Alternating Current 35d34

Physics Alternating Current

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Magnetism and Magnetic Fields  magnetic field: north experiences definition on the north magnet.

a region of space where a magnetic monopole a force. The direction of the field is by the direction of the force end of a The figure shows the lines of magnetic fields from a bar magnet form closed lines. By convention, the field direction is taken to be outward from the North pole and in to the South pole of the magnet. 2

The earth behaves magnetically almost as if a bar magnet were located near its center. The axis of this fictitious bar magnet does not coincide with the earth’s rotational axis: the two axes are currently about 11.5 0 apart.

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Magnetic Field of Current

The magnetic field lines around a long wire which carries an electric current form concentric circles around the wire. The direction of the magnetic field is perpendicular to the wire and is in the direction the fingers of your right hand would curl if you wrapped them aroundinthe wire withField your thumb in the direction Charges a Magnetic of the current. Use your right hand to determine the direction of for

on a moving positively charged particle a magnetic field. With the fingers from so to north (the same direction as the field), and the thumb pointing in the direction of the velocity of the particle, the pa points in the direction of the force on the particle. For a negative use your left hand.

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Electromagnetic Induction MAGNETIC

FLUX

  Magnetic

flux may be thought of as an amount of magnetic field ing through an area. The following diagram depicts a magnetic field directed away from the observer. A particular area of the field has been enclosed in a rectangle.

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There are several ways of increasing magnet flux. One way is to use a larger rectangle.

Or, another is to use a stronger field:

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Magnetic flux therefore depends on field strength, B, and on area, A. Twice as much of either one gives you twice as much magnetic flux. Also, magnetic flux depends on the angle between B and A. The greatest amount of magnetic flux is when A and B are perpendicular.

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Reducing the angle between A and B reduces the amount of f assing through the area. Therefore the magnitude of magne ux, ƒ, is

ƒ = Bq0vsinθ

Eqn. 1

where f is the magnitude of the magnetic fo a positive test charge q0

B is magnitude of the magnetic field at any point in space

v is the velocity of the charge and makes a an angle θ (0 < θ < 1800) with the direction of the magnetic field. SI Unit of Magnetic Field: newton-second/coulomb-me = 1 Tesla (T) 8

Nikola Tesla, born July 9/10, 1856, Smiljan, Austria-Hungary Croatia] died Jan. 7, 1943, New York City, N.Y., U.S.

Serbian-American inventor and engineer, discovered and patented the rotating magnetic field, field the basis of most alternating current machinery. He also developed the three-phase system of electric power transmission. He emigrated to the United States in 1884 and sold the patent rights to his system of alternating-current dynamos, transformers, and motors to George Westinghouse. In 1891 he invented the Tesla coil, an induction coil widely used in radio. The unit of magnetic field strength is the N/(C-m), called a tesla, a tribute to this great man.

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One tesla is the strength of the magnetic field in which a unit test charge, travelling perpendicular to the magnetic field with a speed of one meter per second, experiences a force of one newton. Because a coulomb per second is an ampere, the tesla is often written as: 1 Tesla = N/(A-m) Another convenient unit to use is the Gauss: 1 gauss = 10-4 tesla 10

The magnetic field is defined in of the magnetic force and a moving test charge. The photo shows an aurora borealis (“northern lights”) display over silhouetted trees. Charged particles from the sun are captured by the earth’s magnetic field. When the particles collide with the gas molecules in the upper atmosphere, curtains of colorful light are often formed.

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The following two conditions must be met for a charge to experience a magnetic force when placed in a magnetic field. 1. The charge must be moving, for no magnetic force acts on a stationary charge. 2. The velocity of a moving charge must have a component that is perpendicular to the direction of the magnetic field.

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Example 1.

Magnetic Forces on Charged Particles

A proton in a particle accelerator has a speed of 5.0 x 106 m/s. The proton encounters a magnetic field whose magnitude is 0.40 T and whose direction makes an angle of θ = 30.00 with respect to the proton’s velocity. Find (a) the magnitude and direction of the magnetic force on the proton and (b) the acceleration of the proton. (c) What would be the force and acceleration if the particle were an electron instead of a proton? Reasoning For both the proton and the electron, the magnitude of the magnetic force are given by equation 1 but have opposite directions, because the charges have opposite signs. In either case, the acceleration is given by Newton’s second law, which applies to the magnetic force just as it does to any force. In using the second law, we must take into the fact that the masses of the proton and the electron are different.

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Solution (c)The positive charge on a proton is 1.60 x 10-19 C. The magnitude of the magnetic force is given by f = q0vBsin θ:

f = (1.60 x 10-19 C)(5.0 x 106 m/s)(0.40 T)(sin 300) = 1.60 x 10-13N Ans.) (b) The proton’s acceleration follows directly from Newton’s second law. Since the only force acting on the proton is the magnetic force f, it is the net force. Thus…

a = f/mP = (1.6 x 10-13 N)/1.67 x 10-27 kg = 9.6 x 1013 m/s2 Ans (c) The magnitude of the magnetic force on the electron is the same as that on the proton, since both have the same speed and charge magnitude. However, the direction of the force on the electron is opposite to that on the proton, or downward. a = f/mE = (1.6 x 10-13 N)/(9.11 x 10-31 kg = 1.9 x 1017 m/s2 Ans

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EXAMPLE 2. .An alpha particle (two protons and two neutrons) traveling east at 2.0 x 105 m/s enters a magnetic field of 0.20 T pointing straight up. What is the force acting on the alpha particle? Solution With the fingers of the right hand pointing straight up, and the thumb pointing east, the palm points south. F = qvBsinø = (2 x 1.6 x 10-19 C)(2.0 x 105 m/s)(0.20 T)sin90º = 1.28 x 10-14 N [S]. Example 3. An electron traveling to the left, moves into a magnetic field directed toward the observer. Trace the path of the particle, assuming it eventually leaves the field. 15

Solution The moment the electron enters the field, it experiences a force perpendicular to its velocity. The electron follows a circular path until it leaves the field.

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To understand why the path is circular consider two points on the circumference labeled 1 and 2. When the positively charged particle is at point 1, the magnetic force f is perpendicular to the velocity v and points directly upward. When the particle reaches point 2, the magnetic force still remains perpendicular to the velocity but is now directed to the left in the drawing.

The magnetic force always remains perpendicular to the velocity and is directed toward the center of the circular path.

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