Induction -->Inductors Back to Circuits for a bit …. 11-11-06 Induction - Fall 2006

What the heck are we doing? Today 7:30 AM we had our problem review session Continue on with Induction & Inductors Watch those piling up WebAssigns! Monday More of the same, Wednesday EXAMINATION #3 After Holiday Complete the remaining six chapters in the syllabus. 11-11-06 Induction - Fall 2006

The field at A & B are the same. The loop is pushed into a region where the magnetic field is into the page. The motion creates an induced current in the loop which in turn produces a magnetic field at A and B. The field at A & B are the same. The field at A is stronger than the one at B The field at B is stronger than the one at A. None of these. 11-11-06 Induction - Fall 2006

Today we will consider a Coil 11-11-06 Induction - Fall 2006

We will base our discussion on Faraday’s Law Lentz 11-11-06 Induction - Fall 2006

Consider the following: There are N turns in the solenoid. There is a current flowing in the direction shown. The power supply is set to 20 volts and the resistor is 10 ohms. The coil wire has negligible resistance. 11-11-06 Induction - Fall 2006

The steady state current (20 volts, 10 ohms) in the circuit is 2 amperes Less than 2 amperes More than 2 amperes Need more info. 11-11-06 Induction - Fall 2006

The DIRECTION of the magnetic field in the coil is B From A to B From B to A Not enough information is given. 11-11-06 Induction - Fall 2006

Back to the coil diagram … Recall from the last discussion that the magnetic field in the coil is given by: n = #turns per unit length Coil is infinitely long Sort of 11-11-06 Induction - Fall 2006

Back to the coil diagram … The Flux through a single turn of the coil is BA or m0niA. Now, let’s increase the applied voltage linearly at a rate of DV/Dt. The current will change at a rate DI/Dt. 11-11-06 Induction - Fall 2006

Back to the coil diagram … For the single coil (as Dt0) FARADAY Says: 11-11-06 Induction - Fall 2006

Looking into the coil from the end with the red arrow, the emf around the coil induced current will be B In a clockwise direction In a counterclockwise direction 11-11-06 Induction - Fall 2006

So … the induced emf B Will create a current that will oppose the change in the current. The induced emf will therefore oppose the applied voltage (also an emf) from the power supply. 11-11-06 Induction - Fall 2006

So for the single coil 11-11-06 Induction - Fall 2006

Definition of Inductance L UNIT of Inductance = 1 henry = 1 T- m2/A FB is the flux near the center of one of the coils making the inductor 11-11-06 Induction - Fall 2006

An inductor in the form of a solenoid contains 420 turns, is 16 An inductor in the form of a solenoid contains 420 turns, is 16.0 cm in length, and has a cross-sectional area of 3.00 cm2. What uniform rate of decrease of current through the inductor induces an emf of 175 μV? 11-11-06 Induction - Fall 2006

11-11-06 Induction - Fall 2006

Look at the following circuit: Switch is open NO current flows in the circuit. All is at peace! Let's close the switch.... 11-11-06 Induction - Fall 2006

At the INSTANT that the switch is closed, the current through the resistor is: Zero E/R Can’t tell 11-11-06 Induction - Fall 2006

Don’t care .. we will be out by then! Three years after the switch is closed, the current through the resistor is: E/R Zero Don’t care .. we will be out by then! 11-11-06 Induction - Fall 2006

Graph? IR E/R Probably looks something Like this. time 11-11-06 Induction - Fall 2006

Close the circuit… After the circuit has been closed for a long time, the current settles down. Since the current is constant, the flux through the coil is constant and there is no Emf. Current is simply E/R (Ohm’s Law) 11-11-06 Induction - Fall 2006

Close the circuit… When switch is first closed, current begins to flow rapidly. The flux through the inductor changes rapidly. An emf is created in the coil that opposes the increase in current. The net potential difference across the resistor is the battery emf opposed by the emf of the coil. 11-11-06 Induction - Fall 2006

Looking at the math 11-11-06 Induction - Fall 2006

Just as we did with the capacitor, we can solve this equation and we get: 11-11-06 Induction - Fall 2006

The growth 11-11-06 Induction - Fall 2006

Death of the current: 11-11-06 Induction - Fall 2006

Graph 11-11-06 Induction - Fall 2006

Consider the Solenoid Again… n turns per unit length 11-11-06 Induction - Fall 2006

Inductance & Geometry Depends only on geometry just like C and is independent of current. 11-11-06 Induction - Fall 2006

Max Current Rate of increase = max emf VR=iR ~current 11-11-06 Induction - Fall 2006

Solve the loop equation. 11-11-06 Induction - Fall 2006

IMPORTANT QUESTION Switch closes. No emf Current flows for a while It flows through R Energy is conserved (i2R) WHERE DOES THE ENERGY COME FROM?? 11-11-06 Induction - Fall 2006

For an answer Return to the Big C We move a charge dq from the (-) plate to the (+) one. The (-) plate becomes more (-) The (+) plate becomes more (+). dW=Fd=dq x E x d +q -q E=e0A/d +dq 11-11-06 Induction - Fall 2006

The calc The energy is in the FIELD !!! 11-11-06 Induction - Fall 2006

What about POWER?? power to circuit power dissipated by resistor Must be dWL/dt 11-11-06 Induction - Fall 2006

So Energy stored in the Capacitor 11-11-06 Induction - Fall 2006

WHERE is the energy?? l 11-11-06 Induction - Fall 2006

Remember the Inductor?? ????????????? 11-11-06 Induction - Fall 2006

So … 11-11-06 Induction - Fall 2006

ENERGY IN THE FIELD TOO! 11-11-06 Induction - Fall 2006

IMPORTANT CONCLUSION A region of space that contains either a magnetic or an electric field contains electromagnetic energy. The energy density of either is proportional to the square of the field strength. 11-11-06 Induction - Fall 2006

END OF TOPIC 11-11-06 Induction - Fall 2006