63MJ.

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SI Unit: joule per cubic meter, J/m3.

Ex. Calculate the energy density in a 5.0T field.

um = (5)2/(8x10-7) = 10 MJ/m3.

energy density

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RL Circuits

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0 IRdt

dIL

IRdt

dIL

L

dt

IR

dI

RdudI

RdIdu

IRu

/

integrate from t = 0 to t

integrate from I = 0 to I

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RL Circuits

• “charging”

• “discharging”

• tau = L/R seconds

/1 teR

I

/teR

I

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Inductors have resistive and inductive voltage drops

Ex. Current flow is increased through an inductor with L = 800mH and resistance r = 20 ohms at a rate of 30 amperes/second. The voltage when I = 2A is:

V = - (0.8)(30) – (2)(20) = -24 – 40 = -64 volts.

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FL = qvd x B FR = qEperpendicularFup = qv x B