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emf = L Δ I Δ t , size 12{"emf"= - L { {ΔI} over {Δt} } } {}

where L size 12{L} {} is the self-inductance of the device. A device that exhibits significant self-inductance is called an inductor    , and given the symbol in [link] .

Two straight lines connected by three half-circles adjacent to each other.
The minus sign is an expression of Lenz’s law, indicating that emf opposes the change in current. Units of self-inductance are henries (H) just as for mutual inductance. The larger the self-inductance L size 12{L} {} of a device, the greater its opposition to any change in current through it. For example, a large coil with many turns and an iron core has a large L size 12{L} {} and will not allow current to change quickly. To avoid this effect, a small L size 12{L} {} must be achieved, such as by counterwinding coils as in [link] .

A 1 H inductor is a large inductor. To illustrate this, consider a device with L = 1 . 0 H size 12{L=1 "." 0`H} {} that has a 10 A current flowing through it. What happens if we try to shut off the current rapidly, perhaps in only 1.0 ms? An emf, given by emf = L ( Δ I / Δ t ) size 12{"emf"= - L \( ΔI/Δt \) } {} , will oppose the change. Thus an emf will be induced given by emf = L ( Δ I / Δ t ) = ( 1 . 0 H ) [ ( 10 A ) / ( 1 . 0 ms ) ] = 10,000 V . The positive sign means this large voltage is in the same direction as the current, opposing its decrease. Such large emfs can cause arcs, damaging switching equipment, and so it may be necessary to change current more slowly.

There are uses for such a large induced voltage. Camera flashes use a battery, two inductors that function as a transformer, and a switching system or oscillator to induce large voltages. (Remember that we need a changing magnetic field, brought about by a changing current, to induce a voltage in another coil.) The oscillator system will do this many times as the battery voltage is boosted to over one thousand volts. (You may hear the high pitched whine from the transformer as the capacitor is being charged.) A capacitor stores the high voltage for later use in powering the flash. (See [link] .)

The figure describes an inductor L which is connected in parallel to a capacitor C through a variable switch. There is a cell of voltage V placed parallel to the capacitor. The ends of switch can be removed from the capacitor and connected to Cell V for charging. This variable connection is shown as dashed arrows.
Through rapid switching of an inductor, 1.5 V batteries can be used to induce emfs of several thousand volts. This voltage can be used to store charge in a capacitor for later use, such as in a camera flash attachment.

It is possible to calculate L size 12{L} {} for an inductor given its geometry (size and shape) and knowing the magnetic field that it produces. This is difficult in most cases, because of the complexity of the field created. So in this text the inductance L size 12{L} {} is usually a given quantity. One exception is the solenoid, because it has a very uniform field inside, a nearly zero field outside, and a simple shape. It is instructive to derive an equation for its inductance. We start by noting that the induced emf is given by Faraday’s law of induction as emf = N ( Δ Φ / Δ t ) size 12{"emf"= - N \( ΔΦ/Δt \) } {} and, by the definition of self-inductance, as emf = L ( Δ I / Δ t ) size 12{"emf"= - L \( ΔI/Δt \) } {} . Equating these yields

emf = N Δ Φ Δ t = L Δ I Δ t . size 12{"emf"= - N { {ΔΦ} over {Δt} } = - L { {ΔI} over {Δt} } } {}

Solving for L size 12{L} {} gives

L = N Δ Φ Δ I . size 12{L=N { {ΔΦ} over {ΔI} } } {}

This equation for the self-inductance L size 12{L} {} of a device is always valid. It means that self-inductance L size 12{L} {} depends on how effective the current is in creating flux; the more effective, the greater Δ Φ size 12{ΔΦ} {} / Δ I size 12{ΔI} {} is.

Let us use this last equation to find an expression for the inductance of a solenoid. Since the area A of a solenoid is fixed, the change in flux is Δ Φ = Δ ( B A ) = A Δ B . To find Δ B , we note that the magnetic field of a solenoid is given by B = μ 0 nI = μ 0 NI size 12{B=μ rSub { size 8{0} } ital "nI"=μ rSub { size 8{0} } { { ital "NI"} over {ℓ} } } {} . (Here n = N / size 12{n=N/ℓ} {} , where N is the number of coils and is the solenoid’s length.) Only the current changes, so that Δ Φ = A Δ B = μ 0 NA Δ I size 12{ΔΦ=AΔB=μ rSub { size 8{0} } ital "NA" { {ΔI} over {ℓ} } } {} . Substituting Δ Φ into L = N Δ Φ Δ I size 12{L=N { {ΔΦ} over {ΔI} } } {} gives

Practice Key Terms 6

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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