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Mutual Inductance AP Physics C Montwood High School R. Casao.

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Presentation on theme: "Mutual Inductance AP Physics C Montwood High School R. Casao."— Presentation transcript:

1 Mutual Inductance AP Physics C Montwood High School R. Casao

2 The magnetic flux through a circuit varies with time because of varying currents in a nearby circuit. This gives rise to an induced EMF through a process known as mutual induction because it depends on the interaction of two circuits. Consider the two closely wound coils shown in cross- section. The current I 1 in coil 1, which has N 1 turns, creates magnetic field lines, some of which pass through coil 2, which has N 2 turns.

3 The corresponding flux through coil 2 produced by coil 1 is represented by Φ 21. The mutual inductance M 1 on 2 of coil 2 with respect to coil 1 is the ratio of N 2 ·Φ 1 on 2 and the current I 1 : The mutual inductance depends on the geometry of both circuit elements and on their orientation with respect to one another.

4 As the circuit separation increases, the mutual inductance decreases since the magnetic flux linking the circuits decreases. If the current I 1 varies with time, the EMF induced in coil 2 by coil 1 is given by:

5 If the current I 2 varies with time, the induced EMF in coil 1 due to coil 2 is given by:

6 These results are similar in form to the expression for the self-induced EMF; The EMF induced by mutual induction in one coil is always proportional to the rate of current change in the other coil. If the rates at which the currents change with time are equal (that is, if dI 1 /dt = dI 2 /dt), then EMF 1 = EMF 2. Therefore, M 2 on 1 = M 1 on 2 = M, and Mutual inductance unit: Henry

7 Mutual Inductance of Two Solenoids An electric toothbrush has a base designed to hold the toothbrush handle when not in use. The handle has a cylindrical hole that fits loosely over a matching cylinder on the base. When the handle is placed on the base, a changing current in a solenoid inside the base cylinder induces a current in a coil inside the handle.

8 This induced current charges the battery in the handle. We can model the base as a solenoid of length l with N B turns, carrying a source current I, and having a cross-sectional area A. The handle coil contains N H turns. Find the mutual inductance of the system. The base solenoid carries a source current I, the magnetic field in its interior is:

9 The magnetic flux Φ BH through the handle’s coil is caused by the magnetic field of the base coil and is equal to B·A. Mutual inductance:

10 Calculating Mutual Inductance In one form of Tesla coil (a high-voltage generator), a long solenoid with length l and cross-sectional area A is closely wound with N 1 turns of wire. A coil with N 2 turns surrounds it at its center. Find the mutual inductance.

11 Mutual inductance occurs because a current in one of the coils sets up a magnetic field that causes a flux thru the other coil. Using To determine the mutual inductance, we need to know either the flux thru each turn of the outer coil due to current I 1 in the solenoid or the flux thru each turn of the solenoid due to a current I 2 in the outer coil.

12 The magnetic field for a solenoid is: B = μ o ·(N/ l) ·I 1 The flux thru a cross section of the solenoid is B 1 ·A A very long solenoid produces no magnetic field outside of its coil, so the flux through each turn of the outer, surrounding coil is equal to the flux thru the solenoid, no matter what the cross-sectional are of the outer coil.

13 Suppose l = 0.5 m, A = m 2, N 1 = 1000 turns, and N 2 = 10 turns:

14 EMF Due to Mutual Induction In the previous example, suppose the current I 2 in the outer, surrounding coil is given by I 2 = 2 x 10 6 A/s·t. –At time t = 3 μs, what average magnetic flux through each turn of the solenoid is caused by the current in the outer surrounding coil? –Given the value of the mutual inductance from the previous problem, M = 25 x H, determine the flux thru each turn of the solenoid caused by a given current I 2 in the outer coil. –At t = 3 μs = 3 x s, the current in the outer coil (coil 2) is I 2 = 2 x A/s· 3 x s = 6 A. –The average flux through each turn of the solenoid (coil 1) is:

15 This is an average value for the flux; the flux can vary considerably between the center and the ends of the solenoid. What is the induced EMF in the solenoid?


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