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CHAPTER inductance Self-Inductance Energy in a Magnetic Field

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**example Current clockwise; force up Current counterclockwise; force up**

Week 9, Day 2 example A coil moves up from underneath a magnet with its north pole pointing upward. The current in the coil and the force on the coil: Current clockwise; force up Current counterclockwise; force up Current clockwise; force down Current counterclockwise; force down Class 22

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**Answer: 3. Current is clockwise; force is down**

Week 9, Day 2 Answer: 3. Current is clockwise; force is down The clockwise current creates a self-field downward, trying to offset the increase of magnetic flux through the coil as it moves upward into stronger fields (Lenz’s Law). The I dl x B force on the coil is a force which is trying to keep the flux through the coil from increasing by slowing it down (Lenz’s Law again). Class 22

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**example Bout v Current CW, Force Left, No Torque**

Week 09, Day 1 example v Bout A rectangular wire loop is pulled thru a uniform B field penetrating its top half, as shown. The induced current and the force and torque on the loop are: Current CW, Force Left, No Torque Current CW, No Force, Torque Rotates CCW Current CCW, Force Left, No Torque Current CCW, No Force, Torque Rotates CCW No current, force or torque Class 20

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**Bout v Answer: 5. No current, force or torque**

Week 09, Day 1 v Bout Answer: 5. No current, force or torque The motion does not change the magnetic flux, so Faraday’s Law says there is no induced EMF, or current, or force, or torque. Of course, if we were pulling at all up or down there would be a force to oppose that motion. Class 20

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Self-Inductance When the switch is closed, the current does not immediately reach its maximum value Faraday’s law can be used to describe the effect

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Self-inductance, cont. (a) A current in the coil produces a magnetic field directed to the left. (b) If the current increases, the coil acts as a source of emf directed as shown by the dashed battery. (c) The induced emf in the coil changes its polarity if the current decreases.

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**Self-inductance occurs when the changing flux through a circuit arises from the circuit itself**

As the current increases, the magnetic flux through a loop due to this current also increases The increasing flux induces an emf that opposes the current As the magnitude of the current increases, the rate of increase lessens and hence the induced emf decreases This opposing emf results in a gradual increase in the current

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**Self Inductance Define: Self Inductance**

An inductor is a device that produces a uniform magnetic field when a current passes through it. A solenoid is an inductor. The magnetic flux of an inductor is proportional to the current. For each coil (turn) of the solenoid: Φper coil = A•B Φsol = N(A•B) = NAB = NA(µ0NI/ℓ) = (Au0N2/ℓ)Isol Define: Self Inductance

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**Inductance of a Solenoid**

The magnetic flux through each turn is Therefore, the inductance is This shows that L depends on the geometry of the object

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Inductance Units

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Self-inductance, cont. The self-induced emf is given by Faraday’s law and must be proportional to the time rate of change of the current L is a proportionality constant called the inductance of the device The negative sign indicates that a changing current induces an emf in opposition to that change

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**Inductor has a large inductance (L) and consist of closely wrapped coil of many turns**

Inductance can be interpreted as a measure of opposition to the rate of change in the current Remember resistance R is a measure of opposition to the current As a circuit is completed, the current begins to increase, but the inductor produces an emf that opposes the increasing current Therefore, the current doesn’t change from 0 to its maximum instantaneously

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**Inductance of a Solenoid**

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**Potential difference across an inductor**

An inductor is a conducting coil. An inductor in a circuit resists change in current with an induced potential. An inductor stores energy in a magnetic field. The “strength” of an inductor is determined by is “inductance”, represented by the letter L. For the ideal inductor, R = 0, therefore potential difference across the inductor also equals zero, as long as the current is constant. What happens if we increase the current?

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**Potential difference across an inductor**

Increasing the current increases the flux. An induced magnetic field will oppose the increase by pointing to the right. The induced current is opposite the solenoid current. The induced current carries positive charge to the left and establishes a potential difference across the inductor. Induced current Induced field Potential difference

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**Potential difference across an inductor**

The potential difference across the inductor can be found using Faraday’s Law: Where Φm = Φper coil Φsol = N Φper coil We defined Φ = LI dΦsol/dt = L |dI/dt| Induced current Induced field Potential difference

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**Potential difference across an inductor**

If the inductor current is decreased, the induced magnetic field, the induced current and the potential difference all change direction. Note that whether you increase or decrease the current, the inductor always “resists” the change with an induced current.

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**The sign of potential difference across an inductor**

∆VL = -L dI/dt ∆VL decreases in the direction of current flow if current is increasing. ∆VL increases in the direction of current flow if current is decreasing. ∆VL is measured in the direction of current in the circuit

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**Comparison of R and L in a simple circuit**

e=-L(DI/Dt) e=-IR R is a measure of opposition to the current L is a measure of opposition to the rate of change in current

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example Electrons are going around a circle in a counterclockwise direction as shown. At the center of the circle they produce a magnetic field that is: e A. into the page B. out of the page C. to the left D. to the right E. zero A

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Physics C 4/12/2017 example : How much current flows through the resistor? How much power is dissipated by the resistor? 50 cm B = 0.15 T 3 W v = 2 m/s Bertrand

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Physics C Inductor, L 4/12/2017 L e eL i When switch is first closed, eL opposes emf of cell. Bertrand

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**eL Inductor, L e When switch is opened, eL supports emf of cell. L i**

Physics C Inductor, L 4/12/2017 L e i eL When switch is opened, eL supports emf of cell. Bertrand

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**Energy Stored in Magnetic Field**

Energy density = Magnetic Energy per unit volume (J/m3) Energy density = B2 / (2 m0) Example: 1 Tesla field Energy density = (1 T)2 / [8p·10-7T ·m /A] = 3.98 ·105 T·A / m = 3.98 ·105 T·A ·m / m2 = 3.98 ·105 N / m2 = 3.98 ·105 J / m3

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Physics C 4/12/2017 example : A coil has an inductance of 3.00 mH and the current changes from A to 1.5 A in a time of s. Find the magnitude of the average induced emf in the coil during this time. Bertrand

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example

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example What e.m.f. will be induced in a 10 H inductor in which current changes from 10A to 7A in 9x10-2 s ? Solution: L= 10H, I1= 10A, I2= 7A, dt= 9x10-2s 19x10 s ε= -L dI/dt, = -L (I2-I1)/dt = -10 (7-10)/ 9x10-2 = Volt

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Example i The current in a 10 H inductor is decreasing at a steady rate of 5 A/s. If the current is as shown at some instant in time, what is the magnitude and direction of the induced EMF? Magnitude = (10 H)(5 A/s) = 50 V Current is decreasing Induced emf must be in a direction that OPPOSES this change. So, induced emf must be in same direction as current (a) 50 V (b) 50 V

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Example : Find the inductance of a uniformly wound solenoid having N turns and length . Assume that is much longer than the radius of the windings and that the core of the solenoid is air. We can assume that the interior magnetic field due to the source current is uniform and given by Equation where n = N/ is the n umber of turns per unit length. The magnetic flux through each turn is : where A is the cross-sectional area of the solenoid. Using this expression and Equation we find that : Because N = n , we can express the result in the form : V = volume of the solenoid = A

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**Example : Calculating Inductance and emf**

Calculate the inductance of an air-core solenoid containing 300 turns if the length of the solenoid is 25.0 cm and its cross-sectional area is 4.00 cm2. Calculate the self-induced emf in the solenoid if the current through it is decreasing at the rate of 50.0A/s. Solution for (a) Using Equation (32.4), we obtain : Solution for (b) Using Equation (32.1) and given that dI/dt = -50.0A/s, we obtain :

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example An inductor is made by tightly winding 0.30 mm diameter wire around a 4.0 mm diameter cylinder. What length cylinder has an inductance of 0.01 mH?

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**example A 10.0 A current passes through a 10 mH inductor coil.**

What potential difference is induced across the coil if the current drops to zero in 5 ms?

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**Energy in Inductors and Magnetic Fields**

A magnetic field stores considerable energy. Therefore, an inductor, which operates by creating a magnetic field, stores energy. Let’s consider how much energy UL is stored in an inductor L carrying current I:

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example A solenoid of radius 2.5cm has 400 turns and a length of 20 cm. Find (a) its inductance and (b) the rate at which current must change through it to produce an emf of 75mV.

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example

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example

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**Example: Current I increases uniformly from 0 to 1 A. in **

0.1 seconds. Find the induced voltage across a 50 mH (milli-Henry) inductance. i + - EL Apply: Substitute: Negative result means that induced EMF is opposed to both di/dt and i.

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example a) At equilibrium (infinite time) how much energy is stored in the coil? E = 12 V L = 53 mH

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example

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example What is the magnetic energy stored in a3-mH inductor when the current through it is 4 mA? 24 × 10-9 joule or 2.4 × 10-8 joule

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example What happens to the energy stored by an inductor when the current through it is doubled? Its energy is quadrupled (i.e. 4 times of original)

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example The current through a 4-mH inductor is varying as follows: I = 2t What will be the induced emf at t = 1 second? ε = -24 mV

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Example Calculate the inductance of a solenoid with 100 turns, a length of 5.0 cm, and a cross sectional area of 0.30 cm2. L = m0N2A l L = (4p X 10-7 T m/A)(100)2(3 X 10-5m2) (0.05 m) L = 7.5 X 10-6 H or 7.5 mH

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Example The same solenoid is now filled with an iron cores (m = 4000 m0). Calculate the inductance L = (4000)(7.5 X 10-6H) L = H or 30 mH

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**Energy stored in a Magnetic Field**

Physics C 4/12/2017 Energy stored in a Magnetic Field UB = ½ L I2 UB : energy stored in magnetic field L: inductance in Henrys I: current in amperes Bertrand

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Inductance (review) Increasing current in a coil of wire will generate a counter emf which opposes the current. Applying the voltage law allows us to see the effect of this emf on the circuit equation. The fact that the emf always opposes the change in current is an example of Lenz's law. The relation of this counter emf to the current is the origin of the concept of inductance. The inductance of a coil follows from Faraday's law.

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Inductance (review) Inductance of a coil: For a fixed area and changing current, Faraday's law becomes Since the magnetic field of a solenoid is then for a long coil the emf is approximated by

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Inductance (review) From the definition of inductance we obtain

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**Energy in a Magnetic Field**

The energy stored in the inductor Battery Power Inductor Power Resistor Power For energy density, consider a solenoid

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