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Making connections: conservation of energy

Lenz’s law is a manifestation of the conservation of energy. The induced emf produces a current that opposes the change in flux, because a change in flux means a change in energy. Energy can enter or leave, but not instantaneously. Lenz’s law is a consequence. As the change begins, the law says induction opposes and, thus, slows the change. In fact, if the induced emf were in the same direction as the change in flux, there would be a positive feedback that would give us free energy from no apparent source—conservation of energy would be violated.

Calculating emf: how great is the induced emf?

Calculate the magnitude of the induced emf when the magnet in [link] (a) is thrust into the coil, given the following information: the single loop coil has a radius of 6.00 cm and the average value of B cos θ size 12{B"cos"θ} {} (this is given, since the bar magnet’s field is complex) increases from 0.0500 T to 0.250 T in 0.100 s.

Strategy

To find the magnitude of emf, we use Faraday’s law of induction as stated by emf = N Δ Φ Δ t , but without the minus sign that indicates direction:

emf = N Δ Φ Δ t .

Solution

We are given that N = 1 size 12{N=1} {} and Δ t = 0 . 100 s , but we must determine the change in flux Δ Φ size 12{ΔΦ} {} before we can find emf. Since the area of the loop is fixed, we see that

Δ Φ = Δ ( BA cos θ ) = A Δ ( B cos θ ). size 12{ΔΦ=Δ \( BA"cos"θ \) =AΔ \( B"cos"θ \) } {}

Now Δ ( B cos θ ) = 0 . 200 T size 12{Δ \( B"cos"θ \) =0 "." "200"`T} {} , since it was given that B cos θ size 12{B"cos"θ} {} changes from 0.0500 to 0.250 T. The area of the loop is A = πr 2 = ( 3 . 14 . . . ) ( 0 . 060 m ) 2 = 1 . 13 × 10 2 m 2 size 12{A=πr rSup { size 8{2} } = \( 3 "." "14" "." "." "." \) \( 0 "." "060"`m \) rSup { size 8{2} } =1 "." "13" times "10" rSup { size 8{ - 2} } `m rSup { size 8{2} } } {} . Thus,

Δ Φ = ( 1.13 × 10 2 m 2 ) ( 0.200 T ). size 12{ΔΦ= \( 1 "." "13" times "10" rSup { size 8{ - 2} } " m" rSup { size 8{2} } \) \( 0 "." "200"" T" \) } {}

Entering the determined values into the expression for emf gives

Emf = N Δ Φ Δ t = ( 1.13 × 10 2 m 2 ) ( 0 . 200 T ) 0 . 100 s = 22 . 6 mV. size 12{E=N { {ΔΦ} over {Δt} } = { { \( 1 "." "13" times "10" rSup { size 8{ - 2} } " m" rSup { size 8{2} } \) \( 0 "." "200"" T" \) } over {0 "." "100"" s"} } ="22" "." 6" mV"} {}

Discussion

While this is an easily measured voltage, it is certainly not large enough for most practical applications. More loops in the coil, a stronger magnet, and faster movement make induction the practical source of voltages that it is.

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Phet explorations: faraday's electromagnetic lab

Play with a bar magnet and coils to learn about Faraday's law. Move a bar magnet near one or two coils to make a light bulb glow. View the magnetic field lines. A meter shows the direction and magnitude of the current. View the magnetic field lines or use a meter to show the direction and magnitude of the current. You can also play with electromagnets, generators and transformers!

Faraday's Electromagnetic Lab

Section summary

  • Faraday’s law of induction states that the emf induced by a change in magnetic flux is
    emf = N Δ Φ Δ t size 12{"emf"= - N { {ΔΦ} over {Δt} } } {}

    when flux changes by Δ Φ size 12{ΔΦ} {} in a time Δ t size 12{Δt} {} .

  • If emf is induced in a coil, N is its number of turns.
  • The minus sign means that the emf creates a current I size 12{I} {} and magnetic field B size 12{B} {} that oppose the change in flux Δ Φ size 12{ΔΦ} {} —this opposition is known as Lenz’s law.

Conceptual questions

A person who works with large magnets sometimes places her head inside a strong field. She reports feeling dizzy as she quickly turns her head. How might this be associated with induction?

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A particle accelerator sends high-velocity charged particles down an evacuated pipe. Explain how a coil of wire wrapped around the pipe could detect the passage of individual particles. Sketch a graph of the voltage output of the coil as a single particle passes through it.

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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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