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Energy In a Magnetic Field AP Physics C Montwood High School R. Casao.

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Presentation on theme: "Energy In a Magnetic Field AP Physics C Montwood High School R. Casao."— Presentation transcript:

1 Energy In a Magnetic Field AP Physics C Montwood High School R. Casao

2 The induced EMF set up by an inductor prevents a battery from establishing an instantaneous current. Therefore, a battery has to do work against an inductor to create a current. Part of the energy supplied by the battery goes into joule heat dissipated in the resistor, while the remaining energy is stored in the inductor. Beginning with Kirchhoffs equation for an RL circuit in which the current is increasing, multiply each term by the current I and rearrange:

3 The rate at which energy is supplied by the battery, EMF·I, is equal to the sum of the rate at which joule heat is dissipated in the resistor, I 2 ·R, and the rate at which energy is stored in the inductor, L·I·dI/dt. The equation is an expression of the law of conservation of energy. If U m is the energy stored in the inductor at any time, then the rate dU m /dt at which energy is stored in the inductor is:

4 To find the total energy stored in the inductor, rewrite the equation as: This equation represents the energy stored as magnetic energy in the field of the inductor when the current is I. After the current has reached its final steady state value I, dI/dt = 0 and no more energy is input into the inductor.

5 When there is no current, the stored energy U m is 0 J; when the current is I, the stored energy U m is 0.5·L·I 2. When the current decreases from I to zero, the inductor acts as a source that supplies a total amount of energy 0.5·L·I 2 to the external circuit. If we interrupt the circuit suddenly by opening a switch or yanking a plug from a wall socket, the current decreases very rapidly, the induced EMF is very large, and the energy may be dissipated in an arc across the switch contacts. This large EMF is the electrical equivalent to the large force exerted by a car running into a brick wall and stopping very suddenly.

6 It is important not to confuse the behavior of resistors and inductors where energy is concerned. Energy flows into a resistor whenever a current passes through it, whether the current is steady or varying; this energy is dissipated in the form of heat. Energy flows into an ideal, zero-resistance inductor only when the current in the inductor increases. The energy is not dissipated; it is stored in the inductor and released when the current decreases. When a steady current flows through an inductor, there is no energy flow in or out.

7 The energy in an inductor is actually stored in the magnetic field within the coil, just as the energy of a capacitor is stored in the electric field between the plates. You can also determine the energy per unit volume, or energy density, stored in a magnetic field. Consider a solenoid whose inductance is given by L = μ o ·n 2 ·A· l. The magnetic field of the solenoid is given by B = μ o ·n 2 ·I.

8 Substituting:

9 Because A· l is the volume of the solenoid, the energy stored per unit volume in a magnetic field is:

10 Although the magnetic energy density equation was derived for a solenoid, it is valid for any region of space in which a magnetic field exists. What Happens to the Energy in the Inductor? Consider the RL circuit shown. Recall that the current in the right-hand loop decays exponentially with time according to the equation: where I o = EMF/R is the initial current in the circuit and τ = L/R is the time constant.

11 Show that all the energy initially stored in the magnetic field of the inductor appears as internal energy in the resistor as the current decays to zero. The rate at which energy is dissipated in the resistor, dU/dt (or the power), is equal to I 2 ·R, where I is the instantaneous current. To find the total energy dissipated in the resistor, integrate the equation from t = 0 s to t =.

12

13 Note that this is equal to the initial energy stored in the magnetic field of the inductor. The Coaxial Cable A long coaxial cable consists of two concentric cylindrical conductors of radii a and b and length l. The inner conductor is assumed to be a thin cylindrical shell.

14 Each conductor carries a current I (the outer one being a return path). Calculate the self-inductance L of the cable. –To obtain L, we must know the magnetic flux through any cross- section between the two conductors. –From Amperes law, it is easy to see that the magnetic field between the conductors is given by

15 –The magnetic field is zero outside the conductors and zero inside the inner hollow cylinder. –The magnetic field is zero outside the conductors because the net current through a circular path surrounding both wires is zero. –The magnetic field inside the inner conductor because there is no current within the hollow inner cylinder. The magnetic field is perpendicular to the shaded rectangular strip of length l and width b – a.

16 Dividing the rectangle into strips of width dr, the area of each strip is l ·dr and the magnetic flux through each strip is B · dA = B· l ·dr. The total magnetic flux through any cross-section is:

17 The self-inductance of the coaxial cable is:

18 Calculate the total energy stored in the magnetic field of the cable.

19 For the homework, remember that the current I as a function of time for an RC circuit that is increasing is:


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