What is the peak emf generated by a 0. It has a 1. How many turns are in the coil? Integrated Concepts This problem refers to the bicycle generator considered in the previous problem. It is driven by a 1. Its turn, 5. What is the field strength needed to produce a A turn, Unreasonable Results A turn coil with a 0. Skip to main content. Search for:. Electric Generators Learning Objectives By the end of this section, you will be able to: Calculate the emf induced in a generator. Calculate the peak emf which can be induced in a particular generator system.
Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Induced Emf and Magnetic Flux.
We will now explore generators in more detail. Consider the following example. The coil is attached to a ballistic galvanometer, a device that measures the total charge that passes through it.
The coil is placed in a magnetic field such that its face is perpendicular to the field. It is then flipped through and the total charge Q that flows through the galvanometer is measured. Because the coil is very small, you can assume that is uniform over it. B is proportional to Q ; b. If the coin turns easily, the magnetic field is perpendicular.
If the coin is at an equilibrium position, it is parallel. The flip coil of the preceding problem has a radius of 3. The total resistance of the coil and ballistic galvanometer is When the coil is flipped through in a magnetic field a change of 0. A V, series-wound motor has a field resistance of 80 and an armature resistance of When it is operating at full speed, a back emf of 75 V is generated.
When the motor is operating at full speed, where are b the current drawn by the motor, c the power output of the source, d the power output of the motor, and e the power dissipated in the two resistances?
A small series-wound dc motor is operated from a V car battery. Under a normal load, the motor draws 4. What is the back emf when the motor is operating normally? Skip to content Electromagnetic Induction. Learning Objectives By the end of this section, you will be able to: Explain how an electric generator works Determine the induced emf in a loop at any time interval, rotating at a constant rate in a magnetic field Show that rotating coils have an induced emf; in motors this is called back emf because it opposes the emf input to the motor.
Electric Generators Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Motional Emf. When this generator coil is rotated through one-fourth of a revolution, the magnetic flux changes from its maximum to zero, inducing an emf. A generator with a single rectangular coil rotated at constant angular velocity in a uniform magnetic field produces an emf that varies sinusoidally in time.
Note the generator is similar to a motor, except the shaft is rotated to produce a current rather than the other way around. The emf of a generator is sent to a light bulb with the system of rings and brushes shown. The graph gives the emf of the generator as a function of time, where is the peak emf.
The period is where f is the frequency. Split rings, called commutators, produce a pulsed dc emf output in this configuration. The steam produced by burning coal impacts the turbine blades, turning the shaft, which is connected to the generator. Back Emf Generators convert mechanical energy into electrical energy, whereas motors convert electrical energy into mechanical energy. The coil of a dc motor is represented as a resistor in this schematic.
The back emf is represented as a variable emf that opposes the emf driving the motor. Circuit representation of a series-wound direct current motor. Summary An electric generator rotates a coil in a magnetic field, inducing an emf given as a function of time by where A is the area of an N -turn coil rotated at a constant angular velocity in a uniform magnetic field The peak emf of a generator is.
What is the emf induced? Figure We recognize this situation as the same one in Example According to the diagram, the projection of the surface normal vector to the magnetic field is initially , and this is inserted by the definition of the dot product.
The magnitude of the magnetic field and area of the loop are fixed over time, which makes the integration simplify quickly. We are given that , , , , and. The area of the loop is. This is a practical average value, similar to the used in household power. The emf calculated in Example What is the emf at any given instant?
It varies with the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf on a rotating rectangular coil of width and height in a uniform magnetic field, as illustrated in Figure Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experience forces parallel to the wire, causing currents.
But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires. Motional emf is given to be , where the velocity is perpendicular to the magnetic field. Here the velocity is at an angle with , so that its component perpendicular to is see Figure Thus, in this case, the emf induced on each side is , and they are in the same direction.
The total emf around the loop is then. This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coil rotates at a constant angular velocity. The angle is related to angular velocity by , so that. Now, linear velocity is related to angular velocity by. A bar magnet rests outside a wire coil connected to an ammeter showing no current.
The magnet moves into the coil of wire and the ammeter registers positive current flow. The magnet is stationary within the coil of wire, there is no current flow. The magnet moves out of the coil of wire and the ammeter registers negative current flow.
0コメント