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14.2 Self-inductance and inductors  (Page 4/6)

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L = N Φ m I = μ 0 N 2 h 2 π ln R 2 R 1 .

As expected, the self-inductance is a constant determined by only the physical properties of the toroid.

Check Your Understanding (a) Calculate the self-inductance of a solenoid that is tightly wound with wire of diameter 0.10 cm, has a cross-sectional area of 0.90 cm 2 , and is 40 cm long. (b) If the current through the solenoid decreases uniformly from 10 to 0 A in 0.10 s, what is the emf induced between the ends of the solenoid?

a. 4.5 × 10 −5 H ; b. 4.5 × 10 −3 V

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Check Your Understanding (a) What is the magnetic flux through one turn of a solenoid of self-inductance 8.0 × 10 −5 H when a current of 3.0 A flows through it? Assume that the solenoid has 1000 turns and is wound from wire of diameter 1.0 mm. (b) What is the cross-sectional area of the solenoid?

a. 2.4 × 10 −7 Wb ; b. 6.4 × 10 −5 m 2

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Summary

  • Current changes in a device induce an emf in the device itself, called self-inductance,
    ε = − L d I d t ,

    where L is the self-inductance of the inductor and d I / d 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-inductance and inductance is the henry (H), where 1 H = 1 Ω · s .
  • The self-inductance of a solenoid is
    L = μ 0 N 2 A l ,

    where N is its number of turns in the solenoid, A is its cross-sectional area, l is its length, and μ 0 = 4 π × 10 −7 T · m/A is the permeability of free space.
  • The self-inductance of a toroid is
    L = μ 0 N 2 h 2 π ln R 2 R 1 ,

    where N is its number of turns in the toroid, R 1 and R 2 are the inner and outer radii of the toroid, h is the height of the toroid, and μ 0 = 4 π × 10 −7 T · m/A is the permeability of free space.

Conceptual questions

Does self-inductance depend on the value of the magnetic flux? Does it depend on the current through the wire? Correlate your answers with the equation N Φ m = L I .

Self-inductance is proportional to the magnetic flux and inversely proportional to the current. However, since the magnetic flux depends on the current I , these effects cancel out. This means that the self-inductance does not depend on the current. If the emf is induced across an element, it does depend on how the current changes with time.

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Would the self-inductance of a 1.0 m long, tightly wound solenoid differ from the self-inductance per meter of an infinite, but otherwise identical, solenoid?

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Discuss how you might determine the self-inductance per unit length of a long, straight wire.

Consider the ends of a wire a part of an RL circuit and determine the self-inductance from this circuit.

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The self-inductance of a coil is zero if there is no current passing through the windings. True or false?

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How does the self-inductance per unit length near the center of a solenoid (away from the ends) compare with its value near the end of the solenoid?

The magnetic field will flare out at the end of the solenoid so there is less flux through the last turn than through the middle of the solenoid.

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Problems

An emf of 0.40 V is induced across a coil when the current through it changes uniformly from 0.10 to 0.60 A in 0.30 s. What is the self-inductance of the coil?

0.24 H

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The current shown in part (a) below is increasing, whereas that shown in part (b) is decreasing. In each case, determine which end of the inductor is at the higher potential.

Figure a shows current flowing through a coil from left to right. Figure b shows current flowing through a coil from right to left.
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What is the rate at which the current though a 0.30-H coil is changing if an emf of 0.12 V is induced across the coil?

0.4 A/s

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When a camera uses a flash, a fully charged capacitor discharges through an inductor. 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?

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A coil with a self-inductance of 2.0 H carries a current that varies with time according to I ( t ) = ( 2.0 A ) sin 120 π t . Find an expression for the emf induced in the coil.

ε = 480 π sin ( 120 π t − π / 2 ) V

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A solenoid 50 cm long is wound with 500 turns of wire. The cross-sectional area of the coil is 2.0 cm 2 What is the self-inductance of the solenoid?

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A coil with a self-inductance of 3.0 H carries a current that decreases at a uniform rate d I / d t = −0.050 A/s . What is the emf induced in the coil? Describe the polarity of the induced emf.

0.15 V. This is the same polarity as the emf driving the current.

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The current I(t) through a 5.0-mH inductor varies with time, as shown below. The resistance of the inductor is 5.0 Ω . Calculate the voltage across the inductor at t = 2.0 ms , t = 4.0 ms , and t = 8.0 ms .

The graph of current in amperes versus time in milliseconds. The current starts from 0 at 0 milliseconds, increases with time and reaches just over 6 amperes at roughly 3 milliseconds. It decreases sharply till about 6 milliseconds, then decreases at a slightly slower rate till it reaches 0 at 12 milliseconds.
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A long, cylindrical solenoid with 100 turns per centimeter has a radius of 1.5 cm. (a) Neglecting end effects, what is the self-inductance per unit length of the solenoid? (b) If the current through the solenoid changes at the rate 5.0 A/s, what is the emf induced per unit length?

a. 0.089 H/m; b. 0.44 V/m

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Suppose that a rectangular toroid has 2000 windings and a self-inductance of 0.040 H. If h = 0.10 m , what is the ratio of its outer radius to its inner radius?

Figure shows the cross section of a toroid. The inner radius of the ring is R1 and the outer radius is R2. The height of the rectangular cross section is h. A small section of thickness dr is located at the center of the rectangular cross section. This is at a distance r from the center of the ring. The area within the rectangular cross section with thickness dr and height h is highlighted and labeled da. Field lines and current i flowing through the toroid are shown.
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What is the self-inductance per meter of a coaxial cable whose inner radius is 0.50 mm and whose outer radius is 4.00 mm?

L l = 4.16 × 10 −7 H/m

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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