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 Demo “Self-Inductance in a closed loop” We must keep the shape and size of the loop fixed. BI (From the Biot-Savart Law) BI d dI dt dt B B L 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: dI ε L dt L (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: dI d L N dt dt L An equivalent definition of L uses the integral of the above where is the flux through each loop produced by current I in each loop: LI N To find L we use: LN I Quiz A flat circular coil with 10 turns (each loop identical) has inductance L1 . A second coil, of the same size, shape and current passing through the conductor but with 20 turns, would have inductance: A) 2 L1 B) ½ L1 C) 4 L1 D) ¼ L1 E) L1 Also: dI L dt L dI / dt 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 2r I I l Given: N = 300 turns r = 1 cm l = 25cm Calculate L Solution Example 2: Long solenoid B 2r I I l Given: N = 300 turns r = 1 cm l = 25cm dI/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): I o I B 2r Show that o b L ln a 2 Solution Kirchhoff’s Loop Rule terms (again) Voltage change in going along path from left to right: - resistor, V IR - capacitor, q V - inductor, dI V L dt I q q C 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 20mH Energy How much work is done by an external emf to increase the current from 0 to Ifinal ? I dI Kirchoff’s law gives: L 0 dt (Recall, P=VI) + εL= LdI/dt - dI I LI dt Power supplied by external emf = power absorbed by inductor dU dI LI dt dt Power absorbed by inductor is: L dU LIdI L The total potential energy stored is: If U L I dI LI 1 2 L 2 f 0 U L LI 1 2 2 Find the potential energy of an inductor with L=400mH and a final current of 10A.