the wires are separated by distance a and carry currents i 1 and i 2 in the same direction. wire 2,...

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• The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2 , sets up a magnetic field B 2 at the position of wire 1. - The direction of the magnetic field B 2 is perpendicular to the wire.

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Page 1: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

• The wires are separated by distance a and carry currents I1 and I2 in the same direction.

• Wire 2, carrying current I2, sets up a magnetic field B2 at the position of wire 1.

- The direction of the magnetic field B2 is perpendicular to the wire. - F1 = F2 on 1 = I1·(L x B2) - Angle q between L and B2

is 90.

Page 2: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2
Page 3: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

Force between two parallel current-carrying straight wires

1. Parallel wires with current flowing in the same direction, attract each other.

2. Parallel wires with current flowing in the opposite direction, repel each other.

Page 4: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

• When the currents are in opposite directions, the magnetic forces again equal in magnitude but are opposite in direction and the wires repel each other.

Page 5: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

Recall:

Page 6: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

The Magnetic Field inside a Current Carrying Solenoid

• The magnetic field inside a coil, loop of wire, or a solenoid, is uniform and proportional to the current, the number of loops of wire N, and the length of the solenoid l.

• The is can be expressed with the following equation:

Where n is the number of loops per unit length:

To determine the direction of the field we need to either use the first or second right hand rule.

Page 7: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

Direction of magnetic field inside a solenoid:

The magnetic field near the center of a single coil carrying a current I.

The magnetic field near the center of a set of three coils all carrying a current I.

A solenoid approximates many current-carrying coils spaced much more closely than the coil diameter.

2nd RIGHT-HAND RULE

Page 8: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

To determine the direction of the magnetic field inside and around a solenoid you can use the first

or second right hand rule.• 2nd RIGHT-HAND RULE• If you grasp the coil

with the right hand and curl your fingers around the loops in the direction of the conventional current, your thumb points towards the North pole of the electromagnet.

Page 9: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

Examples:

• Ex 1: A 0.400 m long solenoid has 6 720 turns of wire. A current of 14.5 A flows in the solenoid. What is the direction and size of the magnetic field inside the solenoid?

(Picture shown above right)

• Ex 2: An electron inside the solenoid travels perpendicular to the axis of the solenoid with a speed of 6.50 x105 m/s. What is the magnitude and direction of the magnetic force acting on the electron?

Page 10: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

Electromagnetism and Cathode Ray Tubes

• Instead of having an electric field create a force on an electron to deflect its path, a magnetic field can also be used.

• Unlike using charged parallel plates to cause deflection, the velocity of the electron doesn’t change the deflection caused by a magnetic field.

• By increasing its velocity, an electron will spend less time in the magnetic field, but the force that it will experience will be greater.

Page 11: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

Ex 3: A solenoid placed beneath a cathode ray tube as shown below produces a magnetic field of 0.011 T on the electron

beam causing it to hit the screen at position 1.

a) The electron beam travels at 4.7 x 107 m/s. What is the acceleration of the electrons that make up the beam? (9.1 x 1016 m/s2)b) What direction is the magnetic field caused by the solenoid? (down)c)What two changes can be made to the current in the solenoid in order for the beam to now strike the screen at position #2? (reduce current, change direction)

Page 12: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

• Ex 4: The diagrams to the right show two views of an electron beam in a cathode ray tube. When the accelerating voltage is 100.0 V, the deflection of the beam is zero. Which way will the beam be deflected if the accelerating voltage is changed to 400.0 V? Explain your reasoning using the principles of physics

Page 13: The wires are separated by distance a and carry currents I 1 and I 2 in the same direction. Wire 2, carrying current I 2, sets up a magnetic field B 2

Practice:

1. Protons travelling at 5.2 X 105 m/ s enter at right angles to a magnetic field. The field is produced by a 0.36 m long solenoid. A current of 6.3 A flows through the 520 turns of wire of the solenoid. What is the magnetic field in the solenoid?2. What is the strength of the magnetic field in the center of a solenoid 15 cm in length if the solenoid has 5.0 x 102 turns of wire and carries a current of 5.0 A? 3. The magnetic field at the center of a solenoid of length 0.25 m is 1.2 x 10-2 T. The current in the windings is 7.5 A. How many windings does the solenoid have? (1.3 x 103)