Self Inductance and RL Circuits Self Induction Self Inductance and RL Circuits B(t) ~ i R e a b L I I

Lecture Outline Concept of Self-Inductance Definition of Self-Inductance Calculation of Self-Inductance for Simple Cases RL Circuits Energy in Magnetic Field Text Reference: Chapter 32

\ magnetic field produced in the area enclosed by the loop. Self-Inductance Consider the loop at the right. X X X X X X X X X X switch closed Þ current starts to flow in the loop. \ magnetic field produced in the area enclosed by the loop. \ flux through loop changes \ emf induced in loop opposing initial emf Self-Induction: the act of a changing current through a loop inducing an opposing current in that same loop. Just as a variable current can induce an EMF on another coil, it can induce an EMF on itself (back EMF).

Recall that for a solenoid Self-Inductance We can put solenoids in a circuit: R e a b L I Because of the back EMF the current in this circuit does not instantaneously reach steady state. This is due to the presence of the solenoid L. Flux through one loop } Recall that for a solenoid } Flux through the whole solenoid is where, Thus, or, Inductance Faradays Law then takes form,

r << l Self-Inductance The inductance of an inductor ( a set of coils in some geometry ..eg solenoid, toroid) then, like a capacitor, can be calculated from its geometry alone if the device is constructed from conductors and air. If extra material (eg iron core) is added, then we need to add some knowledge of materials as we did for capacitors (dielectrics) and resistors (resistivity) Archetypal inductor is a long solenoid, just as a pair of parallel plates is the archetypal capacitor. r << l

Self-Inductance Units of Inductance are Henrys:

(a) L2 < L1 (b) L2 = L1 (c) L2 > L1 Lecture 19, CQ Consider the two inductors shown: Inductor 1 has length l, N total turns and has inductance L1. Inductor 2 has length 2l, 2N total turns and has inductance L2. What is the relation between L1 and L2? (a) L2 < L1 (b) L2 = L1 (c) L2 > L1

(a) L2 < L1 (b) L2 = L1 (c) L2 > L1 Lecture 19, CQ Consider the two inductors shown: Inductor 1 has length l, N total turns and has inductance L1. Inductor 2 has length 2l, 2N total turns and has inductance L2. What is the relation between L1 and L2? (a) L2 < L1 (b) L2 = L1 (c) L2 > L1 To determine the self-inductance L, we need to determine the flux FB which passes through the coils when a current I flows: L º FB / I. To calculate the flux, we first need to calculate the magnetic field B produced by the current: B = m0(N/l)I ie the B field is proportional to the number of turns per unit length. Therefore, B1 = B2. But does that mean L1 = L2? To calculate L, we need to calculate the flux. Since B1 = B2 , the flux through any given turn is the same in each inductor. However, there are twice as many turns in inductor 2; therefore the flux through inductor 2 is twice as much as the flux through inductor 1!!! Therefore, L2 = 2L1.

RL Circuits

RL Circuits At t=0, the switch is closed and the current I starts to flow. I a b L Loop rule: Note that this eqn is identical in form to that for the RC circuit with the following substitutions: RC: RC®RL: Þ \ Þ

(a) I¥ = 0 (b) I¥ = e / 2R (c) I¥ = 2e / R (a) I0 = 0 (b) I0 = e / 2R Lecture 18, ACT 2 a b e R L I 1) At t=0 the switch is thrown from position b to position a in the circuit shown: What is the value of the current I a long time after the switch is thrown? (a) I¥ = 0 (b) I¥ = e / 2R (c) I¥ = 2e / R A long time after the switch is thrown, the current approaches an asymptotic value. ie as t ® ¥, dI/dt ® 0. As dI/dt ® 0 , the voltage across the inductor ® 0. \ I¥ = e / 2R. (a) I0 = 0 (b) I0 = e / 2R (c) I0 = 2e / R 2) What is the value of the current I0 immediately after the switch is thrown? Immediately after the switch is thrown, the rate of change of current is as large as it can be (we’ve been assuming it was ¥!) The inductor limits dI/dt to be initially equal to e / L. ie the voltage across the inductor = e; the current then must be 0!

L Þ I RL Circuits We therefore write: To find the current I as a fcn of time t, we need to choose an exponential solution which satisfies the boundary condition: a b L I Þ We therefore write: The voltage drop across the inductor is given by:

Sign Error! Current Max = e/R 63% Max at t=L/R RL Circuit (e on) I t Sketch curves ! L/R t I 2L/R e/R VL t e Current Max = e/R 63% Max at t=L/R Voltage on L Max = e/R 37% Max at t=L/R Sign Error!

L I RL Circuits Loop rule: The appropriate initial condition is: After the switch has been in position a for a long time, redefined to be t=0, it is moved to position b. a b L I Loop rule: The appropriate initial condition is: The solution then must have the form:

Sign Error! Current Max = e/R 37% Max at t=L/R RL Circuit (e off) I t Sketch curves ! L/R t 2L/R I e/R Current Max = e/R 37% Max at t=L/R -e VL t Voltage on L Max = -e 37% Max at t=L/R Sign Error!

e on e off L/R 2L/R L/R 2L/R e/R e/R I I t t e VL VL t -e t

Can you determine from this the rule for adding inductors in series? Lecture 18, ACT 3 At t=0, the switch is thrown from position b to position a as shown: Let tI be the time for circuit I to reach 1/2 of its asymptotic current. Let tII be the time for circuit II to reach 1/2 of its asymptotic current. What is the relation between tI and tII? I II (a) tII < tI (b) tII = tI (c) tII > tI To answer this question, we must determine the time constants of the two circuits. To determine the time constants of the two circuits, we just need to write down the loop eqns: I: Can you determine from this the rule for adding inductors in series? II:

L Þ I Energy of an Inductor Start with loop rule: a b How much energy is stored in an inductor when a current is flowing through it? a b L I Start with loop rule: Multiply this equation by I: From this equation, we can identify PL, the rate at which energy is being stored in the inductor: We can integrate this equation to find an expression for U, the energy stored in the inductor when the current = I: Þ

Where is the Energy Stored? Claim: (without proof) energy is stored in the Magnetic field itself (just as in the Capacitor / Electric field case). To calculate this energy density, consider the uniform field generated by a long solenoid: l r N turns The inductance L is: Energy U: We can turn this into an energy density by dividing by the volume containing the field: