unit ii · web viewuse faraday's law in integral form and lenz's law to calculate...

Unit VIII – ELECTROMAGNETIC INDUCTION

References:PHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 31, 32

FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed., Ch. 31

Unit ObjectivesWhen you have completed Unit VIII, you should be able to:

1. Calculate the magnetic flux and/or the time rate of change of the magnetic flux for an area in a magnetic field.

2. Use Faraday's Law in integral form and Lenz's Law to calculate the magnitude and direction of the induced EMF (AKA: E or voltage) and current

a. in a loop of wire being moved into or out of a uniform magnetic field.b. in a loop of wire placed in a spatially uniform magnetic field whose magnitude is given

as a function of time.c. in a loop of wire rotating at constant speed about an axis perpendicular to a uniform

magnetic field.

3. Analyze the forces that act on induced currents and solve simple problems involving the mechanical consequences of electromagnetic induction.

4. Apply the definition of inductance, Ampere's Law, and Faraday's Law to solenoids and toroids to:

a. calculate the inductance.b. relate the induced EMF to the time rate of change of the current in the inductor or the

time rate of change of the magnetic flux.

5. Write the differential equation for an RL circuit and state the solution of this equation.

6. Calculate the currents, potential differences, stored energies, and power dissipation in simple RL circuits.

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Before jumping into electromagnetic induction let’s review a few things.

For answers & solutions see “In-Text Fill-in Answers” beginning on page VIII-27

1. In the sketch at the right a (+) and a (–) charge are entering a

uniform magnetic field with a velocity . The force on the (–)

charge is _______________ with the force having a magnitude (direction?)

of F = _______________ where is the angle between and at any instant. Similarly, the force on the (+) charge is

_______________ and has a magnitude given by F = _______________ . (VIII-2 #1) (direction?)

2. In the sketch at the right a length of wire carrying a conventional current I is at rest in a uniform magnetic field

. The direction of the forcce on the hunk of wire is

_______________ and has a magnitude given by

_______________ where is the angle between the current element I and the magnetic field . (VIII-2 #2)

3. If you need to, look back in the material on Gauss’ Law (Unit II)

and review the idea of the flux of a vector. Since the -field is a

vector quantity we can talk about the flux of or the magnetic

flux, . In the sketch, if there is a magnetic field (it may or

may not be uniform) passing through a teensy weensy area the magnetic flux through the teensy weensy area is given by

( is the angle between and )

and the flux through the entire area A is

where means evaluate the integral over the entire area A. The simplest case is for the

situation where is uniform and perpendicular to area . In this case it can be easily shown that the expression above reduces to

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Unit VIII – ELECTROMAGNETIC INDUCTIONVIII-3


Units – The MKSA unit for magnetic flux is the WEBER (abbrev. Wb). Note that since

, magnetic field can be expressed in units of where

The is used in many texts as the primary unit of magnetic field, so you were hereby warned.

Example VIII–1: A square loop, 0.2 m on a side, rotates with a constant angular speed of rad/s about the y-axis as shown at the

right. It is rotating in a uniform -field that is parallel to the x-axis. At

t = 0 the loop is in the y-z plane. | | = 1.5 T.

a. Express the magnitude of the through the loop as a function of

, the angle between the area vector and the -field.

Solution:

b. Express the magnitude of as a function of time and find the time rate of change of the flux.

Solution: Since the loop is rotating at constant speed of = rad/s, then (rad) = t = t and

Example VIII–2: In the sketch at the right the straight wire carries a current of 1.0 A. Calculate the magnetic flux through a square loop located as shown due to the magnetic field of the wire.

a. The -field at a distance x from a long straight wire carrying a current I is given by the equation:

B = ____________________,

where, at the location of the loop its direction is ____________________. (VIII-3 #1)

b. The area dA of the cross-hatched area is given by

dA = ____________________. (VIII-3 #2)

Thus the magnetic flux through dA is

__________________________, (VIII-3 #3)

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In this example = ______o or ______o (VIII-3 #4 ) depending on the choice of direction of

and thus the magnitude of can be written simply as

__________________________. (VIII-3 #5)

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c. The total flux through the entire loop can be found by adding (aka integrating) the flux through each teensy weensy dA. Writing the integral:

(VIII-4 #1)

where the limits of integration are from x = __________ m to x = __________ m. (VIII-4 #2)

Completing the integration and plugging in the known values we get:

= ____________________ _________ (VIII-4 #3) units

Introduction to Faraday’s Law of Induction – Consider the following figures.

Suppose we cause a conducting wire to move to the right with constant velocity through a

uniform -field as shown in Figure A. In the wire there are free electrons that are pulled along to the right with the wire. These electrons are charged particles moving to the right in a

-field directed into the page therefore, they must experience a magnetic force. If we let qe represent the charge on an electron, the magnetic force on an electron in terms of its charge,

speed, and the strength of the -field is

(VIII-4 #4)

This force causes the electrons to move toward one end of the wire causing the other end to become positive. This separation of charge creates and electric-field within the wire directed

__________ward (VIII-4 #5). This -field exerts a force on the electrons in the __________ward

direction (VIII-4 #6) while at the same time the -field is exerting a force on them in the

__________ward direction (VIII-4 #7). Again if qe is the charge on the electron and E the

magnitude of the -field, in terms of qe and E, the magnitude of the electric force on the

electron at any instant is = _____________ (VIII-4 #7).

When the magnitudes of these forces become equal then the motion of the electrons along the length of the wire ceases. That is, when

If you get stuck see Unit VII problem #12

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(VIII-4 #8)

the charge on each end of the wire becomes constant as long as the wire continues to move at

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constant velocity. Thus we can say that the wire has a difference in potential between its ends. Saying it another way, we have induced an EMF (or voltage, AKA E) between its ends and the wire is effectively the same as a battery of EMF = E as illustrated in Figure B. Note that since FB depends on v of our wire, the E of our wire will increase if v is increased and decrease if v is decreased.

Now let’s consider a conducting loop of wire moving at constant velocity perpendicular to a

uniform -field. In particular let’s look at the three positions shown in the sketch below.

At position corner c becomes (+) and d (–) because the free electrons in section cd experience a magnetic force

downward. (Those in section ab don’t feel a -force

because they are not in the - field) The loop

then at position is equivalent to:

and conventional current flows counter-clockwise in the loop. Note that in this case the

magnetic flux is increasing because the area of the loop containing any -field is increasing.

At position corners a and d both become (–) and b and c (+) due to the magnetic force on the free electrons in sides ab and cd. This situation is equivalent to:

where the EMF of the batteries have the same polarity. Since the -force is trying to push electrons counter-clockwise in side ab and clockwise in side cd the net result is zero current. The current will not be zero if is not uniform. You should be able to easily show yourself

why this is so. Note that if the is uniform, the magnetic flux is constant.

At position a becomes (–) and b (+) and this situation then is equivalent to:

and thus conventional current flows clockwise in the loop. Note that in this case the magnetic flux is decreasing because the area of the loop containing any -field is decreasing.

From these three positions of our loop we can draw the following conclusions:

1. A current will be induced in the loop only when the magnetic flux is changing. That is, when

2. When the magnetic flux increases the current flows in one direction and when it decreases the current flows in the opposite direction.

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3. Changing the velocity of the loop changes the magnetic force on the electrons (and thus the E of our imaginary “battery”), and changes the and thus the magnitude of the induced current changes also.

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Now let’s determine how the EMF across our “equivalent battery” depends on . The loop shown in the sketch is pushed to the right with a force . Note that once the loop begins to move and the current I begins to flow side cd becomes a current

carrying wire in a -field and thus experiences a magnetic force

(the I force). Use your right hand rule to show yourself that is directed to the left as

shown in the sketch. Part of bc and part of ad are also current carrying wires in the -field. What is the net effect on the loop of these two forces? (VIII-6 #1) It should be obvious then, that if the loop moves to the right at constant velocity , a force must be applied to the right with a magnitude that is equal to the magnetic force on cd but in the opposite direction. You should expect to have to exert this force since energy is required to cause the current to flow and, therefore, work must be done on the loop by some external agent (you or me or…?) to supply this energy.

The difference in potential (or EMF or E) that is produced across cd is, by definition, the work done per unit charge in moving (+) charges from d to c [or we could also say moving (–) charges

from c to d]. The work done by moving the loop a distance dx is

Now |I | = I where I = and dq is the net charge transferred from d to c (or c to d) in the time dt. Hence, the work done can be written as

and the EMF across cd is

E

The last two forms of the EMF is because dx/dt = v and also is the change in the

area of the loop containing the -field in time dt. Notice that this expression can be written

E = In other words, “the induced difference in potential is equal to the time rate of change of magnetic flux.” More formally the induced voltage or induced EMF in a loop is given by

E

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Note that a (–) sign was snuck into the equation. It has to be introduced to give the right “sense” or polarity to our equivalent battery. For now, just consider the magnitude of the

to give the magnitude of E. We’ll see later that the polarity is easy to determine.

It is always a good idea to keep the units of different quantities in mind. So let’s practice a little

using E = . The MKSA unit of EMF is the ______________, of magnetic flux is

the _________________, and of time is the _________________. (VIII-7 #1) The units of the left-hand side of the equation then are volts and the right-hand side are Webers/sec. in the pace below, show that a volt is the same as a Wb/s.

(VIII-7 #2)

Example VIII–3: Given the situation shown at the right where B = 0.5 Wb/m2, v = 2 m/s and the total resistance of the loop is 0.5 .

a. Which direction does conventional current flow in the loop due to the induced EMF? (Ans: CCW)

b. What is the induced voltage (or EMF) in the loop?

Solution: E = – (___________)(___________)(___________) = ___________ (VIII-7 #3)

c. If the loop was out of the -field and at rest, what would have to be the polarity of a 0.1 V battery to give the current the same direction as in (a)?

d. What is the magnitude of the current?

Solution: Invision the circuit as shown

and apply Ohm’s Law: I= E/R = (_________)(_________)

= ____________ (VIII-7 #4)

e. If the induced current has the magnitude and direction found in (a) and (d), find the

magnitude and direction of the net force due to the -field on the loop. (Ans: N, to the right)

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Solution: The magnetic force on a wire of length , carrying a current I, perpendicular to a

magnetic field is given by

= (__________)(__________) (__________) (VIII-7 #5)

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Thus the magnitude and direction of the magnetic force on side ab is:

= (__________)(__________)(__________)sin (______), ___________ direction

= ________________, ____________ (VIII-8 #1) magnitude direction

At this point try problems 1 through 7 at the back of this unit.

In our derivation of E notice that (1) was constant, (2) varied with time and (3) we considered one loop of wire. Now let’s generalize a bit. It turns out that

anything we can do to change the value of will result in an induced E in a loop. That is,

we can change only (in magnitude and/or direction), change only (in magnitude and/or

direction), or change both at the same time in such a way changes. So to stress this, we can write Faraday’s Law like so:

E To illustrate the above equation here is a bunch of situations where an EMF is induced due to a dA/dt and a dB/dt.

1. Increasing or decreasing the magnitude of the -field threading through a loop of constant area.

2. Changing the area through which passes.

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3. Use any combination in #1 and #2 to cause to change.

Now for another slight modification. Suppose we have 3 or 4 or N identical but separate loops that are close enough together that we can say the magnetic flux through each is the same. If

we change , then the same E will be induced in each loop and we can imagine each loop having an imaginary “battery” of EMF E. Now what if the loops are all connected. That is,

what if they are made of one continuous piece of wire? Again changing induces an E in each loop (or turn) of wire. Again imagining our little batteries existing in each loop, all of our “batteries” are in series. Therefore, the Etotal across the ends of our long wire having N loops (a solenoid for example) will equal NE, where E is the EMF induced in one loop. So Faraday’s law becomes for N loops or turns of wire:

Eacross ends

Lenz’s Law

Let’s now describe a sure fire way of figureing out the polarity of the induced E. It was mentioned that we would do this at the top of page VIII-7 and that it had something to do with that pesky (–) sign in Faraday’s Law.

Consider the situation at the right. A conducting rod slides on a hunk of wire bent as shown. The rod and wire forms a closed conducting loop

that lies in a uniform -field. When the rod is pushed by an external

force ( ) to the left, a conventional current flows clockwise in the loop

due to the force on the conducting electrons in the rod. The induced current in the

rod causes the rod to experience a magnetic force to the right. Thus, to keep the

rod moving at constant velocity and must be equal in magnitude but opposite in direction. The point is that work must continually be done if the rod is to keep moving. The

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energy is dissipated as heat due to the resistance in the wire and rod, i.e. energy is conserved

in the system. Now let’s notice something. As the rod moves to the left is

___________________ because _____________ (is/are) decreasing. The magnetic field (increasing, decreasing) ( )created by the induced current is directed _____________ the page within the loop thereby

(into, out of)

having the effect of ___________________ the magnitude of the -field in the center of the (increasing, decreasing) (VIII-9 #1)

loop. The effect then, of the current flowing clockwise is to increase while the area of the loop is decreasing thereby reducing the amount of change in . I has to flow clockwise or energy would not be conserved. You should show yourself that if the rod is moved to the right the current would flow counter-clockwise and produce a magnetic field that would be in a direction that would reduce or oppose the change in . That is, the polarity of the induced EMF (which determines the direction of the induced current) will be such that it opposes the change in magnetic flux. Thus, the meaning of the (–) in E is that the polarity of E

opposes any change in .

The stuff above is all summarized in Lenz’ Law:

The polarity of an induced EMF (E) is such that the induced current (I) produces

a -field that opposes the any change in the flux of the magnetic field .

Applying Lenz’ Law we have a quick and easy way of determining the direction of the induced I and in turn, the polarity of E.

Example VIII–4: Two conducting loops lie on a table. Loop 1 contains a battery and a switch S. Determine the direction of the induced current in loop 2 and the polarity of an imaginary battery that, when inserted in loop 2 as shown, would result in a current in the same direction when: (a) S is initially closed, (b) S has been closed for a while, and (c) S is opened.

Solutions:

a. The instant the switch is closed, current begins to flow clockwise in loop 1 thereby producing a -field directed out of the page everywhere outside loop 1. Therefore, the -field within loop 2 goes from nothing before S is closed to something out of the page after it is closed. then increases as the switch is closed and Lenz’s Law says the current will

flow in loop 2 in a direction that will oppose this increase. Since the area of the loop is constant the increasing magnetic field of loop 1 is causing the change in in loop 2. Now the question reduces to: If is increasing out of the page then the current in loop 2 will

VIII-15


produce a magnetic field within loop 2, , which will oppose the change in , i.e., will

be into the page. I then must flow clockwise to produce a -field into the page within loop 2. To produce this clockwise current our imaginary battery would have to be (+) on the left and (–) on the right.

b. When S has been closed awhile, I, is ____________________ and the -field due to (increasing, decreasing, steady)

loop 1 within loop 2 is _______________ the page and ____________________. (into, out of, zero) (increasing, decreasing, steady)

Thus, within loop 2 is ____________________. and is _______________ (increasing, decreasing, steady) (into pg, out of pg, zero) and the current in loop 2 is _______________.

(CW, CCW, zero) (VIII-10 #1)

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c. The instant the switch is opened, the current in loop 1 goes from somethin’ CW to nothin’ and hence the -field of loop 1 that passes through loop 2 goes from something out of the page to nothing. Thus, in loop 2 is ____________________ due to the decreasing

(increasing, decreasing, steady)

-field of loop 1. The -field produced by the induced current in loop 2 must be in a direction to oppose the change in and, therefore, the change in . Thus the -field in loop 2 due to its own current must be directed _______________ the page and is produced (into, out of, zero) by a current flowing _______________ in loop 2. Our imaginary battery would have to be

(CW, CCW, zero) connected with its (+) terminal on the_______________.

(right, left) (VIII-11 #1)

In summary, for this example, when this switch is closed current flows momentarily clockwise, becomes zero, then as the switch is opened it flows momentarily counterclockwise.

Now we have two ways to determine the polarity of induced EMF and direction of flow of

induced current: (1) by using the force and (2) by using Lenz’s law. However (1) can only be used when a portion of the loop is moving, (2) can be used in all cases.

For practice using Lenz’s Law try problems 8 through 10 at the back of this unit.

Induced EMF due to Rotation

Consider a wire coil of N turns rotating in a uniform -field at a

constant angular speed . Since the angle between vectors and

is changing, then the product is changing, and thus there is a

change in , resulting in an induced E in the coil. Let’s find the E for this situation.

is a vector representing the area of the coil and is ________________ to the plane of the

coil. Therefore, in the sketch above, the angle between and is _________. (VIII-11 #2)

Now for N turns of wire in the coil, the E induced is given by :

E = (VIII-11 #3)

Where . The angle however is related to the

angular speed and time by the relation = ______ (VIII-11 #4). Hence, can be written in terms of B, A, , and t as

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= ____________________ (VIII-11 #5)

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and since the induced E in the coil is given by

E = ,

the induced E of the revolving coil is E = ____________________ (VIII-12 #1)

If the ends of the coil are connected, at the instant the coil is in the position shown in the

sketch on page VIII-11, which way does the current flow? This way?

Or this way? (VIII-12 #2)

Determine the direction both ways: using and Lenz’s Law.

At this point try problem 11 at the back of this unit.

So far we have just dealt with situation where the -field is constant and the area of a loop is changing. That is, in Faraday’s Law:

E

We have just been using E because . Now let’s consider the case where

is constant and .

Example VIII–5 The conducting loop at the right is in a uniform -field that is increasing at a constant rate of 0.02 T/s.

a. What is the direction of the induced current in the loop?

Solution:

Since the -field is increasing and is constant, the change in must be due to .

Lenz’s Law says that the induced current will be in a direction that produces a that

opposes the change in or, in this case, the current will be in a

in a direction that will produce a -field that is ___________ the page thereby opposing (into, out of)

the change in . Hence, to produce this -field the current in the loop must flow

___________. (VIII-12 #3)

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(CW, CCW)

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b. What is the magnitude of the induced E in the loop? Solution: From Faraday’s Law:

E

Now

And units

Thus, |E| = = __________ ______ (VIII-13 #1) units

c. If the total resistance of the loop is 100 , what current flows in the loop? Solution:

From Ohm’s law: V = E = RI I = E/R.

Therefore, I = = _____________ (VIII-13 #2)

At this point try Worksheet VIII.1 and problems 12 through 14 at the back of this unit.

Faraday’s Law in general FormIn a previous unit we found that in electrostatic situations (those situations where charges are at rest or where constant currents are involved) the difference in potential between two points

a and b in an -field was given by the line integral

and since was a conservative field when the integral is evaluated around a closed path then,

However this is NOT true when there is a changing magnetic flux in the region of interest. Let’s see why.

Recall that the difference in potential between two points a and b, , is the work done per unit of charge moving a charge from a to b, i.e.,

,

since is defined as the electric field, .

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Now let’s consider the integral around a closed path in a region where there is a changing magnetic flux. As shown in the sketch at the right, suppose that the magnitude of is increasing in the region enclosed by a wire loop. The result of this will be an induced EMF in the wire and a current will flow counter-clockwise. The induced EMF is a difference in potential V and has the meaning that it is the work done per unit change in moving the charge around the loop once. That is,

EMF = E = .

AHA! Look what this equation says: the thing that is making the charges go ‘round the loop is

an -field that is being set up in the loop and is pushing the charges around. Recall that

E . Therefore,

E .

So we see that if there is a changing in a region, this changing creates and -field which in turn does electrical work on any charges present in the region. Notice also that since

in this situation then the -field created by the changing is a NON-conservative -field.

Let’s restate a few things:

1. If charged particles are stationary (meaing = 0) or are moving in a constant current

(meaning is constant), then

( is a conservative field)

2. If charges are accelerating or currents changing ( constant), then

E Faraday’s Law in general form

where is the magnetic flux through the area enclosed by the path.

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As with Faraday’s Law in differential form (E ) just do your calculations ignoring the

(–) and depend on Lenz’s law to give the direction for I (and therefore , since I and are in the same direction).

Example VIII–6: The wire loop sketched at the right is circular, has a

radius R and lies in a -field whose magnitude is changing with time

according to the relation where is a constant.

a. Use Lenz’s Law to show yourself that the induced current in the loop,

and therefore the direction of the -field set up in the loop is clockwise.

b. Determine the magnitude of the -field in the loop.

Solution:

Taking the direction of as the same as , then and thus, noting that

will be constant in the circular loop, if we integrate around the loop

Eqn (1) (VIII-15 #1)

The magnitude of the time rate of change of is

,

where the area of the loop is constant and equals . Since then

Eqn (2) (VIII-15 #2)

Now , so Eqn (1) must equal Eqn (2):

(VIII-15 #3)

Solving for E, the magnitude of in the loop is

(VIII-15 #4)

and its direction is clockwise.

Notice an important thing. Faraday’s Law in integral form (or differential form for that matter) has no term, constant, etc. that has to do with a piece of wire. For example, if we would have worked out Example VIII-6 for some r where r < R = radius of the wire loop, we would have gotten

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,and a final answer of . This says there is an -field at a distance r from the center of the loop even though there ain’t a wire there! In other words, any time there is a changing magnetic flux in a region of space there is an -field in that region even though the space is empty! Putting a wire loop and attaching a current meter to it is merely a convenient way to verify that the -field exists by letting the -field push the charges around in the loop in the form of a current.

Self-Induction Consider the loop shown in Figure A. The magnetic field passing through the loop could be caused by a current in another loop, by an electromagnet, by a permanent magnet, etc. If we tweek the source of the -field thereby causing it to increase in magnitude, then, in accordance with Lenz’s Law a current will be induced in the ______________ direction. If is caused to decrease, the current

(CW? CCW?) (VIII-16 #1) will flow in the opposite direction. In either case, the magnitude of the induced EMF will be given by Faraday’s law:

|E| (or |E| if the loop has N turns)

A battery inserted at the bottom of the loop oriented will produce a current in the same direciton as when the -field is increased. (VIII-16 #2)

Now consider the situation in Figure B, where there are no -fields in the vicinity of the loop (as long as switch S is open). When S is closed the battery causes a current to flow clockwise. This current causes a -field to be established directed into the page within the loop and out of the page outside the loop. But notice – as the switch is closed the -field within the loop is going from nothin’ to somethin’ into the page. The point is it’s increasing into the page just like it was in Figure A above! In other words, its OWN changing -field is going to induce an EMF and current. The magnitude of the induced EMF will as usual be given by Faraday’s Law and the direction of the current will be in the opposite direction (CCW) to the current due to the battery. (Show yourself this is true using Lenz’s law.) The process by which an EMF is induced in a loop due to its OWN -field changing is called SELF-INDUCTION. The loop “looks” like Figure C right after the switch is closed where is the current due to the real battery

where is the induced current. As soon as grows from zero to its steady value the -field within the loop stops changing, making

become constant ( ) and thus Ei (and ) becomes zero.

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Using the same line of reasoning you should be able to show yourself that when S is opened goes from somethin’ to nothin’ – causing the -field to collapse to zero – causing Ei to be

induced this time with a polarity that is the same as that of the battery – causing to flow in the same direction as . We can say (correctly) that the self-induced current will always be in

the direction that opposes any change in the current in the loop. That is, if is increasing,

is in the opposite direction to , if is decreasing is in the same direction as . This, of course, follows from Lenz’s Law.

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Now let’s see what the magnitude of the self-induced EMF depends on.

Recall that when and are parallel as they are in the case of our loop.

From previous work we know that the magnitude of (in a loop, solenoid toroid, etc.) is directly proportional to the current. Thus,

N = A(kI) = LI (again N = the total number of turns of wire)

where Ak, which is a constant product, has been replaced by a single constant of proportionality L. L is a constant that is independent of the current and only depends on the size and shape of the loop. The constant L is called the “self-inducatance” or simply the “INDUCTANCE” of our loop (or solenoid, or toroid, etc.). Since the EMF induced equals the time rate of change of the magnetic flux, then

Ei .

As we would expect then, the self-induced EMF is directly proportional to the time rate of change of the current. Notice that, as we saw before, if I is increasing (dI/dt is +) the polarity of E is (–) meaning opposite to EB in Figure C. And if I is decreasing (dI/dt is –), Ei has the same polarity as EB.

Units of Inductance-The inductance of a loop or coil (henceforth called an “INDUCTOR” is either

or L = – E/

The MKSA units of: are ________________

I are ________________

dI/dt are ________________

E are ________________ (VIII-17 #1)

So the MKSA units of Inductance can be either

(VIII-17 #2)

As usual, we give these complicated units a special, simpler name – By definition:

1 Henry (abbreviated: H) =

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(Named after Joe Henry – famous American Physicist and ex-high school teacher that did a lot of work on electromagnetic induction)

Example VIII–7: A solenoid of length has N turns and a circular cross-section of radius r. Calculate its inductance, L.

Solution: If I is the current at the some instant flowing in the coil, L is given by

The magintude of the -field in a

solenoid of N turns, having length and carrying a current I is (see Unit VII)

B = oNI/ ,

Thus

= BA = Br2 .

And, therefore,

L = .

Notice L only depends on the geometry (area = r2, length = ) and the number of turns in the solenoid. This is generally the case. Another thing that will affect L would be if we fill the center of the solenoid with some magnetic material like iron, etc. This changes the constant o (remember o is the “magnetic permeability” of a vacuum) to the characteristic for the magnetic material. In these notes we will only consider the case where the solenoid or toroid is

filled with nothing but air ( ).

In the example above calculate the inductance of a solenoid in units of mH, where N = 1000

turns, r = 1 cm and = 20 cm.

At this point try problems 15 & 16 at the back of this unit.

Don’t get confused: on page VII-12 the B-field inside a solenoid was derived to be onI where n represents the # turns per unit length. Therefore, n =N/ .

VIII-28


LR CircuitsThe situation with an “LR” circuit results in a time varying current just as in the case of the “RC” circuits presented back at the end of Unit V. The reason for the time varying current is due to the characteristics of the inductor we went over in the preceeding pages. Let’s approach LR circuits in the same way we did RC circuits. As we proceed you should compare the results (equations, graphs, etc.) with those arrived at for RC circuits noting the similarities and differences.

Consider the situation in the sketch at the right where the symbol is the symbol for an inductor.

Case 1 – When the switch S is switched to a the current flowing from the battery through R and L goes from nothin’ to somethin’. That is, I increases until it reaches a steady value. What effect does the inductor have? Since the current is changing there is a

difference in potential across the inductor ( ) that opposes the EMF of the battery. That is x is “+” relative to y in the diagram. As

time goes on the current reaches a steady value (i.e., dI/dt = 0),

goes to 0, and becomes equal to E. Let’s show what happens mathematically. When the switch is at a the circuit is as shown at the

right.

Applying Kirchoff’s 2nd rule (K-2):

Now we have to be careful not to screw up the signs. We’ve shown that = (the EMF across L) = –L(dI/dt). Recall that the (–) sign means that V has a polarity that is opposite to that of the battery if the current is increasing (i.e., dI/dt is +) and a polarity that is the same as that of the battery if the current is decreasing (dI/dt is –). If this is puzzling to you, go back to review pages VIII-16 and 17. Applying K-2:

Eand

IR + LdI/dtor

E = IR + LdI/dt.

Another way is to just consider the energy supplied to or lost by the charges as they move around the loop. Note the polariies on the circuit diagram above. The battery supplies energy to the charges so E is a V increase. The resistor removes energy from the charges and

dissipates it as heat so is a V decrease. The inductor in this case is like a battery hooked up backwards (relative to the real battery) as if it is a battery being charged. Therefore,

VIII-29


it is also removing energy from the charges so is a V decrease also. The amount of decrease is LdI/dt. So looking at it this way we also get

E = IR + LdI/dt.

To find how I varies with time let’s rearrage the equation to look like: so thateverything involving I is on the left and everything involving t on the right.

Now lets integrate it.

or

Recalling from Algebra II & page V-21 loge (x) = ln (x) = y then x = ey.

Thus, can be written as .Solving for I:

Let’s look at what this expression tells us. When the switch S is switched to terminal a at t = 0 the current is initially zero ‘cause

As time goes on (t ) I approaches the maximum value ‘cause

In other words as t the circuit is just as if the inductor weren’t present and = IR = E.That is, the circuit becomes

If I is plotted against time the graph looks like:

Left integral: Let u = E – IR du = – RdI

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Notice that for a given value of R, the larger L is (i.e., the greater the inductance) the longer it takes for I to reach its steady state value of E/R. An inductor is sometimes called a “choke” because it “chokes off” the current preventing it from instantaneously becoming equal to E/R as soon as the switch is closed.

We can also sketch graphs of vs t and vs t.

Since = RI I the Since dI/dt, the magnitude

magnitude vs t looks like: of vs t looks like:

(VIII-21 #1) The equation of as a function of t is The equation of as a function of t is

(VIII-21 #2) Remember these expressions developed up to this point are for the circuit described on page VIII-19 where the switch S is switched to terminal a. Now let’s see what happens when it is switched from a to b.

Case 2 – Suppose switch S has been connected to terminal a for a long time and we quickly switch it to terminal b. This effectively removes the battery from the circuit thereby causing the current through R and L to decay from E/R to zero. But since the current is changing a VL is induced in the inductor with a polarity that opposes the change in I. Previously, when S was connected to a the current was flowing through the inductor from x to y. When S is switched to b the current decreases. The inductor then assumes a polarity that tries to prevent this decrease. The circuit effectively becomes:

==

VIII-32


Applying Kirchoff’s loop rule (K-2) to find I as a function of t:

Let’s use the same separation of variables technique we used at the bottom of page VIII-19 and VIII-20:

(VIII-22 #1)

Integrating both sides we get:

(VIII-22 #2)

which becomes

(VIII-22 #3)

Again recalling loge (x) = ln (x) = y then x = ey, we arrive at the expression for I as a function of time:

Since at any time t, then

(VIII-22 #4)

Also since , then

(VIII-22 #5)

The graphs for these functions look like:

Where Io = the current at t = 0and I = the current at time t

VIII-33


(VIII-22 #6)

(VIII-23 #1)

The LR Time Constant

has the same meaning as the for an RC circuit as discussed in Unit V (see page V-24).

is defined as the time required for the exponential term in the equation for I (or or )

to become . This occurs when t = L/R. is a measure of how fast the current and voltages change in an LR circuit. As an example consider the current flowing in the LR circuit in Case 1 on page VIII-19. The current was given by

I = (E/R) .

At the instant t = 1 the current flowing in the circuit is

I = (E/R) = (E/R) 0.63(E/R).

That is, at t = 1 the current has reached 63% of its maximum value of (E/R).

In Case 2 the current is given by . At t = 1 after the switch is switched to terminal b (see page VIII-22) the current has decayed exponentially to

.

That is, the current has decreased to roughly 37% of its value at t = 0.

Energy storage in an inductorIn Case 1 we said the inductor caused the energy of the charges to decrease as they moved through it. What happened to the energy? It was stored as potential energy in the inductor. In Case 2 the inductor increased the energy of the charges. Where did this energy come from? The energy in Case 2 was just the amount stored in Case 1. The situation is completely analogous to storing potential energy in a spring by compressing it, then getting the energy back as the spring expands.

How can the energy stored be calculated? Recall that instantaneous power is

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IV, where U = potential energy stored or dissipated.

VIII-35


For our inductor

Thus,

{ which reduces to

.

Assuming U = 0 at t = 0 and integrating

Where U is the potential energy of the inductor at any time t and I is the current at that instant.

Example VIII–8 In the circuit sketched at the right an inductor having inductance L and resistance r is connected in parallel with a resistance R and a battery having an EMF, E. After switch S has been closed a long time it is suddenly opened.

a. Find the following currents in terms of E, L, R, and r.i. I through the inductor before the switch is opened.

Solution: Since the current anywhere in the circuit is steady and the circuit acts as if L were not present except for its resistance

r. The circuit then is

effectively and thus E.

Therefore, since E = Ir, the current E/r.

ii. I through L immediately after S is opened.

Solution: Since and Io = I thru r just before S is opened = E/r and t = 0

= (E/r) = E/r

iii. I through R before S is opened. Solution: The situation is the same as in (i) and VR = E = IR. Therefore, IR = E/R.

iv. I through R at t = after S is opened. Solution: Since r, R, and L are all in series

(E/r) = 0.37(E/r)

the product of the current through and the potential differenceacross the inductor at any given time t.

Ignoring the sign of because thatjust tells us whether the energy is gained or lost.

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b. Find immediately after S is opened.

Solution: at t = 0. = E/r (see a - ii) and thus E.c. Write the differential equation for the current and derive the expression for the current in L

as a function of time after the switch is opened. Sketch the graph of I vs t.

Solution: Applying K-2 to the circuit after the switch is opened:

Putting everything involving I on one side of “=” and everything involving t on the other and integrating we get:

or

= (E/r)

Graph of I vs t

d. Find the total amount of energy dissipated after the switch is opened.

Solution: The energy dissipated is equal to the total energy stored in the inductor before the switch is opened. This energy is

(E/r)2

At this point try problems 17 through 20 at the back of this unit.

When you have finished problems 17 - 20 of the “End of Unit Problems” you will have finished the Electricity and Magnetism Workbook for this course. I hope you learned as much as I did as I wrote this workbook.

VIII-38


In-Text Fill-In Answers

Page Fill-In # Fill-In Answer

VIII-2 1 (–) charge – direction: down, magnitude: (+) charge - direction: up, magnitude:

2 F on wire is to the right, magnitude:

VIII-3 1 B = , into page

2

3

4

5

VIII-4 1

2 limits: x = 0.1 m to 0.2 m

3

4 down toward the bottom of the page. (Remember electrons are negative LEFT-hand rule)

5 Upper end is (+) & lower end is (–) is directed downward toward thebottom of the page

6 is up toward the top of the page

7

8

VIII-6 1 The current carrying wire force on bc is down toward the bottom of the

page and the force on ad is up toward the top of the page. Since the

forces are equal in magnitude these two forces add to zero. Therefore

their net effect on the loop is zero.

VIII-40




VIII-7 1 Volt, weber, sec.

2

3 E = -- (0.05 Wb/m2)(0.1 m)(2 m/s) = -- 0.1 V

4 = E/R = (0.1 V)/(0.5 ) = 0.2 A

5

VIII-91 As the rod moves to the left is decreasing because is decreasing.

The magnetic field created by the induced current is directed into the page within the loop thereby having the effect of increasing the

magnitude of the -field in the center of the loop.

VIII-10 1 When S has been closed awhile, I, is steady and the -field due to

loop 1 within loop 2 is out of the page and steady. Thus, within loop

2 is steady and is zero and the current in loop 2 is zero.

VIII-11 1 Thus, in loop 2 is decreasing due to the decreasing -field of loop 1.

The -field produced by the induced current in loop 2 must be in a

direction to oppose the change in and, therefore, the change in .

Thus the -field in loop 2 due to its own current must be directed

out of the page and is produced by a current flowing CCW in loop 2. Our imaginary battery would have to be connected with its (+) terminal on the right.

2 perpendicular,

E

4

5

VIII-42


VIII-12 1 E

2 This way

3 into, CW

VIII-43




VIII-13 1

|E|

2 I = (0.6 mV)/(102) = 6 A

VIII-15 1

2

3

4

VIII-16 1 CCW

2

VIII-17 1 The MKSA units of:

are

I are A = C/s

dI/dt are A/s = C/s2

E are

2

VIII-44


VIII-18 1



VIII-21 1

2 RI = I (see p. VIII-20)

= E Ans.

= E Ans.

VIII-22 1

2

3

4

5

Note: Ans. same as VR

VIII-45


6

VIII-46




VIII-23 1

VIII-47


End of Unit Problems

1. A straight wire carrying a current I and a circular conducting loop lie in the same plane as shown at the right.

a. There are a number of things that can be done with this set up that would cause a change in magnetic flux within the loop thereby inducing an E in the loop. List as many as you can.

b. There are several ways that the loop can be moved without changing

. Describe as many as you can.

2. As shown in the sketch at the right, a conducting wire slides along a

section of wire with a constant velocity . Describe the magnitude and direction of a magnetic field that would cause the current to flow in the direction of shown.

3. An airplane with the wingspan of 70 meters is flying horizontally at 1000 km/h directly toward magnetic north. If the vertical component of the Earth’s magnetic field at the airplanes position is 20 T downward,

a. What is the induced difference in potential between its wingtips? (Ans: 0.4 V)

b. Which wing tip is at the higher potential? (Ans: the western one)

c. Why can’t this potential difference be used as a source of power?

4. A loop of mass m and having a total resistance R falls vertically

through a uniform magnetic field as shown.

a. What is the direction of the induced current in the loop? (CW)

b. As it emerges from the -field the loop reaches a constant terminal velocity. Show that the magnitude of the of this

terminal velocity is given by .

c. Show that while the loop is traveling at terminal velocity, the rate at which heat is being

dissipated in the loop is .

VIII-48



5. As shown in the sketch at the right, two wheels connected by a 1.5 m axle are pushed along to horizontal rails at a constant speed of 3 m/s. The rails, wheels, and axle are conductors and form a complete circuit with the resistor. R is the only sizable resistance in the system. As shown there is a uniform magnetic field of 0.08 T directed vertically downward.

a. Find the induced E in the circuit. Which wheel is it the higher potential? Ans: 0.36 V, right wheel)

b. What is the magnitude and direction of current flow? (Ans: 0.9 A, clockwise)c. At what rate is thermal energy being dissipated in the resistor? (Ans: 320 mW)d. What force is required to keep the wheels and axle moving at constant speed?

(Ans: 0.1 N, in the direction of )e. Find the rate at which work is being done by the force found in (d). Compare your

answer with the energy dissipated in the resistor. [see answer for (c)]. (Ans: 320 mW, power in = power dissipated)

f. After the axle/wheels roll past the resistor, does the current in the resistor reverse direction or continue to flow in the same direction? (Ans: reverse direction)

6. A conducting rod of length L rotates about point P at one of its ends with a constant angular speed in a plane perpendicular to a uniform magnetic field.

a. Show that the induced dE across a teensy-weensy hunk of the rod of length dr that is a distance that is a distance r from P is given by dE = – Brdr.

b. By integration, show that the potential difference across the ends of the rod, , is given by E = = – BL2/2.

Suppose instead the rod slides along a circular conducting track as shown at the right. A resistor R is in electrical contact with the track and with the end of the rod at point P. Assume the resistance of the rod and track to be negligible.c. Show that the current through R is given by I = BL2/(2R).

Does I flow upward through R or downward? (downward)d. Show that the power dissipated in R is given by

P = (BL2)2/(4R).e. Show that the torque required to keep the rod traveling with constant is given by

, out of the page.

VIII-49


End of Unit Problems7. A conducting elastic band is stretched to form a circle 10 cm in radius

and is placed with its plane perpendicular to a uniform 0.80 T magnetic field as shown at the right. The instant after it is released its radius begins to shrink at a rate of 80 cm/s. Find the induced E in the band at this instant. (Ans: 0.40 V)

8. Two conducting loops have common axes and lie and planes parallel to one another as shown. The larger loop contains a battery and switch S. An observer sites along their common axis.

a. Using Lenz’s Law to determine the direction of current flow in the smaller loop as seen by the observer the instant S is closed. the instant S is opened. (Ans: CCW, CW)

b. If the smaller loop were replaced by the incomplete loop shown at the right, determine the polarity of the difference in potential induced in the loop as seen by and by the observer as S is closed. as S is opened. [Ans: left terminal (+), right terminal (+)]

c. Describe the direction of the force of the smaller loop in (a) as S is closed. (Ans: repelled from larger loop)

9. A bar magnet is dropped through a horizontal conducting loop as shown at the right.

a. Use Lenz's Law to describe the current induced in the loop as the magnet approaches, passes through and recedes away from the loop.

b. Use the principle of conservation of energy to describe the speed of the magnet as it approaches, passes through and recedes away from the loop.

10. A small solenoid carrying a steady current is moved toward and away from a conducting loop as shown.

a. Use Lenz's Law to determine the direction of the induced current in the loop as seen by the observer shown when the solenoid is moving toward the loop. away from the loop. (Ans: CW, CCW)

b. What is the direction of the magnetic force on the loop as the solenoid approaches the loop? as it moves away from the loop? (Ans: to the right, to the left)

VIII-50


End of Unit Problems11. A rectangular loop of N turns, length a and width b is

rotated with an angular speed in a uniform magnetic field as shown at the right.

a. Show that an induced voltage given by

E = NabB)sin t = Eo sin t

appears across the terminals of the loop as it is rotated. (This is the principle of an alternating current generator.)

b. Design a loop that will produce an Emax = 1.5 V when rotated at 60 rev/s (remember

is in rad/s) in a -field of 0.5 Wb/m2.c. What is the magnitude and direction of the current in R the instant the loop is in the

position shown. (Ans: zero)

12. The -field through a one turn loop of wire having a radius of 10

cm and a total resistance of 10 changes with time as shown in the graph of the right. Plot

a. the E induced in the loop as a function of time.b. the rate at which heat is dissipated in the loop as a function of time.

13. A rectangular loop of wire having sides a L and L/3 and resistance R lies in a uniform -field that is perpendicular to its plane and whose magnitude varies with time according to

the equation B = Bocost where Bo and are constants. Recall from your work in SHM that where T is the period of the oscillation.a. Let the direction of at t = 0 be into the page. Make a sketch of the loop indicating the

direction of the -field and the direction of the induced current between t = 0 and t = 3T/4.

b. Show that the magnitude of the induced current as a function of time is given by

I .

c. Describe the direction of the magnetic force on the loop (the I force) associated with the induced current during one complete cycle.

d. Show that the total amount of charge that passes a point between t = 0 and t = T/2 is given by

.

e. Show that the rate at which thermal energy is being dissipated at time t is given by

VIII-51



14. In the sketch at the right, a rectangular wire loop and a long straight

wire lie in the same plane. The resistance in the loop is 2 .

a. Show that for a steady current I in the straight wire, the magnetic

flux through the loop is given by .

b. If the current in the long wire decreases uniformly from 10 A to 2 A in 2 seconds, find the magnitude and direction of the induced current in the loop. Let L = 30 cm. (Ans: 0.13 A, CCW)

15. Suppose an inductor is wound on a cylindrical bar having a rectangular cross-section measuring 1 cm by 2 cm. When wound the inductor has 10 turns of wire per centimeter of

length. Ignore any divergence of the -field at the ends of the inductor.

a. Find through a single turn near the center of the inductor as a function of the

current I. [Ans:

I]b. Roughly how long should the inductor be if it’s inductance is 150 H? (Ans: 0.6 m)c. What will be the rate of change of magnetic flux, , when a V of 3 mV is

applied across the inductor? (Ans: 5 Wb/s)

16. A toroidal inductor having 1000 turns of wire is wound on a form having an inner radius ro = 19.5 cm and a square cross-section of 1 cm on a side. A portion of the toroid is shown in

the sketch of the right.

a. Find the magnitude of the -field inside the inductor as a function of the radius r from

the center of the toroid and the current I. [Ans:

]

b. Find the inductance L of the toroid. (Neglect the variation of B with r inside the windings, i.e., let r = .20 cm.) (Ans: 0.1 mH)

c. Find the induced E if the current in the inductor is increasing at a rate of 10 A/sec. (Ans: 1

mV)

17. A 6 H inductor is connected in series with a 1 k resistor.

a. How long after a 10 V battery is connected across the pair will it take for the current in the resistor to reach 80% of its maximum value? (Ans: 9.7 ns)

b. What is the current in the resistor when ? (Ans: I = 6.3 mA)

VIII-52



18. Given the circuit shown at the right, calculate the following quantities (i) right after the switch S is closed and (ii) after the switch has been closed a long time.

a. I1 [Ans: (i) 2 A, (ii) 2 A]b. I2 [Ans: (i) zero, (ii) 1 A]c. I [Ans: (i) 2 A, (ii) 3 A]d. V across R2 [Ans: (i) zero, (ii) 10 V]e. V across L [Ans: (i) 10 V, (ii) zero]f. dI2 /dt [Ans: (i) 2 A/s, (ii) zero]

19. The current in a 12 mH inductor having negligible resistance varies with time as shown in the graph at the right. Sketch a graph of the potential difference across the inductor as a function of time.

20. A 3 H inductor is connected in series with a 10 resistor. A 3 V battery is connected

across the combination. At after the battery is connected,

a. What is the rate at which energy is being supplied by the battery? (Ans: 0.57 W)b. What is the rate at which energy is being dissipated in the resistor? (Ans: 0.36 W)c. What is the rate at which energy is being stored in the -field of the inductor?

(Ans: 0.21 W)

VIII-54

unit ii · web viewuse faraday's law in integral form and lenz's law to calculate...

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