Self Inductance Self Inductance A variable power supply

Self Inductance

Self Inductance A variable power supply is connected to a loop. The current in the loop creates a magnetic field. What happens when the power supply dial is turned down reducing the current, or turned up increasing the current?

Self Induction R The current does not go from zero to ε/R in the circuit immediately after the switch is closed: 1. as the current flows through, magnetic flux through the loop is set up 2. this is opposed by induced emf in the loop which opposes the change in net magnetic flux 3. by Lentz’s law, the induced E-field opposes the current flow

Self Induction R As a time-varying current flow through the conductor, the same thing happens: 1. as the changing current flows through, the magnetic flux through the loop changes 2. this is opposed by induced emf in the loop which opposes the change in net magnetic flux 3. by Lentz’s law, the induced E-field opposes the current flow

“Self-Inductance in a closed loop” We must keep the shape and size of the loop fixed. (From the Biot-Savart Law) The self-inductance L is the proportionality constant. It depends on the shape of the loop, that is, its geometry. The self-induced ems is proportional to the rate of change of the current: (definition of L) Unit for L: 1 henry (H) = 1 (V • s)/A

From Faraday’s Law for N loops and our definition of inductance, L: An equivalent definition of L uses the integral of the above where F is the flux through each loop produced by current I in each loop: To find L we use:

Quiz A flat circular coil with 10 turns (each loop identical) has inductance L 1. A second coil, of the same size, shape and current passing through the conductor but with 20 turns, would have inductance: A) 2 L 1 B) ½ L 1 C) 4 L 1 D) ¼ L 1 E) L 1

Also: We can see that inductance is the measure of the opposition to the change in current, I. (Recall that resistance, R, is the opposition to current, I: R=V/I)

Example 1: Long solenoid B 2 r I I l Given: N = 300 turns r = 1 cm l = 25 cm Calculate L

Solution

Example 2: Long solenoid B 2 r I I l Given: N = 300 turns r = 1 cm l = 25 cm d. I/dt = -50 A/s Calculate the self-induced emf

Solution

Example 3: Self-Inductance of a coaxial cable b a r In the gap (a < r < b): Show that I

Solution

Kirchhoff’s Loop Rule (again) Voltage change in going along path from left to right: - resistor, - capacitor, - inductor, I q q I Path direction same as current

Example 4 What is the current 2 ms after the switch is closed? Hint: write down Kirchoffs loop rule and solve the differential equation for I 12 V 20 m. H

Energy How much work is done by an external emf to increase the current from 0 to Ifinal ? I + εL= - Ld. I/dt - Kirchoff’s law gives: (Recall, P=VI) Power supplied by external emf = power absorbed by inductor

Power absorbed by inductor is: The total potential energy stored is: Find the potential energy of an inductor with L=400 m. H and a final current of 10 A.

Examples of Inductors