Induction is the process in which an emf is induced by changing magnetic flux. Many examples have been discussed so far, some more effective than others. Transformers, for example, are designed to be particularly effective at inducing a desired voltage and current with very little loss of energy to other forms. Is there a useful physical quantity related to how “effective” a given device is? The answer is yes, and that physical quantity is called inductance.
Mutual inductance is the effect of Faraday’s law of induction for one device upon another, such as the primary coil in transmitting energy to the secondary in a transformer. See Figure 1, where simple coils induce emfs in one another.
In the many cases where the geometry of the devices is fixed, flux is changed by varying current. We therefore concentrate on the rate of change of current, ΔI/ΔtΔI/Δt size 12{ΔI} {}, as the cause of induction. A change in the current I1I1 size 12{I rSub { size 8{1} } } {} in one device, coil 1 in the figure, induces an emf2emf2 size 12{"emf" rSub { size 8{2} } } {} in the other. We express this in equation form as
emf2=−MΔI1Δt,emf2=−MΔI1Δt, size 12{"emf" rSub { size 8{2} } = - M { {ΔI rSub { size 8{1} } } over {Δt} } } {}
(1)where MM size 12{M} {} is defined to be the mutual inductance between the two devices. The minus sign is an expression of Lenz’s law. The larger the mutual inductance MM size 12{M} {}, the more effective the coupling. For example, the coils in Figure 1 have a small MM size 12{M} {} compared with the transformer coils in (Reference). Units for MM size 12{M} {} are (V⋅s)/A=Ω⋅s(V⋅s)/A=Ω⋅s size 12{ \( V cdot s \) "/A"= %OMEGA cdot s} {}, which is named a henry (H), after Joseph Henry. That is, 1 H=1Ω⋅s1 H=1Ω⋅s size 12{1`H=1` %OMEGA cdot s} {}.
Nature is symmetric here. If we change the current I2I2 size 12{I rSub { size 8{2} } } {} in coil 2, we induce an emf1emf1 size 12{"emf" rSub { size 8{1} } } {} in coil 1, which is given by
emf1=−MΔI2Δt,emf1=−MΔI2Δt, size 12{"emf" rSub { size 8{1} } = - M { {ΔI rSub { size 8{2} } } over {Δt} } } {}
(2)where MM size 12{M} {} is the same as for the reverse process. Transformers run backward with the same effectiveness, or mutual inductance MM size 12{M} {}.
A large mutual inductance MM size 12{M} {} may or may not be desirable. We want a transformer to have a large mutual inductance. But an appliance, such as an electric clothes dryer, can induce a dangerous emf on its case if the mutual inductance between its coils and the case is large. One way to reduce mutual inductance MM size 12{M} {} is to counterwind coils to cancel the magnetic field produced. (See Figure 2.)
Self-inductance, the effect of Faraday’s law of induction of a device on itself, also exists. When, for example, current through a coil is increased, the magnetic field and flux also increase, inducing a counter emf, as required by Lenz’s law. Conversely, if the current is decreased, an emf is induced that opposes the decrease. Most devices have a fixed geometry, and so the change in flux is due entirely to the change in current ΔIΔI size 12{ΔI} {} through the device. The induced emf is related to the physical geometry of the device and the rate of change of current. It is given by
emf=−LΔIΔt,emf=−LΔIΔt, size 12{"emf"= - L { {ΔI} over {Δt} } } {}
(3)where LL 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 Figure 3.
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
LL 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
LL size 12{L} {} and will not allow current to change quickly. To avoid this effect, a small
LL size 12{L} {} must be achieved, such as by counterwinding coils as in
Figure 2.
A 1 H inductor is a large inductor. To illustrate this, consider a device with L=1.0 HL=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(ΔI/Δt)emf=−L(ΔI/Δt) size 12{"emf"= - L \( ΔI/Δt \) } {}, will oppose the change. Thus an emf will be induced given by emf=−L(ΔI/Δt)=(1.0 H)[(10 A)/(1.0 ms)]=10,000 Vemf=−L(ΔI/Δt)=(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 Figure 4.)
It is possible to calculate LL 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 LL 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(ΔΦ/Δt)emf=−N(ΔΦ/Δt) size 12{"emf"= - N \( ΔΦ/Δt \) } {} and, by the definition of self-inductance, as emf=−L(ΔI/Δt)emf=−L(ΔI/Δt) size 12{"emf"= - L \( ΔI/Δt \) } {}. Equating these yields
emf=−NΔΦΔt=−LΔIΔt.emf=−NΔΦΔt=−LΔIΔt. size 12{"emf"= - N { {ΔΦ} over {Δt} } = - L { {ΔI} over {Δt} } } {}
(4)Solving for LL size 12{L} {} gives
L=NΔΦΔI.L=NΔΦΔI. size 12{L=N { {ΔΦ} over {ΔI} } } {}
(5)This equation for the self-inductance LL size 12{L} {} of a device is always valid. It means that self-inductance LL size 12{L} {} depends on how effective the current is in creating flux; the more effective, the greater ΔΦΔΦ size 12{ΔΦ} {}/ ΔIΔI size 12{ΔI} {} is.
Let us use this last equation to find an expression for the inductance of a solenoid. Since the area
A
A
of a solenoid is fixed, the change in flux is
Δ
Φ
=
Δ
(
B
A
)
=
A
Δ
B
Δ
Φ
=
Δ
(
B
A
)
=
A
Δ
B
.
To find
Δ
B
Δ
B
, we note that the magnetic field of a solenoid is given by B=μ0nI=μ0NIℓB=μ0nI=μ0NIℓ size 12{B=μ rSub { size 8{0} } ital "nI"=μ rSub { size 8{0} } { { ital "NI"} over {ℓ} } } {}. (Here n=N/ℓn=N/ℓ size 12{n=N/ℓ} {}, where
N
N
is the number of coils and
ℓ
ℓ
is the solenoid’s length.) Only the current changes, so that ΔΦ=AΔB=μ0NAΔIℓΔΦ=AΔB=μ0NAΔIℓ size 12{ΔΦ=AΔB=μ rSub { size 8{0} } ital "NA" { {ΔI} over {ℓ} } } {}. Substituting
Δ
Φ
Δ
Φ
into L=NΔΦΔIL=NΔΦΔI size 12{L=N { {ΔΦ} over {ΔI} } } {} gives
L=NΔΦΔI=Nμ0NAΔIℓΔI.L=NΔΦΔI=Nμ0NAΔIℓΔI. size 12{L=N { {ΔΦ} over {ΔI} } =N { {μ rSub { size 8{0} } ital "NA" { {ΔI} over {ℓ} } } over {ΔI} } } {}
(6)This simplifies to
L=μ0N2Aℓ(solenoid).L=μ0N2Aℓ(solenoid). size 12{L= { {μ rSub { size 8{0} } N rSup { size 8{2} } A} over {ℓ} } } {}
(7)This is the self-inductance of a solenoid of cross-sectional area
A
A
and length
ℓ
ℓ
. Note that the inductance depends only on the physical characteristics of the solenoid, consistent with its definition.
Calculate the self-inductance of a 10.0 cm long, 4.00 cm diameter solenoid that has 200 coils.
Strategy
This is a straightforward application of L=μ0N2AℓL=μ0N2Aℓ size 12{L= { {μ rSub { size 8{0} } N rSup { size 8{2} } A} over {ℓ} } } {}, since all quantities in the equation except LL size 12{L} {} are known.
Solution
Use the following expression for the self-inductance of a solenoid:
L=μ0N2Aℓ.L=μ0N2Aℓ. size 12{L= { {μ rSub { size 8{0} } N rSup { size 8{2} } A} over {ℓ} } } {}
(8)The cross-sectional area in this example is A=πr2=(3.14...)(0.0200 m)2=1.26×10−3m2A=πr2=(3.14...)(0.0200 m)2=1.26×10−3m2 size 12{A=πr rSup { size 8{2} } = \( 3 "." "14" "." "." "." \) \( 0 "." "0200"`m \) rSup { size 8{2} } =1 "." "26" times "10" rSup { size 8{ - 3} } `m rSup { size 8{2} } } {},
N
N
is given to be 200, and the length
ℓ
ℓ
is 0.100 m. We know the permeability of free space is μ0=4π×10−7T⋅m/Aμ0=4π×10−7T⋅m/A. Substituting these into the expression for
L
L
gives
L
=
(4π×10−7 T⋅m/A)(200)2(1.26×10−3 m2)0.100 m
=
0.632 mH.
L
=
(4π×10−7 T⋅m/A)(200)2(1.26×10−3 m2)0.100 m
=
0.632 mH.
(9)
Discussion
This solenoid is moderate in size. Its inductance of nearly a millihenry is also considered moderate.
One common application of inductance is used in traffic lights that can tell when vehicles are waiting at the intersection. An electrical circuit with an inductor is placed in the road under the place a waiting car will stop over. The body of the car increases the inductance and the circuit changes sending a signal to the traffic lights to change colors. Similarly, metal detectors used for airport security employ the same technique. A coil or inductor in the metal detector frame acts as both a transmitter and a receiver. The pulsed signal in the transmitter coil induces a signal in the receiver. The self-inductance of the circuit is affected by any metal object in the path. Such detectors can be adjusted for sensitivity and also can indicate the approximate location of metal found on a person. (But they will not be able to detect any plastic explosive such as that found on the “underwear bomber.”) See Figure 5.