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Exploration 29.5: Self-Inductance


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This animation shows a cross section of a solenoid (think of a long tube cut lengthwise down the cylinder and then looking at the edge) so that the black dots represent the current-carrying wires coming into and out of the screen. Restart. The arrows show the direction and magnitude of the magnetic field. You can drag the black dot around to measure the field in different spots (position is given in centimeters, the magnetic field strength is given in millitesla, 10-3 T, and current is given in amperes). You can either change field by varying the current in the wires with the slider or you can choose to change the current linearly as a function of time.

Faraday's law tells us that when a loop is in a changing magnetic field, an induced emf in the loop will result. But what if the loop itself has a changing current? With a changing current, the loop has a changing magnetic field. Wouldn't it make sense, then, for there to be an induced emf and an induced current to oppose the changing flux? The answer is that there are: If the current is changed in a current loop, there is a self-induced back emf. The measure of the back emf produced when a current is changed in a loop is called its self-inductance, or simply inductance, represented by L and measured in henries, H (1 H = 1 T·m2/A). From Faraday's law, emf = - dΦ/dt, the self-inductance is the back emf = - L (dI/dt).

Run the change field by varying the current in the wires with the slider. Instead of considering a loop, we will look at a solenoid (it is easier to calculate the magnetic field inside a long solenoid).

  1. For the solenoid above, adjust the current with the slider and determine how the magnetic field varies with current.
  2. For this solenoid (given the value of the magnetic field at the current chosen), how many loops per meter are there?

Run change the current linearly as a function of time.

  1. What is the emf?
  2. Using the equation above, what is the inductance, L?
  3. Using Faraday's law and the equation above, show that L = (Φ/I) N for an inductor with N loops.
  4. Therefore, show that the inductance, L, of a solenoid is μ0N2A/(length), where N is the number of loops, A is the cross-sectional area and length is the length of the solenoid (so that N/length is the number of loops per meter).
  5. If this solenoid is 2 m long, calculate the inductance and compare it to your answer in (d) above.

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Exploration authored by Anne J. Cox.

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