Faraday’s Law and Lenz's Law Electromagnetic Induction Simulation

Quiz

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Q1) Faraday’s law says that

A. an emf is induced in a loop when it moves through an electric field
B. the induced emf produces a current whose magnetic field opposes the original change
C. the induced emf is proportional to the rate of change of magnetic flux

Answer) C.

Q2)  If a coil is shrinking in a B field pointing into the page, in what direction is the induced current?

A. clockwise
B. counter-clockwise
C. no induced current

 Answer) A.
Downward flux is decreasing, so need to create downward B field.

Faraday’s Law Electromagnetic Induction Simulation(Virtual Experiment)

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Faraday and Henry both made the same discovery. Electric current can be produced in a wire by simply moving a magnet into or out of a wire coil. No battery or other voltage source was needed only the motion of a magnet in a coil or in a single wire loop as shown in Figure 1.

 FIGURE 1. When the magnet is plunged into the coil, voltage is induced in the coil and charges in the coil are set in motion.

They discovered that voltage was induced by the relative motion of a wire with respect to a magnetic field. The production of voltage depends only on the relative motion of the conductor with respect to the magnetic field. Voltage is induced whether the magnetic field of a magnet moves past a stationary conductor, or the conductor moves through a stationary magnetic field as shown in Figure 2. The results are the same for the same relative motion.

FIGURE 2. Voltage is induced in the wire loop whether the magnetic field moves past the wire or the wire moves through the magnetic field.

Faraday’s Law of Electromagnetic Induction

To see how an emf can be induced by a changing magnetic field, consider a loop of wire connected to a sensitive ammeter, as illustrated in Figure 3. When a magnet is moved toward the loop, the galvanometer needle deflects in one direction, arbitrarily shown to the right in Figure 3a. When the magnet is brought to rest and held stationary relative to the loop (Fig. 3b), no deflection is observed. When the magnet is moved away from the loop, the needle deflects in the opposite direction, as shown in Figure 3c. Finally, if the magnet is held stationary and the loop is moved either toward or away from it, the needle deflects. From these observations, we conclude that the loop detects that the magnet is moving relative to it and we relate this detection to a change in magnetic field. Thus, it seems that a relationship exists between current and changing magnetic field.

These results are quite remarkable in view of the fact that a current is set up even though no batteries are present in the circuit! We call such a current an induced current and say that it is produced by an induced emf.

Figure 3 (a) When a magnet is moved toward a loop of wire connected to a sensitive ammeter, the ammeter deflects as shown, indicating that a current is induced in the loop. (b) When the magnet is held stationary, there is no induced current in the loop, even when the magnet is inside the loop. (c) When the magnet is moved away from the loop, the ammeter deflects in the opposite direction, indicating that the induced current is opposite that shown in part (a). Changing the direction of the magnet’s motion changes the direction of the current induced by that motion.

Now let us describe an experiment conducted by Faraday and illustrated in Figure 4. A primary coil is connected to a switch and a battery. The coil is wrapped around an iron ring, and a current in the coil produces a magnetic field when the switch is closed. A secondary coil also is wrapped around the ring and is connected to a sensitive ammeter. No battery is present in the secondary circuit, and the secondary coil is not electrically connected to the primary coil. Any current detected in the secondary circuit must be induced by some external agent.

Initially, you might guess that no current is ever detected in the secondary circuit. However, something quite amazing happens when the switch in the primary circuit is either opened or thrown closed. At the instant the switch is closed, the galvanometer needle deflects in one direction and then returns to zero. At the instant the switch is opened, the needle deflects in the opposite direction and again returns to zero.

Figure 4. Faraday’s experiment. When the switch in the primary circuit is closed, the ammeter in the secondary circuit deflects momentarily. The emf induced in the secondary circuit is caused by the changing magnetic field through the secondary coil.

Finally, the galvanometer reads zero when there is either a steady current or no current in the primary circuit. The key to understanding what happens in this experiment is to first note that when the switch is closed, the current in the primary circuit produces a magnetic field in the region of the circuit, and it is this magnetic field that penetrates the secondary circuit. Furthermore, when the switch is closed, the magnetic field produced by the current in the primary circuit changes from zero to some value over some finite time, and this changing field induces a current in the secondary circuit.

As a result of these observations, Faraday concluded that an electric current can be induced in a circuit (the secondary circuit in our setup) by a changing magnetic field. The induced current exists for only a short time while the magnetic field through the secondary coil is changing. Once the magnetic field reaches a steady value, the current in the secondary coil disappears. In effect, the secondary circuit behaves as though a source of emf were connected to it for a short time. It is customary to say that an induced emf is produced in the secondary circuit by the changing magnetic field.

The experiments shown in Figures 3 and 4 have one thing in common: in each case, an emf is induced in the circuit when the magnetic flux through the circuit changes with time. In general,

The emf induced in a circuit is directly proportional to the time rate of change of the magnetic flux through the circuit.

This statement, known as Faraday’s law of electromagnetic induction, can be written

Faraday’s Law $\ E = -N \frac {d\Phi_{B}}{dt}$ where $\Phi_{B} =\int \mathbf{B}\cdot d\mathbf{A}$ is the magnetic flux through the circuit in N loops.

Lenz’s Law

The SI unit for the induced emf is the volt, V. The minus sign in the above Faraday’s law of induction is due to the fact that the induced emf will always oppose the change. It is also known as the Lenz’s law and it is stated as follows,

Lenz’s Law The current from the induced emf will produce a magnetic field, which will always oppose the original change in the magnetic flux.

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Two Bar Magnet Magnetic Field Lines Simulation

Quiz

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Q1) The force between two unlike magnetic poles

A. increases linearly with the separation distance between them
B. increases as the square of the separation distance between them
C. decreases as the separation distance between them
D. decreases as the square of the separation distance between them

Answer) D.

Q2)  An iron rod becomes magnetic when.

a. positive ions accumulate at _me end and negative ions at the other end.
b. its atoms are aligned having plus charges on one side and negative charges on the other.
c. the net spins of its electrons are in the same direction.
d. its electrons stop moving and point in the same direction.
e. none of these.

 Answer) C.

Two Bar Magnet Magnetic Field Lines Simulation(virtual experiment)

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Magnetic Fields Around Permanent Magnets

When you experimented with two magnets, you noticed that the forces between magnets, both attraction and repulsion, occur not only when the magnets touch each other, but also when they are held apart. In the same way that long-range electric and gravitational forces can be described by electric and gravitational fields, magnetic forces can be described by the existence of fields around magnets. These magnetic fields are vector quantities that exist in a region in space where a magnetic force occurs.

 Figure 1 The magnetic field of a bar magnet shows up clearly in three dimensions when the magnet is suspended in glycerol with iron filings (a). It is, however, easier to set up a magnet on a sheet of paper covered with iron filings to see the pattern in two dimensions (b).

The presence of a magnetic field around a magnet can be shown using iron filings. Each long, thin, iron filing becomes a small magnet by induction.

Just like a tiny compass needle, the iron filing rotates until it is parallel to the magnetic field. Figure 1a shows filings in a glycerol solution surrounding a bar magnet. The three-dimensional shape of the field is visible. In Figure 1b, the filings make up a two-dimensional plot of the field, which can help you visualize magnetic field lines. Filings also can show how the field can be distorted by an object. 

Magnetic field lines 

Note that magnetic field lines, like electric field lines, are imaginary. They are used to help us visualize a field, and they also provide a measure of the strength of the magnetic field. The number of magnetic field lines passing through a surface is called the magnetic flux. The flux per unit area is proportional to the strength of the magnetic field. As you can see in Figure 1, the magnetic flux is most concentrated at the poles; thus, this is where the magnetic field strength is the greatest. The direction of a magnetic field line is defined as the direction in which the north pole of a compass points when it is placed in the magnetic field. Outside the magnet, the field lines emerge from the magnet at its north pole and enter the magnet at its south pole, as illustrated in Figure 2. What happens inside the magnet? 

 Figure 2 Magnetic field lines can be visualized as closed loops leaving the north pole of a magnet and entering the south pole of the same magnet.

There are no isolated poles on which field lines can start or stop, so magnetic field lines always travel inside the magnet from the south pole to the north pole to form closed loops.

Figure 3 The magnetic field lines indicated by iron filings on paper clearly show that like poles repel (a) and unlike poles attract (b). The iron filings do not form continuous lines between like poles. Between a north and a south pole, however, the iron filings show that field lines run directly between the two magnets.

What kinds of magnetic fields are produced by pairs of bar magnets? You can visualize these fields by placing two magnets on a sheet of paper, and then sprinkling the paper with iron filings. Figure 3a shows the field lines between two like poles. In contrast, two unlike poles (north and south) placed close together produce the pattern shown in Figure 3b. The filings show that the field lines between two unlike poles run directly from one magnet to the other. 

Forces on objects in magnetic fields 

Magnetic fields exert forces on other magnets. The field produced by the north pole of one magnet pushes the north pole of a second magnet away in the direction of the field line. The force exerted by the same field on the south pole of the second magnet is attractive in a direction opposite the field lines. The second magnet attempts to line up with the field, just like a compass needle.

When a sample made of iron, cobalt, or nickel is placed in the magnetic field of a permanent magnet, the field lines become concentrated within the sample. Lines leaving the north pole of the magnet enter one end of the sample, pass through it, and leave the other end. Thus, the end of the sample closest to the magnet’s north pole becomes the sample’s south pole, and the sample is attracted to the magnet.

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A Bar Magnet Magnetic Field Lines Simulation(Virtual Experiment)

Quiz

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Q1) The magnetic field lines of a bar magnet

A. begin on a north pole and end on a north pole
B. begin on a south pole and end on a south pole
C. begin on a north pole and end on a south pole
D. begin on a south pole and end on a north pole

Answer) C.

Q2)  When I cut a magnet into two pieces I get

A. An isolated north and south magnetic pole
B. Two smaller magnets
C. The two pieces are no longer magnets

 Answer) B.

Bar Magnet Magnetic Field Lines Simulation(Virtual Experiment)

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Magnetic Fields

Place a sheet of paper over a bar magnet and sprinkle iron filings on the paper. The filings will tend to trace out an orderly pattern of lines that surround the magnet. The space around a magnet, in which a magnetic force is exerted, is filled with a magnetic field. The shape of the field is revealed by magnetic field lines. Magnetic field lines spread out from one pole, curve around the magnet, and return to the other pole, as shown in Figure 1.

FIGURE 1. Iron filings trace out a pattern of magnetic field lines in the space surrounding the magnet.

Magnetic Field lines

Magnetic fields exist in three dimensions surrounding a magnet and are more intense at the poles. Magnetic fields are invisible, but can be represented in diagrams with magnetic field lines. Magnetic field lines

• point from the north pole to the south pole outside a magnet, and from the south pole to the north pole inside a magnet

• never cross one another

• are closer together where the magnetic field is stronger

Drawing Field lines

A compass can be used to map the direction of the field lines around a magnet. The compass needle will align itself along the direction of the field. Figure 2 shows the field lines around different magnets.

Figure 2 (a) A magnetic field around a bar magnet with mini compasses on the field lines, (b) A magnetic field around a horseshoe magnet with mini compasses on the field lines, (c) A magnetic field around Earth with mini compasses on the field lines

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Complex Resistor Combinations Simulation

Quiz

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Q1) You have three 100 resistors available. How would you connect these three resistors to produce a 150 equivalent resistance? You must use all of the resistors.

A. Connect all three resistors in parallel.
B. Connect all three resistors in series.
C. Connect one resistor in series with two resistors in parallel.
D. Connect two resistors in series with the third resistor in parallel to the first two.

Answer) C.

Q2)  A circuit is constructed as follows: four resistors in parallel connected in series with three resistors in parallel connected in series with two resistors in parallel. All of the resistors have the same value. How does the equivalent resistance of this circuit compare to the resistance of a single resistor?

A. The equivalent resistance is less than a single resistor.
B. The equivalent resistance is the same as a single resistor.
C. The equivalent resistance is greater than a single resistor.
D. The equivalent resistance cannot be determined without the resistance value of a single resistor.

 Answer) C.

Resistors in parallel total resistance Simulation

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RESISTORS COMBINED BOTH IN PARALLEL AND IN SERIES

Series and parallel circuits are not often encountered independent of one another. Most circuits today employ both series and parallel wiring to utilize the advantages of each type.

A common example of a complex circuit is the electrical wiring typical in a home. In a home, a fuse or circuit breaker is connected in series to numerous outlets, which are wired to one another in parallel. An example of a typical household circuit is shown in Figure 1.

Figure 1. (a) When all of these devices are plugged into the same household circuit, (b) the result is a parallel combination of resistors in series with a circuit breaker.

As a result of the outlets being wired in parallel, all the appliances operate independently; if one is switched off, any others remain on.Wiring the outlets in parallel ensures that an identical potential difference exists across any appliance.

This way, appliance manufacturers can produce appliances that all use the same standard potential difference.

To prevent excessive current, a fuse or circuit breaker must be placed in series with all of the outlets. Fuses and circuit breakers open the circuit when the current becomes too high. A fuse is a small metallic strip that melts if the current exceeds a certain value. After a fuse has melted, it must be replaced. A circuit breaker, a more modern device, triggers a switch when current reaches a certain value. The switch must be reset, rather than replaced, after the circuit overload has been removed. Both fuses and circuit breakers must be in series with the entire load to prevent excessive current from reaching any appliance. In fact, if all the devices in Figure 1 were used at once, the circuit would be overloaded. The circuit breaker would interrupt the current.

Fuses and circuit breakers are carefully selected to meet the demands of a circuit. If the circuit is to carry currents as large as 30 A, an appropriate fuse or circuit breaker must be used. Because the fuse or circuit breaker is placed in series with the rest of the circuit, the current in the fuse or circuit breaker is the same as the total current in the circuit. To find this current, one must determine the equivalent resistance.

When determining the equivalent resistance for a complex circuit, you must simplify the circuit into groups of series and parallel resistors and then find the equivalent resistance for each group by using the rules for finding the equivalent resistance of series and parallel resistors.

Work backward to find the current in and potential difference across a part of a circuit Now that the equivalent resistance for a complex circuit has been determined, you can work backward to find the current in and potential difference across any resistor in that circuit. In the household example, substitute potential difference and equivalent resistance in $\Delta V = IR$ to find the total current in the circuit. Because the fuse or circuit breaker is in series with the load, the current in it is equal to the total current. Once this total current is determined, $\Delta V = IR$ can again be used to find the potential difference across the fuse or circuit breaker.

There is no single formula for finding the current in and potential difference across a resistor buried inside a complex circuit. Instead, $\Delta V = IR$ and the rules reviewed in Table must be applied to smaller pieces of the circuit until the desired values are found.

Resistors in parallel equivalent resistance Simulation

Quiz

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Q1) A current of 5 ampere flows into a circuit where two identical resistors are attached in parallel. What current would flow through one of the resistors?

A. Both resistors carry 5 A
B. each resistor will carry 2.5 A
C. It is impossible to tell unless you know the size of the resistance of each resistor
D. The resistor would carry less than 5 A but you cant tell how much energy was lost before

Answer) B.

Q2)  Four resistors having equal values are wired as a parallel circuit. How does the equivalent resistance of the circuit compare to the resistance of a single resistor?

A. The equivalent resistance is greater than the resistance of any single resistor.
B. The equivalent resistance is the same as the resistance of any single resistor.
C. The equivalent resistance is one-fourth the resistance value of a single resistor.
D. The equivalent resistance is one-half the resistance value of a single resistor.

 Answer) C.

Resistors in parallel equivalent resistance Simulation

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RESISTORS IN PARALLEL

As discussed above, when a single bulb in a series light set burns out, the entire string of lights goes dark because the circuit is no longer closed.What would happen if there were alternative pathways for the movement of charge, as shown in Figure 1.

A wiring arrangement that provides alternative pathways for the movement of a charge is a parallel arrangement. The bulbs of the decorative light set shown in the schematic diagram in Figure 1 are arranged in parallel with each other.

Figure 1. These decorative lights are wired in parallel. Notice that in a parallel arrangement there is more than one path for current.

Resistors in parallel have the same potential differences across them 

To explore the consequences of arranging resistors in parallel, consider the two bulbs connected to a battery in Figure 13(a). In this arrangement, the left side of each bulb is connected to the positive terminal of the battery, and the right side of each bulb is connected to the negative terminal. Because the sides of each bulb are connected to common points, the potential difference across each bulb is the same. If the common points are the battery’s

terminals, as they are in the figure, the potential difference across each resistor is also equal to the terminal voltage of the battery. The current in each bulb, however, is not always the same.

Figure 2. (a) This simple parallel circuit with two bulbs connected to a battery can be represented by (b) the schematic diagram shown on the right.

The sum of currents in parallel resistors equals the total current 

In Figure 2, when a certain amount of charge leaves the positive terminal and reaches the branch on the left side of the circuit, some of the charge moves through the top bulb and some moves through the bottom bulb. If one of the bulbs has less resistance, more charge moves through that bulb because the bulb offers less opposition to the flow of charges.

Because charge is conserved, the sum of the currents in each bulb equals the current $\ I$ delivered by the battery. This is true for all resistors in parallel.

$\ I =  I_{1} + I_{2} + I_{3}.....$

The parallel circuit shown in Figure 2 can be simplified to an equivalent resistance with a method similar to the one used for series circuits. To do this, first show the relationship among the currents.

$\ I =  I_{1} + I_{2}$

Then substitute the equivalents for current according to $\Delta V = IR$.

$\frac{\Delta V}{R_{eq}}=\frac{\Delta V_{1}}{R_{1}} + \frac{\Delta V_{2}}{R_{2}}$

Because the potential difference across each bulb in a parallel arrangement equals the terminal voltage ($\Delta V$ = $\Delta V_{1}$ = $\Delta V_{2}$), you can divide each side of the equation by $\Delta V$ to get the following equation.

$\frac{1}{R_{eq}}=\frac{1}{R_{1}} + \frac{1}{R_{2}}$

An extension of this analysis shows that the equivalent resistance of two or more resistors connected in parallel can be calculated using the following equation.

RESISTORS IN PARALLEL $\frac{1}{R_{eq}}=\frac{1}{R_{1}} + \frac{1}{R_{2}}$..... The equivalent resistance of resistors in parallel can be calculated using a reciprocal relationship.

Notice that this equation does not give the value of the equivalent resistance directly. You must take the reciprocal of your answer to obtain the value of the equivalent resistance.

Because of the reciprocal relationship, the equivalent resistance for a parallel arrangement of resistors must always be less than the smallest resistance in the group of resistors.

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Resistors in series total resistance Simulation

Quiz

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Q1) When three resistors are combined in series the total resistance of the combination is

A. greater than any of the individual resistance values.
B. less than any of the individual resistance values.
C. the average of the individual resistance values.

Answer) A.

Q2)  Several resistors are wired in series. What is true about this circuit?

A. The sum of the currents through each of the resistors is equal to the total circuit current.
B. The total circuit current is the same as the current through any one of the resistors.
C. The voltage across any resistor is the same as the voltage of the power supply.
D. The current in any single resistor is determined by its resistance and the voltage of the power supply.


 Answer) B.

Resistors in series total resistance Simulation

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RESISTORS IN SERIES 

In a circuit that consists of a single bulb and a battery, the potential difference across the bulb equals the terminal voltage. The total current in the circuit can be found using the equation $\Delta V = IR$. 

What happens when a second bulb is added to such a circuit, as shown in Figure 1. When moving through this circuit, charges that pass through one bulb must also move through the second bulb. Because all charges in the circuit must follow the same conducting path, these bulbs are said to be connected in series.

Resistors in series carry the same current 

Light-bulb filaments are resistors; thus, Figure 1(b) represents the two bulbs in Figure 1(a) as resistors. Because charge is conserved, charges cannot build up or disappear at a point. For this reason, the amount of charge that enters one bulb in a given time interval equals the amount of charge that exits that bulb in the same amount of time. Because there is only one path for a charge to follow, the amount of charge entering and exiting the first bulb must equal the amount of charge that enters and exits the second bulb in the same time interval. 

Because the current is the amount of charge moving past a point per unit of time, the current in the first bulb must equal the current in the second bulb. This is true for any number of resistors arranged in series. When many resistors are connected in series, the current in each resistor is the same. 

The total current in a series circuit depends on how many resistors are present and on how much resistance each offers. Thus, to find the total current, first use the individual resistance values to find the total resistance of the circuit, called the equivalent resistance. Then the equivalent resistance can be used to find the current.

Figure 1. These two light bulbs are connected in series. Because light-bulb filaments are resistors, (a) the two bulbs in this series circuit can be represented by (b) two resistors in the schematic diagram shown on the right.

The equivalent resistance in a series circuit is the sum of the circuit’s resistances

As described in Section 1, the potential difference across the battery, ΔV, must equal the potential difference across the load, $\Delta V_{1}$ + $\Delta V_{2}$ , where $\Delta V_{1}$ is the potential difference across $\ R_{1}$ and $\Delta V_{2}$ is the potential difference across $\ R_{2}$.

 $\Delta V = \Delta V_{1} +  \Delta V_{2}$

According to $\ V = IR$, the potential difference across each resistor is equal to the current in that resistor multiplied by the resistance.

 $\Delta V = I_{1} R_{1} + I_{2} R_{2}$

Because the resistors are in series, the current in each is the same. For this reason, $\ I_{1}$and $\ I_{2}$ can be replaced with a single variable for the current, $\ I$.

 $\Delta V = I ( R_{1} + R_{2})$

Figure 2. (a) The two resistors in the actual circuit have the same effect on the current in the circuit as (b) the equivalent resistor.

Finding a value for the equivalent resistance of the circuit is now possible.

If you imagine the equivalent resistance replacing the original two resistors, as shown in Figure 2, you can treat the circuit as if it contains only one resistor and use $\Delta V = IR$ to relate the total potential difference, current, and equivalent resistance.

 $\Delta V = I(R_{eq})$

Now set the last two equations for V equal to each other, and divide by the current.

 $\Delta V = I(R_{eq}) = I(R_{1} + R_{2})$

$\ R_{eq} = R_{1} + R_{2}$

Thus, the equivalent resistance of the series combination is the sum of the individual resistances. An extension of this analysis shows that the equivalent resistance of two or more resistors connected in series can be calculated using the following equation.

RESISTORS IN SERIES $\ R_{eq} = R_{1} + R_{2} + R_{3} . . .$ Equivalent resistance equals the total of individual resistances in series.

Because $\ R_{eq} $ represents the sum of the individual resistances that have been connected in series, the equivalent resistance of a series combination of resistors is always greater than any individual resistance.

To find the total current in a series circuit, first simplify the circuit to a single equivalent resistance using the boxed equation above; then use $\Delta V = IR$ to calculate the current.

$\ I=\frac{\Delta V}{R_{eq}}$

Because the current in each bulb is equal to the total current, you can also use  $\Delta V = IR$  to calculate the potential difference across each resistor.

 $\Delta V_{1} = IR_{1}$ and $\Delta V_{2} = IR_{2}$ 

The method described above can be used to find the potential difference across resistors in a series circuit containing any number of resistors.


Series circuits require all elements to conduct
What happens to a series circuit when a single bulb burns out? Consider what a circuit diagram for a string of lights with one broken filament would look like. As the schematic diagram in Figure 3 shows, the broken filament means that there is a gap in the conducting pathway used to make up the circuit. Because the circuit is no longer closed, there is no current in it and all of the bulbs go dark.

Figure 3. A burned-out filament in a bulb has the same effect as an open switch. Because this series circuit is no longer complete, there is no current in the circuit.


Why, then, would anyone arrange resistors in series? Resistors can be placed in series with a device in order to regulate the current in that device. In the case of decorative lights, adding an additional bulb will decrease the current in each bulb. Thus, the filament of each bulb need not withstand such a high current.
Another advantage to placing resistors in series is that several lesser resistances can be used to add up to a single greater resistance that is unavailable. Finally, in some cases, it is important to have a circuit that will have no current if any one of its component parts fails. This technique is used in a variety of contexts, including some burglar alarm systems.

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Electrostatic Induction Simulation of Electroscope

Quiz

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Q1) When you brush your hair and scrape electrons from your hair, the charge of your hair is

A. positive.
B. negative.
C. both A and B
D. neither A nor B

Answer) A.
And if electrons were scraped off the brush onto your hair, your hair would have a negative charge.

Q2) How does charging an object by contact differ from charging an object by induction?

 A. An object charged by contact gains the same type of charge while an object charged by induction gains the opposite type of charge.
 B. An object charged by contact is touched while an object charged by induction is not.
 C. An object charged by contact must be a conductor while an object charged by induction is an insulator.

Answer) A.

Electrostatic Induction Simulation of Electroscope

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Electrically Charged Objects 

Matter is made of atoms, and atoms are made of electrons and protons (and neutrons as well). An object that has equal numbers of electrons and protons has no net electric charge. But if there is an imbalance in the numbers, the object is then electrically charged. An imbalance comes about by adding or removing electrons.

Although the innermost electrons in an atom are bound very tightly to the oppositely charged atomic nucleus, the outermost electrons of many atoms are bound very loosely and can be easily dislodged. 

FIGURE 01. When electrons are transferred from the fur to the rod, the rod becomes negatively charged.

How much energy is required to tear an electron away from an atom varies for different substances. The electrons are held more firmly in rubber than in fur, for example. Hence, when a rubber rod is rubbed by a piece of fur, as illustrated in Figure 01, electrons transfer from the fur to the rubber rod. The rubber then has an excess of electrons and is negatively charged. The fur, in turn, has a deficiency of electrons and is positively charged. If you rub a glass or plastic rod with silk, you’ll find that the rod becomes positively charged. The silk has a greater affinity for electrons than the glass or plastic rod. Electrons are rubbed off the rod and onto the silk.

An object that has unequal numbers of electrons and protons is electrically charged. If it has more electrons than protons, the object is negatively charged. If it has fewer electrons than protons, it is positively charged.


The TriboElectric Series
If we did a study of many materials and put them in order from those with the least desire for  electrons to those with a very strong desire for electrons we would have created a Triboelectric series. If two items from the list are rubbed together, then the item that is higher on the list will end up more positive and the lower one will end up more negatively charged.  For example, if leather were rubbed with wool, the leather becomes positive and the wool negative.  Yet if rubber is rubbed with wool, the rubber becomes negative and the wool positive.
It is important to note that this series is true only if the samples are clean and dry.  The presence of moisture, dirt, or oils may cause some of the items to interact differently.

Table 01. TRIBOELECTRIC SERIES

Positive(+)

Air

Human Hands

Asbestos

Rabbit's Fur

Glass

Human Hair

Mica

Nylon

Wool

Lead

Cat's Fur

Silk

Aluminum

Paper

Cotton

Steel

Wood

Lucite

Sealing wax

Amber

Polystyrene

Polyethylene

Rubber balloon

Sulphur

Hard rubber

Nickel, Copper

Brass, Silver

Gold, Platinum

Sulfur

Acetate, Rayon

Polyester

Celluloid

Polyurethane

Polyethylene

Polypropylene

Vinyl

Silicon

Teflon

Saran Wrap

Negative(-)

Forces on Charged Bodies

The forces that you observed on tape strips also can be demonstrated by suspending a negatively charged, hard rubber rod so that it turns easily, as shown in Figure 02. If you bring another negatively charged rod near the suspended rod, the suspended rod will turn away. The negative charges on the rods repel each other. It is not necessary for the rods to make contact.

Figure 02. A charged rod, when brought close to another charged and suspended rod, will attract or repel the suspended rod.

The force, called the electric force, acts at a distance. If a positively charged glass rod is suspended and a similarly charged glass rod is brought close, the two positively charged rods also will repel each other. If a negatively charged rod is brought near a positively charged rod, however, the two will attract each other, and the suspended rod will turn toward the oppositely charged rod. The results of your tape experiments and these actions of charged rods can be summarized in the following way:

• There are two kinds of electric charges: positive and negative.

• Charges exert forces on other charges at a distance.

• The force is stronger when the charges are closer together.

• Like charges repel; opposite charges attract.

Neither a strip of tape nor a large rod that is hanging in open air is a very sensitive or convenient way of determining charge. Instead, a device called an electroscope is used. An electroscope consists of a metal knob connected by a metal stem to two thin, lightweight pieces of metal foil, called leaves. Figure 03 shows a neutral electroscope. Note that the leaves hang loosely and are enclosed to eliminate stray air currents.

Figure 03. An electroscope is a device used for detecting charges. In a neutral electroscope, the leaves hang loosely, almost touching one another.

Charging by conduction 

When a negatively charged rod is touched to the knob of an electroscope, electrons are added to the knob. These charges spread over all the metal surfaces. As shown in Figure 04a, the two leaves are charged negatively and repel each other; therefore, they spread apart.

The electroscope has been given a net charge. Charging a neutral body by touching it with a charged body is called charging by conduction. The leaves also will spread apart if the electroscope is charged positively. How, then, can you find out whether the electroscope is charged positively or negatively? The type of charge can be determined by observing the leaves when a rod of known charge is brought close to the knob. The leaves will spread farther apart if the rod and the electroscope have the same charge, as shown in Figure 04b. The leaves will fall slightly if the electroscope’s charge is opposite that of the rod, as in Figure 04c.

Figure 04. A negatively charged electroscope will have its leaves spread apart (a). A negatively charged rod pushes electrons down to the leaves, causing them to spread farther apart (b). A positively charged rod attracts some of the electrons, causing the leaves to spread apart less (c).

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Total Internal Reflection of Light Simulation

Quiz

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Q1) Total internal reflection occurs when the angle of incidence is

A. greater than the angle of refraction
B. equal to the critical angle
C. greater than the critical angle
D. greater than 45°

Answer) C.


Q2) Total internal reflection

A. refers to light being reflected from a plane mirror
B. may occur when a fisherman looks at a fish in a lake
C. may occur when a fish looks at a fisherman on a lake

Answer) C.

Total Internal Reflection of Light Simulation

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When the next simulation is not visible, please refer to the following link.

(https://helpx.adobe.com/flash-player/kb/enabling-flash-player-chrome.html)

Total Internal Reflection
When you’re in a physics mood and you’re going to take a bath, fill the tub extra deep and bring a waterproof flashlight into the tub with you. Turn the bathroom light off. Shine the submerged light straight up and then slowly tip it and note how the intensity of the emerging beam diminishes and how more light is reflected from the water surface to the bottom of the tub.

The Critical Angle
At a certain angle, called the critical angle, you’ll notice that the beam no longer emerges into the air above the surface. The critical angle is the angle of incidence that results in the light being refracted at an angle of 90° with respect to the normal. As a result, the intensity of the emerging beam reduces to zero. When the flashlight is tipped beyond the critical angle (48° from the normal in water), the beam cannot enter the air; it is only reflected. The beam is experiencing total internal reflection, which is the complete reflection of light back into its original medium.

FIGURE 1. You can observe total internal reflection in your bathtub. a–d. Light emitted in the water at angles below the critical angle is partly refracted and partly reflected at the surface. e. At the critical angle, the emerging beam skims the surface. f. Past the critical angle, there is total internal reflection.

 Total internal reflection occurs when the angle of incidence is larger than the critical angle. The only light emerging from the water surface is that which is diffusely reflected from the bottom of the bathtub. This procedure is shown in Figure 1. The proportions of light refracted and reflected are indicated by the relative lengths of the solid arrows. The light reflected beneath the surface obeys the law of reflection: The angle of incidence is equal to the angle of reflection. The critical angle for glass is about 43°, depending on the type of glass. This means that within the glass, rays of light that are more than 43° from the normal to a surface will be totally internally reflected at that surface.

Figure 2. (a) This photo demonstrates several different paths of light radiated from the bottom of an aquarium. (b) At the critical angle, $\theta_{c}$ , a light ray will travel parallel to the boundary. Any rays with an angle of incidence greater than $\theta_{c}$  will be totally internally reflected at the boundary.

Snell’s law can be used to find the critical angle. When the angle of incidence, $\theta_{i}$ , equals the critical angle, $\theta_{c}$ , then the angle of refraction, $\theta_{r}$ , equals 90°. Substituting these values into Snell’s law gives the following relation.

$\ n_{i} sin\theta_{c} = n_{r} sin90°$

Because the sine of 90° equals 1, the following relationship results.

CRITICAL ANGLE $\sin\theta_{c} = \frac{n_{r}}{n_{i}}~~~~~for~n_{i}>n_{r}$

Note that this equation can be used only when $\ n_{i}$ is greater than $\ n_{r}$. In other words, total internal reflection occurs only when light moves along a path from a medium of higher index of refraction to a medium of lower index of refraction. If ni were less than $\ n_{r}$, this equation would give $\ sin\theta_{c} > 1$, which is an impossible result because by definition the sine of an angle can never be greater than 1. When the second substance is air, the critical angle is small for substances with large indices of refraction.Diamonds, which have an index of refraction of 2.419, have a critical angle of 24.4°. By comparison, the critical angle for crown glass, a very clear optical glass, where $\ n$ = 1.52, is 41.0°. Because diamonds have such a small critical angle, most of the light that enters a cut diamond is totally internally reflected. The reflected light eventually exits the diamond from the most visible faces of the diamond. Jewelers cut diamonds so that the maximum light entering the upper surface is reflected back to these faces.

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