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## Physics Homework Help

### Magnetic Flux

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

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

Or, another is to use a stronger field.

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.

Reducing the angle between A and B reduces the amount of field passing through the area.

Therefore the magnitude of magnetic flux, φ, is
φ = BAsinø where ø is the angle between B and A.

#### MAGNETIC FLUX PROBLEM

1. What is the magnetic flux through a table top 1.5 m by .75 m inclined 70º to a magnetic field of 1.33 T?

(See bottom of page for answers.)

### ELECTROMAGNETIC INDUCTION AND FARADAY'S LAW

Whenever a conductor moves relative to a magnetic field, electric potential, and therefore current, is induced in the conductor. In more sophisticated language, electric potential is induced in a conductor whenever flux through the conductor changes. A simple generator is a conductor loop rotating in a magnetic field. All functional generators, no matter how large, are based on this concept. The average electric potential produced, regardless of how it is produced, is given by Faraday's Law:
V = -NΔφ/Δt .
Here N is the number of loops, Δφ is the change in flux, and Δt is the time required for the change.
Notice that Δφ may arise out of one or a combination of ΔA, ΔB, or Δsinø.

#### ELECTROMAGNETIC INDUCTION AND FARADAY'S LAW PROBLEMS

2. A 500-turn armature (a rotating solenoid) with a cross-sectional area of 9 cm2 rotates at 120 rpm in a 4.00 T magnetic field. What is the average electric potential produced when the armature rotates from 0º to 90º to the field?

(See bottom of page for answers.)

### LENZ'S LAW

To determine the direction of the current produced when electric potential is induced, we use Lenz's Law: the induced current flows in a direction that opposes the change that induced the current. This is more easily understood through an example.

In the following example the permanent magnet moves to the left. What is the direction of the current through the resistor?

The movement of the north end of the permanent magnet away from the solenoid induces electric potential in the solenoid. To oppose the motion of the magnet, the left end of the solenoid becomes south, attracting the magnet. The attraction is not strong enough to prevent the movement; it just offers resistance to the movement.

Using the right hand rule for solenoids, we point the thumb of the right hand along the direction of the field through the solenoid (ie. to the right). When we "grab" the solenoid with our right hand, the fingers curl upward behind the solenoid and come over top the solenoid and down in front of the solenoid. This is the direction of conventional current flow through the solenoid. (For electron flow use the left hand.) Since the current flows downwards in front of the solenoid, it must travel to the right through the resistor.

#### LENZ'S LAW PROBLEMS

(See bottom of page for answers.)

3. A conductor loop falls down into a magnetic field directed toward the observer (out of the page). What is the direction of conventional current induced in the loop?

4. A conducting loop falls toward a section of conductor carrying conventional current to the right. In what direction does conventional current flow through the loop?

5. A conducting loop moves to the left above a conductor carrying conventional current to the right. What direction does conventional current flow through loop?

6. A conducting loop falling straight down through a uniform magnetic field directed toward the observer. What direction does conventional current flow through loop?

### SELF -INDUCTANCE

As long as the current through a loop does not change, B is constant and flux is constant. However, if current increases or decreases, B changes, and a change in flux is produced. The change in flux is a source of electric potential in the circuit that opposes the original current. This opposing electric potential is called "back-emf." Back-emf reduces the power in the circuit by reducing the current. This power loss can occur in any circuit where there is a change in flux. Back-emf is common in AC circuits and is particularly important in any situation involving motors or generators.

#### SELF INDUCTANCE PROBLEMS

7. Why would an electric motor burn out when it is not turning if it does not burn out while it is turning?

(See bottom of page for answers.)

### MUTUAL INDUCTANCE AND TRANSFORMERS

Transformers make use of mutual inductance. In this process the changing magnetic field produced by the "primary" coil induces electric potential in the "secondary" coil. The primary coil is the one attached to an alternating current power source. An alternating current ("AC") increases then decreases in one direction then reverses, increasing then decreasing in that direction. The changing current creates a changing magnetic field. The changing magnetic field induces electric potential in the nearby secondary coil, which is attached to a load.

Every one who lives in a developed country depends on transformers. Transformers "step down" high voltage power delivered from hydro dams or other far away generators to 240 V delivered to the house. Other devices such as televisions and computer monitors use transformers to step up voltage from the 120 V wall outlets supply to the 2500 V needed by CRT's (television tubes).

Whether the voltage is stepped up or down and by how much depends on the relative number of turns on the two coils. The basis of this wizardry is the Law of Conservation of Energy. Power in an electric circuit is directly proportional to current and voltage:
P = IV .
Ideally the power developed in the secondary coil would be the same as developed in the primary:
Pp = Ps .

Ignoring the heat loss that occurs we will take this as an accurate statement: therefore,
IpVp = IsVs.

Rearranging this equality we are able to compare the relative voltages and currents in the primary and secondary coils:
Ip/Is = Vs/Vp .

What this last equality is saying is that you don't get something for nothing. If you use a transformer to double the voltage, you only get half the current. At best, ignoring heat losses, you get the same power from the secondary as delivered to the primary.

As indicated by Faraday's law, the electric potential induced is directly proportional to the number of loops or >turns in the coil. Therefore,
Ns/Np = Vs/Vp = Ip/Is

In functioning transformers, hundreds or thousands of turns are used. The transformer depicted above has
Np = 4 and Ns = 2. Therefore
Ns/Np = Vs/Vp = 0.5

Therefore, if 120 V was delivered to the primary coil,
0.5 = Vs/(120 V)
and the voltage delivered to the load in the secondary coil would be
Vs = 60 V.
This would be a step-down transformer.

If the source delivered 2 A of current to the primary coil (Ip = 2 A) then
Ns/Np = Ip/Is and
0.5 = (2 A)/Is and
Is = 4 A .

(At best we get half the voltage; twice the current. Nothing gained; nothing lost.)

#### MUTUAL INDUCTANCE AND TRANSFORMERS PROBLEMS

(See bottom of page for answers.)

8. A transformer is used to change a 120 V, 3 A current to 2500 V.
a. What kind of transformer is this?
b. What is the ratio of secondary turns to primary turns?
c. What current would be developed in the secondary coil?

9. A step-up transformer has 100 turns on the primary coil and 500 turns on the secondary coil. If this transformer is to produce an output of 4300 V with a 12 mA current, what input current and voltage are needed?

10. The average emf induced in the secondary coil is 0.12 when the current in the primary coil changes from 3.4 to 1.6 A in 0.14 s. What is the mutual inductance of the coils?

11. A circular coil with 233 turns and a diameter of 23.5 cm rotates about a vertical axis at 1250 rpm. The coil is situated in a magnetic field having a horizontal component of 3.80x10-5 T, and a vertical component of 2.85 x10-5 T. What is the maximum EMF produced in the coil?

12. The current in an air-core solenoid is reduced from 3.99 A to zero over 5.9s. The solenoid has 2000 turns per meter and a cross-sectional area of 0.131 m2. Surrounding the solenoid near the center of its length is a second coil of 50 turns.

a. What is the magnitude of the induced emf in the second coil?

b. If the resistance of the second coil is 0.00409 ohm what is the induced current?

13. A 200-turn air-core solenoid with a cross-sectional area of 100 cm2 has a resistance of 5.0 ohms.  The ends of the wire are joined together to close the circuit.  A 1.1 T magnetic field is directed through the coil perpendicular to its cross-sectional area.  Over a period of 0.1s, the field is reversed.  What average current flows through the coil during that period?

14. If the blade of an electric lawn mower jams and prevents the motor shaft from rotating, the electric motor can burn out. Explain why this happens.

### Answers to Electromagnetic Induction Problems

Selected solutions are printed below.

1. φ = BAsinø = (1.33 T)(1.5 x .75 m2)sin70º = 0.937 Wb (Webers)

2. N = 500

Δt = (0.25 rotation)(1 minute/120 rotations)(60 seconds/minute) = 0.125 s
Δφ = BA(sinøf - sinøi)

= (4.00 T)(9.00 x 10-4 m2)(sin 90º - sin 0º) = 3.6 x 10-3 Wb

V = -NΔφ/Δt = -(500 turns)(3.6 x 10-3 Wb)/(0.125 s) = -14.4 V