23.9 Inductance  (Page 4/10)

 Page 4 / 10

Section summary

• Inductance is the property of a device that tells how effectively it induces an emf in another device.
• Mutual inductance is the effect of two devices in inducing emfs in each other.
• A change in current $\Delta {I}_{1}/\Delta t$ in one induces an emf ${\text{emf}}_{2}$ in the second:
${\text{emf}}_{2}=-M\frac{\Delta {I}_{1}}{\Delta t}\text{,}$
where $M$ is defined to be the mutual inductance between the two devices, and the minus sign is due to Lenz’s law.
• Symmetrically, a change in current $\Delta {I}_{2}/\Delta t$ through the second device induces an emf ${\text{emf}}_{1}$ in the first:
${\text{emf}}_{1}=-M\frac{\Delta {I}_{2}}{\Delta t}\text{,}$
where $M$ is the same mutual inductance as in the reverse process.
• Current changes in a device induce an emf in the device itself.
• Self-inductance is the effect of the device inducing emf in itself.
• The device is called an inductor, and the emf induced in it by a change in current through it is
$\text{emf}=-L\frac{\Delta I}{\Delta t}\text{,}$
where $L$ is the self-inductance of the inductor, and $\Delta I/\Delta t$ is the rate of change of current through it. The minus sign indicates that emf opposes the change in current, as required by Lenz’s law.
• The unit of self- and mutual inductance is the henry (H), where $1 H=1 \Omega \cdot \text{s}$ .
• The self-inductance $L$ of an inductor is proportional to how much flux changes with current. For an $N$ -turn inductor,
$L=N\frac{\Delta \Phi }{\Delta I}\text{.}$
• The self-inductance of a solenoid is
$L=\frac{{\mu }_{0}{N}^{2}A}{\ell }\text{(solenoid),}$
where $N$ is its number of turns in the solenoid, $A$ is its cross-sectional area, $\ell$ is its length, and ${\text{μ}}_{0}=4\pi ×{\text{10}}^{\text{−7}}\phantom{\rule{0.25em}{0ex}}\text{T}\cdot \text{m/A}\phantom{\rule{0.10em}{0ex}}$ is the permeability of free space.
• The energy stored in an inductor ${E}_{\text{ind}}$ is
${E}_{\text{ind}}=\frac{1}{2}{\text{LI}}^{2}\text{.}$

Conceptual questions

How would you place two identical flat coils in contact so that they had the greatest mutual inductance? The least?

How would you shape a given length of wire to give it the greatest self-inductance? The least?

Verify, as was concluded without proof in [link] , that units of $\text{T}\cdot {\text{m}}^{2}/A=\Omega \cdot \text{s}=\text{H}$ .

Problems&Exercises

Two coils are placed close together in a physics lab to demonstrate Faraday’s law of induction. A current of 5.00 A in one is switched off in 1.00 ms, inducing a 9.00 V emf in the other. What is their mutual inductance?

1.80 mH

If two coils placed next to one another have a mutual inductance of 5.00 mH, what voltage is induced in one when the 2.00 A current in the other is switched off in 30.0 ms?

The 4.00 A current through a 7.50 mH inductor is switched off in 8.33 ms. What is the emf induced opposing this?

3.60 V

A device is turned on and 3.00 A flows through it 0.100 ms later. What is the self-inductance of the device if an induced 150 V emf opposes this?

Starting with ${\text{emf}}_{2}=-M\frac{\Delta {I}_{1}}{\Delta t}$ , show that the units of inductance are $\left(\text{V}\cdot \text{s}\right)\text{/A}=\Omega \cdot \text{s}$ .

Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time must the 0.100 A current through a 2.00 mH inductor be switched on or off to induce a 500 V emf?

A large research solenoid has a self-inductance of 25.0 H. (a) What induced emf opposes shutting it off when 100 A of current through it is switched off in 80.0 ms? (b) How much energy is stored in the inductor at full current? (c) At what rate in watts must energy be dissipated to switch the current off in 80.0 ms? (d) In view of the answer to the last part, is it surprising that shutting it down this quickly is difficult?

(a) 31.3 kV

(b) 125 kJ

(c) 1.56 MW

(d) No, it is not surprising since this power is very high.

(a) Calculate the self-inductance of a 50.0 cm long, 10.0 cm diameter solenoid having 1000 loops. (b) How much energy is stored in this inductor when 20.0 A of current flows through it? (c) How fast can it be turned off if the induced emf cannot exceed 3.00 V?

A precision laboratory resistor is made of a coil of wire 1.50 cm in diameter and 4.00 cm long, and it has 500 turns. (a) What is its self-inductance? (b) What average emf is induced if the 12.0 A current through it is turned on in 5.00 ms (one-fourth of a cycle for 50 Hz AC)? (c) What is its inductance if it is shortened to half its length and counter-wound (two layers of 250 turns in opposite directions)?

(a) 1.39 mH

(b) 3.33 V

(c) Zero

The heating coils in a hair dryer are 0.800 cm in diameter, have a combined length of 1.00 m, and a total of 400 turns. (a) What is their total self-inductance assuming they act like a single solenoid? (b) How much energy is stored in them when 6.00 A flows? (c) What average emf opposes shutting them off if this is done in 5.00 ms (one-fourth of a cycle for 50 Hz AC)?

When the 20.0 A current through an inductor is turned off in 1.50 ms, an 800 V emf is induced, opposing the change. What is the value of the self-inductance?

60.0 mH

How fast can the 150 A current through a 0.250 H inductor be shut off if the induced emf cannot exceed 75.0 V?

Integrated Concepts

A very large, superconducting solenoid such as one used in MRI scans, stores 1.00 MJ of energy in its magnetic field when 100 A flows. (a) Find its self-inductance. (b) If the coils “go normal,” they gain resistance and start to dissipate thermal energy. What temperature increase is produced if all the stored energy goes into heating the 1000 kg magnet, given its average specific heat is $\text{200 J/kg·ºC}$ ?

(a) 200 H

(b) $\text{5.00ºC}$

Unreasonable Results

A 25.0 H inductor has 100 A of current turned off in 1.00 ms. (a) What voltage is induced to oppose this? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?

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