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Electromagnetic Induction Inductors. Problem A metal rod of length L and mass m is free to slide, without friction, on two parallel metal tracks. The.

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Presentation on theme: "Electromagnetic Induction Inductors. Problem A metal rod of length L and mass m is free to slide, without friction, on two parallel metal tracks. The."— Presentation transcript:

1 Electromagnetic Induction Inductors

2 Problem A metal rod of length L and mass m is free to slide, without friction, on two parallel metal tracks. The tracks are connected at one end so that they and the rod form a closed circuit (see following slide). The rod has a resistance R, and the tracks have a negligible resistance. A uniform magnetic field is perpendicular to the plane of this circuit. The magnetic field is increasing at a constant rate dB/dt. Initially the magnetic field has a strength B 0 and the rod is at rest at a distance x 0 from the connected end of the rails. Express the acceleration of the rod at this instant in terms of the given quantities. A metal rod of length L and mass m is free to slide, without friction, on two parallel metal tracks. The tracks are connected at one end so that they and the rod form a closed circuit (see following slide). The rod has a resistance R, and the tracks have a negligible resistance. A uniform magnetic field is perpendicular to the plane of this circuit. The magnetic field is increasing at a constant rate dB/dt. Initially the magnetic field has a strength B 0 and the rod is at rest at a distance x 0 from the connected end of the rails. Express the acceleration of the rod at this instant in terms of the given quantities.

3 Problem diagram

4 Answer The magnetic flux [Phi] enclosed by the rod and the tracks at time t = 0 s is given by

5 Answer continued The magnetic field is increasing with a constant rate, and consequently the enclosed magnetic flux is also increasing:

6 Faraday's law of induction can now be used to determine the induced emf:

7 As a result of the induced emf a current will flow through the rod with a magnitude equal to The direction of the current is along the wire, and therefore perpendicular to the magnetic field. The force exerted by the magnetic field on the rod is given by

8 Combining these two equations we get the following equation:

9 Answer to the problem The acceleration of the rod at time t = 0 s is therefore equal to

10 Problem 2 A long solenoid has 300 turns of wire per meter and has a radius of 3.0 cm. If the current in the wire is increasing at a rate of 50 A/s, at what rate does the strength of the magnetic field in the solenoid increase ? A long solenoid has 300 turns of wire per meter and has a radius of 3.0 cm. If the current in the wire is increasing at a rate of 50 A/s, at what rate does the strength of the magnetic field in the solenoid increase ? The solenoid is surrounded by a coil with 120 turns. The radius of this coil is 6.0 cm. What induced emf will be generated in this coil while the current in the solenoid is increasing ? The solenoid is surrounded by a coil with 120 turns. The radius of this coil is 6.0 cm. What induced emf will be generated in this coil while the current in the solenoid is increasing ?

11 a) The magnetic field in a solenoid was discussed in an earlier chapter. If the solenoid has n turns per meter and if I is the current through each coil than the field inside the solenoid is equal to In this problem n = 300 turns/meter and dI/dt = 50 A/s. The change in the magnetic field is thus equal to This equation shows that the magnetic field is increasing at a rate of T/s. Therefore

12 b) Since the magnetic field in the solenoid is changing, the magnetic flux enclosed by the surrounding coil will also change. The flux enclosed by a single winding of this coil is where r in = 3.0 cm is the radius of the solenoid. Here we have assumed that the strength of the magnetic field outside the solenoid is zero. The total flux enclosed by the outside coils is equal to The rate of change of the magnetic flux due to that change in magnetic field is given by As a result of the change in the current in the solenoid an emf will be induced in the outer coil, with a value equal to

13 A changing current in a conductor (like a coil) produces a changing magnetic field. This time- dependent magnetic field can induce a current in a second conductor if it is placed in this field. The emf induced in this second conductor, [epsilon] 2, will depend on the magnetic flux through this conductor:

14 The flux [Phi] B1 depends on the strength of the magnetic field generated by conductor 1, and is therefore proportional to the current I 1 through this conductor:

15 Here, the constant L 21 depends on the size of the two coils, on their separation distance, and on the number of turns in each coil. The constant L 21 is called the mutual inductance of the two coils. Using this constant, we can write this equation

16 The unit of inductance is the Henry (H) and from the last equation we conclude that

17 When the magnetic field generated by a coil changes (due to a change in current) the magnetic flux enclosed by the coil will also change. This change in flux will induce an emf in the coil, and since the emf is due to a change in the current through the coil it is called the self-induced emf. The self-induced emf is equal to In this equation L is called the self inductance of the coil. The self- induced emf will act in such a direction to oppose the change in the current.

18 Problem 3 A long solenoid of radius R has n turns per unit length. A circular coil of wire of radius R' with n' turns surround the solenoid. What is the mutual induction ? Does the shape of the coil of wire matter ? A long solenoid of radius R has n turns per unit length. A circular coil of wire of radius R' with n' turns surround the solenoid. What is the mutual induction ? Does the shape of the coil of wire matter ?

19 The field inside the solenoid is assumed to be that of an infinitely long solenoid and has a strength equal to

20 The flux enclosed by the external coil is equal to

21 The induced emf in the external coil is equal to

22 The mutual inductance L 12 is thus equal to


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