unit ii - mr. greenberg physics - home · web viewfundamentals of physics, halliday, resnick, &...

Unit VII – CALCULATING MAGNETIC FIELDS

References:PHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 30

FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed., Ch. 30

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

1. Use Biot-Savart law toa. find the magnitude and direction of the contribution to the total magnetic field at a

point due to a short segment of current carrying wire.b. derive and apply the expressions for the magnetic field of a long, straight wire or for

the magnetic field of a circular loop at any point along an axis through the center of the loop.

2. Apply the expression for the force between parallel current-carrying wires to determine the magnitude and direction of the force on either wire.

3. Use Ampere's law toa. derive an expression for the magnetic field inside or outside a solid or hollow long

cylinder carrying a current of uniform density.b. derive an approximate expression for the magnetic field inside a very long solenoid or

inside a toroidal solenoid.

4. Apply the superposition principle to determine the magnetic field at a point produced by combinations of the configurations listed above.

VII-1


BIOT–SAVART LAW (aka B-S LAW)

As discussed before, the direction of the net -field at a point in space is in the direction that the north seeking end of a small compass needle points when placed at the point. In the early part of the nineteenth century it was found that when a wire is carrying a current any compass in the vicinity was deflected; the conclusion being that a current-carrying wire must create a magnetic field about itself. It turns out that, more fundamentally, it is moving charges that create the -field, the wire has nothing to do with it. A beam of charged particles moving through empty space will create a magnetic field in the region through which the charges are moving. The -field due to a conventional current in a wire and a beam of (+) charges are shown below.

Notice: wrap your right hand around the wire or beam so that your thumb points in the direction that the (+) charges are moving (i.e., the direction of or I+) and having done this, your

fingers indicate the direction of the -field. (AKA: the right hand rule = RHR).

Since a bunch of (+) charges moving toward the right is equivalent to a bunch of (–) charges moving to the left, notice that grabbing the wire or beam with your left hand and pointing your left thumb in the direction of the movement of the (–) charges, your left fingers give the direction of the -field. All nice and consistent so that the -field is in the same direction no matter whether you use electron current or conventional current. Pretty slick, huh!?

To find the magnetic field at a point due to a moving charge, consider the situation depicted at the right. If a charge +q is moving as shown, the magnitude of the -field at a point P a distance | | from the charge at a particular instant is given by

where k is a constant. In MKSA units and

. is called the “magnetic permeability of a vacuum”

In vector notation:

where is a unit vector in the direction of (from q to P).

VII-2


Sample Exercise VII-1: Given the situation at the right where | | = 1 m,

| | = 1 m/s, and q = –1 C, the magnitude of at point P is __________ T (or

__________ gauss) and its direction is __________. (VII-3 #1)

For answers & solutions see “In-Text Fill-in Answers” beginning on page VII-14

To calculate the -field due to a current in a wire we use a

technique similar to that used in calculating the -field due to a

charged wire back in the first few pages of Unit II. Let be a teensy weensy hunk of a wire carrying a current I. If an amount of charge q moves a distance dL in time t then

Now if we have a TWH of wire carrying (dL) a current I instead of a single charge q, we can

write that the contribution to the total -field at a point P due to this “current element” IdL is

where is in the direction of conventional current flow.

TheTOTAL -field at point P then is the sum of the contributions of all the IdL’s along the whole wire. That is,

This is the Biot-Savart Law for the current carrying wires.

At this point try problem 1 at the back of this unit.

Calculating For Some Simple Geometries Using B-S Law

I. at the center of a circular loop – the contribution of the

current element to the total -field at P in vector notation is:

= ________________ (VII-3 #2)

VII-3


The magnitude of written in terms of (the angle between and ) is

dB = ________________ , where in this situation = _______ (VII-3 #3)

VII-4


The total -field at P then, is found by adding up the contributions of all the little ’s around the loop. Hence,

(VII-4 #1)

Integration gives

(VII-4 #2)

and if I is conventional current in the direction of is _____________. (VII-4 #3)

Let’s plug in some numbers. If I = 10 A and R = 2π cm, then at the center of the loop

is _________ T or _______ G. (VII-4 #4)

Suppose we have a coil containing 10 loops with each loop carrying 10 A. Now the

number of coulombs of charge moving around point P in each second is ___________

and thus is now ___________ T, ___________ . (VII-4 #5) (direction)

At this point try problems 2 & 3 at the back of this unit.

II. on axis through the center of a circular loop.

The magnitude of at P due to the current element is

dB = _____________________ (VII-4 #6)

{If you get stuck, see Example 30.3 on p. 942 in Serway & Beichner}

Verify for yourself that is in the direction shown in the drawing at point P if I is conventional current.

If is integrated around the loop the component of perpendicular to the axis, will

integrate to ______ (VII-4 #7). Thus, when we integrate, we will only need to consider the component of parallel to the axis along D.

The limits of integration will be from

L = _________ to L = _________.

VII-5


Now

but in terms of r and R, sin =_________ (VII-5 #1). Hence,

____________ (VII-5 #2) (Constants)

and integrating dL from L = ________ to L = ________ (VII-5 #3) we get:

(VII-5 #4) (magnitude) (direction)

Putting r in terms of the distance D and the radius of the loop R,

(VII-5 #5) (magnitude) (direction)

Note that if P is at the center of the loop (D = 0) then r = R and BP reduces to the expression we derived in I on p. VII-4 for the -field at the center of a circular loop.

III. due to a long straight wire.

In the situation at the right the contribution of the current element to the total -field at P in vector notation is:

= ____________________ . (VII-5 #6)

Writing the magnitude of in terms of :

dB = ____________________ . (VII-5 #7)

{For an alternative method of setting up the integral see Example 30.1 on p. 940 in Serway & Beichner}

But x, , and are not independent of one another, thus, we must write dB in terms of one variable so it can be integrated. Let’s do it:

i. Write x in terms of D and : x = ____________________. (VII-5 #8)

ii. Differentiating (i) gives: dx = ____________________. (VII-5 #9)

iii. Write R in terms of D and : R = ____________________. (VII-5 #10)

iv. Write an expression involving sine and cosine relating and

____________________ = ____________________. (VII-5 #11)

VII-6


Substituting (i) through (iv) into the expression on the previous page for dB:

dB = _________________________ = _________________________. (VII-6 #1)

Now that we have dB in terms of one variable () we can integrate to find the total at

P due to all the ’s. Assuming the wire to be real long (L ~ ∞), the limits of

integration for are from = __________ to = __________. (VII-6 #2)

Hence

(VII-6 #3)

And if I is conventional current, the direction of is _______________. (VII-6 #4)

At this point try problems 4 through 12 at the back of this unit.

Force Between Parallel Current Carrying WiresIn the sketch at the right, conventional current is to the right in wire #1 thereby producing a -field that is out of the page above wire #1 and into the page below it as shown. If we place a second wire (#2) parallel to #1 and a distance d away from it, then #2 is sitting in the -field of wire #1. Since wire #1 is carrying a current I1, then the magnitude of the -field that wire #2 is sitting in is given by

B1 = ________________. (VII-6 #5)

Now suppose wire #2 is carrying a current I2 also to the right. Since it is a current carrying wire in a magnetic field of strength B1 along its entire length L that is parallel to wire #1, the force it feels due to B1 is

(VII-6 #6) direction

Writing B1 in terms of I1, and d the expression for becomes:

(VII-6 #7) magnitude direction

According to Newton’s 3rd law, wire #1 should (and does) feel a force that is equal in magnitude but opposite in direction to the force on #2. That is,

(VII-6 #8)

VII-7


magnitude direction

VII-8


is, of course, caused by the fact that wire #1 is sitting in the magnetic field created by the current in wire #2.

Note that in the force equations the length, L, of the wire appears. But with two wires which length do we use in the equations for the force if the two wires have different length?

Remembering that in the expression the meaning of L was the length of the wire in

the magnetic field and a little of your vast store of reasoning power you can see that in the force equation for the force between parallel wires, L must be the length of the shorter of the two sections of wire that are parallel to one another.

Using the right (or left) hand rule, you should be able to show yourself that in the situation at the right that no matter whether the currents are due to (+) or (–) charges, if the currents are in the same direction (parallel) the wires will attract one another. If the currents are in opposite direction (anti-parallel), they will repel one another.


AMPERE’S LAW

Ampere’s Law is an alternative formulation of the B-S LAW relating currents to the -fields they produce.

Consider the -field produced by a super long wire carrying a current I into the page as shown in the sketch at the right. The

magnitude of as found using the B-S law is

B =

where d is some perpendicular distance from the wire. Rearranging we get

Notice that: (1) B is the magnitude of the -field at all points that are a distance d from the wire and (2) 2d is the circumference of a circular loop of radius d about the wire.

It turns out that if the product of B and any arbitrary closed path is equal to µ0 times the total current flowing through the path. Formally this idea is expressed as AMPERE’S LAW:

where integrate around a closed loop

VII-9


Ampere’s Law states: If the component of along a path (that is, Bcos , where is the angle

between and ) is multiplied by a teensy weensy hunk of the path ( ) and the product added up (integrated) around a closed path, the result is proportional to the net current enclosed by the path.

Work problem 17 at the back of this unit.Let’s show that the path can be of arbitrary shape and that the line integral is zero is the path does not enclose any current.

Let’s look again at our super long wire carrying current I into the page.

Its -field is given by

B =

Evaluating along path a to c to d to b to e and

back to a, note that from a to c and from d to b since the angle between and

along these sections is 90º. Hence,

{recall that is “arc cd” and “arc bea”}

Now

B2 =

and a little algebraic messing around gives

B1r = B2R.

From the sketch we can see that so

.

Therefore, the line integral becomes

.But

.Thus

VII-10


So you see, we can fiddle around with the path all we want and still get the same result!

VII-11


Now lets evaluate along a path that doesn’t contain any current.

Evaluating along the path a to b to c to d back to a:

Now

1)

2)

3)

4) (VII-9 #1)

Using the same technique as with the previous example, it can be shown that B1r and B2R are related in the following way:

Using this and (1) through (4) above, then

(slick huh!?)

Sure, in this example we chose a simple Mickey Mouse® path, but it can be shown to be true for any arbitrary closed path.

Before going over the following example check out the simple Ampere’s law problem worked out in Example 30.4 on page 947 in Serway & Beichner.

Sample Exercise VII-1: The diagram at the right shows a hollow cylindrical conductor of inner radius a and outer radius b carrying a current I (in the direction indicated) uniformly spread over its cross sectional area.

a) Show that at a distance r from the axis of the cylinder

where a < r < b, that the magnitude of the -field is given by

Check this formula for the limiting cases of a = 0 and r = a. What is the direction of ?

VII-12


b) Make a rough plot of B from r = 0 to r = ∞.

Solution:

a) Applying Ampere’s law let’s integrate around a path within the body

of the cylinder where is constant in magnitude and always in the direction of .

That is, integrate around a circle of radius r. The left hand side becomes:

(Eqn I)

Since I is distributed uniformly over the cross-section of the wire, we must find out how much of current I is within the circle of circumference 2πr. Now

and therefore the current within area πr2 is

Thus the right hand side of Ampere’s Law is

(Eqn II)

Equating Equations I and II:

Solving for B:

(YAY - we did it!)

Direction of : point right thumb in the direction of the current I and wrap your hand

around the cylinder. Your right fingers indicate is counter-clockwise as viewed from

the end of the cylinder (the direction of in the sketch).

VII-13


When a = 0, B reduces to B which is what we would expect for the magnetic field within a solid cylinder.

When r = a, B is zero as we would expect since there is no current inside a path of circumference 2a.

VII-14


b) For r < a, B = 0.

For r > b, the -field is just that for a straight wire:

B

For a < r < b:

where the term a2/r causes a small deviation from B being linear. If we ignore a2/r,

Plotting these three ranges of r give the graph shown above.


-field Inside a Solenoid by Ampere’s LawA solenoid is a tightly wound helix of wire. When a current flows in the

wire a -field is set up roughly the shape shown. The direction of is determined using the right-hand-rule with the conventional current flowing counter-clockwise looking left to right along the axis of the solenoid. If the length of the solenoid is large in comparison to its diameter, we can make two simplifying assumptions that make using

Ampere’s Law to find the -field inside the solenoid a cinch. These

are (1) the -field outside the solenoid close to its surface is zero and (2) the -field inside the solenoid is constant in magnitude and direction (if we stay away from the ends).

Let’s use Ampere’s Law to find -field inside the solenoid. In the drawing at the right we have a section of the solenoid that has

been sliced in half. The and indicate that the conventional current is out of the page and into the page in the wires. Convince

yourself using the right hand rule that is in the correct direction.

Using the assumptions above, evaluate over the path a to b to c to d and back to a.

1)

VII-15


2) (VII-11 #1)

VII-16


3)

4) (VII-12 #1)

Adding your results in (1), (2), (3), and (4)

(VII-12 #2)

Let n equal the number of turns of wire per unit length on the solenoid. Thus, the number of wires each carrying current I within our path is nL and, therefore, the total current enclosed is

_______________ . (VII-12 #3)

So

(VII-12 #4)

Solving for the magnitude of inside the solenoid:

B = __________________ (VII-12 #5)

-Field Inside a ToroidA TOROID is pictured at the right. A toroid is nothing more than a solenoid bent into the shape of a doughnut. The second drawing represents a cross section of the toroid with the current

I out of the page on the outer portion, ( ), and into the page

on the inner portion, ( ). Consider the dotted paths shown. If

we evaluate Ampere’s Law along path 1 we find that inside the “doughnut hole” is _________ (VII-12 #6) because the current enclosed within the path is _________ (VII-12 #7). If we do the

same along path 3 we find that outside the toroid is ________ (VII-12 #8) because the total current enclosed within path 3 is

________ (VII-12 #9). Now we are left with the problem of finding within the coils of the toroid itself. Applying the right-hand rule and noting the symmetry of the toroid, you should be able to

VII-17


convince yourself that, for a given R, the -field is constant in magnitude and tangent to the

circle labeled path 2. Evaluating around path 2:

(VII-12 #10)

VII-18


Now if there are a total of N turns of wire on the entire toroid each carrying a current I, then the total current enclosed within path 2 is __________ (VII-13 #1) and Ampere’s Law simplifies to

(VII-13 #2)

Solving for the magnitude of the -field we find that B inside a toroid is given by

B = _________________. (VII-13 #3)

Now try problems 21 through 25 at the back of this unit.

– End Unit VII –

VII-19


In-Text Fill-In Answers

Page Fill-In # Fill-In Answer

Vii-3 1 10-7 T, 10-3 gauss, Out of page

2

3

VII-4 1 , where the limits of integration are L = 0 to L = 2R

2

3 out of the page

4 10-4 T or 1.0 G

5 100 C, . . . is now 10-2 T, out of the page.

6

7 zero

VII-5 1 sin = R/r

2

3 integrate dL from L = 0 to L = 2R

4

5

6

7

VII-20


VII-5 (Cont’d)

8 x = D tan9 dx = D sec2d

10 R = D/cos = Dsec 11 cos = sin

VII-6 1 simplify

2 limits: from = - /2 to = /2

3

4 , into the page

5

6 , up toward the top of the page

7 , up toward the top of the page

8 , down toward the bottom of the page

VII-9 1 1)

2)

3)

4)

VII-21


VII-11 1 1)

2)

VII-12 1 3)

4)

2

3

4

5

6

7

8

9

10

VII-13 1

2

3

VII-22

Unit VII – CALCULATING MAGNETIC FIELDSVII-23


End of Unit Problems

1. Imagine the commute element in which the current lies in the x-y plane and is directed to the right parallel to the x-axis. What is the magnetic field at the origin due to this current element if located:

a. at the point (0, a, 0) meters? [Ans: ]

b. at the point (a, a, 0) meters? [Ans: ]

c. at the point (a, 0, 0) meters? [Ans: ]

2. The wire shown in the diagram at the right extends infinitely in both directions and carries a current I. What is the magnetic field at the center (point C) of the semicircle arising from:a. each infinite straight segment of wire? [Ans: 0]

b. the semicircular segment of radius R? [Ans: , N2Page]

c. the entire wire? [Ans: , N2Page]

3. Consider the loop of wire sketched at the right. The curved sections are segments of circles of radii a and b. The straight segments are along the radii.Find the magnetic field at point P, assuming the current in the loop is I.

[Ans: , OofPage]

4. A wire carrying is oriented horizontally along a north-south direction. The north seeking end of a compass needle placed directly above the wire is deflected toward the west. What is the direction of flow of conventional current in the wire? [Ans: Southward]

5. A long straight wire carrying a 5 A current of electrons is

perpendicular to a uniform -field of magnitude 10-4 T toward the top of the page as shown in the sketch at the right. Find the

magnitude and direction of the TOTAL -field at points P, Q, R, and S which lie on a circle with a radius of 1 cm around the wire.

[Ans: , 45o out of the plane of the page; ;

, 45o inward from the plane of the page;

VII-24


, toward the top of the page]

End of Unit Problems6. Two long, straight, parallel wires are 1.0 m apart and are perpendicular

to the page as shown in the sketch at the right. The upper wire carries a current of 6 A into the page.

a. What must be the magnitude and direction of the current in the lower wire for the total -field at point P to be zero? [Ans: 2 A N2Page]

b. If the lower wire is carrying the current calculated in (a), what is at point Q? at point S?

[Ans: ]

7. If a +q point charge is moving with a speed v at a distance d from the axis of a long straight wire carrying a current I and is traveling perpendicular to the axis of the wire, show that the magnitude of the force acting on the charge is given by

What is the direction of the force if the charge is moving toward the wire? away from

the wire? [Ans: toward wire then is toward the bottom of the page, away from

the wire then is toward the top of the page]

8. Show that the magnitude of the -field at the center of a square loop having sides of length L and carrying current I is given by

.

(Hint: find the contribution due to one side and multiply by 4)What is the direction of if I is flowing in the direction shown in the diagram? [Ans: OofPg]

9. The magnetic field of the earth is about 0.7 gauss at a certain location. A current is caused to flow in the circular loop so that it exactly cancels the magnetic field of the earth at the center of the loop. If the loop has a

radius of 5 cm and is in the direction shown in the sketch of the right, what is the magnitude and direction of the current flowing in the loop? [Ans: 5.6 A, CCW]

VII-25



10. In the sketch at the right, what must be the magnitude and direction of the current straight wire I1, if the magnetic field is to be zero at the center of the loop carrying current I2?[Ans: 7.85 A, toward the bottom of the page]

11. Two circular coils, each having N turns of wire, have a radius R and are separated by distance 2R as shown in the sketch

of the right. Show that the -field at a point P on the axis of the coils midway between them is given by

, to the right.

Assume that the cross-sectional area of each coil is small in comparison to R2.

12. Recall the flux of the the electric field was given by

where the integral is evaluated over the entire area of interest. Similary the flux of the magnetic field is

.

The long straight wire AB in the sketch of the right carries a current I.a. What is the magnetic field at the shaded area at a perpendicular

distance x from the wire? [Ans: , N2Page]

b. What is the magnetic flux through the shaded area? [Ans: ]c. What is the magnetic flux through the entire rectangular area CDEF in terms of I, L, a,

and b? [Ans: ]

13. A messy loop of flexible wire is placed on the smooth horizontal surface an attached at points a and b to wires that are in turn connected to a power source. If a current I is caused to flow in the direction he indicated in the sketch, will the wire try to expand forming a circular loop or scruntch up into an even smaller mess?

VII-26

Unit VII – CALCULATING MAGNETIC FIELDSVII-27



14. Three parallel wires are shown at the right caring currents I1, I2, and I3, respectively.

a. If D = d, how should the magnitudes and directions of currents I1 and I3 compare if wire #2 is to experience a net magnetic force of zero? [Ans: I1 = I3, same direction]

b. If I1 = 3I3, how should D and d compare if #2 is to experience a net magnetic force of zero? [Ans: D = 3d]

15. Two parallel wires 2 m long and separated by distance of 1 m are caring equal currents. Each experiences of a repulsive magnetic force due to the other of 1.6 x 10-6 N. What is the magnitude of the current in each wire? Are the currents parallel or anti-parallel? [Ans: 2 A, anti-parallel]

16. In the sketch at the right the long wire AB carries a current of 20 A. A rigid rectangular loop whose sides are parallel to the wire AB carries a current of 10 A. Both currents are in the direction indicated.a. When calculating the net force on the loop the force on the top and

bottom sections of the loop due to the magnetic field of AB need not be included. This is also true of the force on each side of the loop due to the magnetic field of the other sides of the loop. Explain why these two statements are true.

b. Find the magnitude and direction of the net magnetic force on the loop. [Ans: 0.7 mN, left]

17. Evaluate around the dotted paths shown below.

18. A large number N of long straight wires are arranged symmetrically on the outer surface of a cylinder of radius R with the wires parallel to the axis of the cylinder. Each wire carries the same current I. Using Ampere’s law, describe the magnetic field inside and outside the cylinder.

19. The sketch at the right shows a cross-section of a long conductor of the type called a coaxial cable. Its dimensions are labeled in the sketch. There are equal but opposite currents I flowing in the two conductors. The currents are uniformly distributed across their areas. Derive expressions for the magnetic field in the ranges:

a. r < c [Ans: B = ] b. c < r < b [Ans: B = ]

c. b < r < a [Ans: B = d. r > a [Ans: B = 0]

VII-28


End of Unit Problems20. Use Ampetre's law to find the magnitude and direction of the magnetic field 4 m from a

long straight coaxial conductor if the inner conductor carries is a current of 4 A in the outer a current of 10 A in the opposite direction. [Ans: 0.3 T, CW if we take the 10 A current as N2Page]

21. A magnetic field 0.07 T exists within a solenoid 50 cm long and 2 cm in diameter.

a. Calculate the magnetic flux B within the solenoid. [Ans: ]b. Calculate the number of terms of wire necessary if the current is 5 A. [Ans: 5570

turns]

22. A solenoid is 30 cm long and is wound with two layers of wire. The inner layer has 300 turns and the outer layer 250 turns. The current is 3 A in the same direction in both lawyers. What is the magnitude of the magnetic field at a point near the center of the solenoid? [Ans: 6.9 mT]

23. A toroid having a square cross-section, 5 cm on edge, and an inner radius of 15 cm has 500 turns and carries a current above 0.8 A. What is the magnitude of the magnetic field at:a. The inner radius of the toroid? [Ans: 533 T]b. The outer radius of the toroid? [Ans: 400 T]

24. The sketch of the right shows a cross-section of an infinite conducting sheet with a current per-unit length emerging out of the page as shown.a. Divide the sheet into teensy-weensy strips of width dx each caring a current of dx.

Show that the magnitude of the magnetic field at any point above or below the sheet is

given by and it’s direction is to the left above the sheet and to the right

below.b. Given that the magnitude of the magnetic field is constant everywhere above or below

the sheet use Ampere’s law to show that .

25. Consider an infinite slab of conducting material as shown at the right carrying a uniform current per unit area in the direction shown.a. What is the direction of the magnetic

field at point A and B within the slab?

[Ans ]b. Use Ampere’s law and symmetry arguments to show that the magnitude of the

magnetic field at point A is given by

VII-29

unit ii - mr. greenberg physics - home · web viewfundamentals of physics, halliday, resnick, &...

Documents