VII-1
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 Objectives
When you have completed Unit VII, you should be able to:
1. Use Biot-Savart law to
a. 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 to
a. 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-2
Unit VII – CALCULATING MAGNETIC FIELDS
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 v 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
.
is called the “magnetic permeability
of a vacuum”
In vector notation:
where
is a unit vector in the direction of
and
(from q to P).
VII-3
Unit VII – CALCULATING MAGNETIC FIELDS
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 Fillin 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
where
-field at a point P due to this “current element” IdL is
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 B For Some Simple Geometries Using B-S Law
I.
B at the center of a circular loop – the contribution of the
current element
to the total B -field at P in vector
notation is:
dB = ________________
(VII-3 #2)
The magnitude of dB written in terms of  (the angle between dL and R ) is
dB = ________________ , where in this situation  = _______
(VII-3 #3)
VII-4
Unit VII – CALCULATING MAGNETIC FIELDS
The total B -field at P then, is found by adding up the contributions of all the little
around the loop. Hence,
’s
The limits of integration will be from
L = _________ to L = _________.
(VII-4 #1)
Integration gives
Bat P = _________________
(VII-4 #2)
and if I is conventional current in the direction of Bat P is _____________.
(VII-4 #3)
Let’s plug in some numbers. If I = 10 A and R = 2π cm, then B 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 Bat P is now ___________ T, ___________ .
(VII-4 #5)
(direction)
At this point try problems 2 & 3 at the back of this unit.
II. B on axis through the center of a circular loop.
The magnitude of dB 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 dB is in the direction shown in the drawing at point P if I is
conventional current.
If dL is integrated around the loop the component of dB perpendicular to the axis, will
integrate to ______ (VII-4 #7). Thus, when we integrate, we will only need to consider the
component of dB parallel to the axis along D.
VII-5
Unit VII – CALCULATING MAGNETIC FIELDS
Now
but in terms of r and R, sin =_________ (VII-5 #1). Hence,
BP = ____________
(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 B -field at the center of a circular loop.
III.
B due to a long straight wire.
In the situation at the right the contribution of the current
element
to the total B -field at P in vector notation is:
dB = ____________________ .
(VII-5 #6)
Writing the magnitude of dB 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 R 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 :
ii.
Differentiating (i) gives:
iii.
Write R in terms of D and :
iv.
Write an expression involving sine and cosine relating  and 
x = ____________________.
(VII-5 #8)
dx = ____________________.
(VII-5 #9)
R = ____________________.
(VII-5 #10)
____________________ = ____________________.
(VII-5 #11)
VII-6
Unit VII – CALCULATING MAGNETIC FIELDS
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 B 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 B is _______________.
(VII-6 #4)
At this point try problems 4 through 12 at the back of this unit.
Force Between Parallel Current Carrying Wires
In the sketch at the right, conventional current is to the right
in wire #1 thereby producing a B -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 B -field of wire #1.
Since wire #1 is carrying a current I1, then the magnitude of the B -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 Fon #2 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)
magnitude
direction
VII-7
Unit VII – CALCULATING MAGNETIC FIELDS
Fon #1 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 B 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.
At this point try problems 13 through 16 at the back of this unit.
AMPERE’S LAW
Ampere’s Law is an alternative formulation of the B-S LAW relating currents to the B -fields
they produce.
Consider the B -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 B 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 B -field at all points that are a distance d from the
wire and (2) 2d 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:
ò B × dL =
where
ò means integrate around a closed loop
Ampere’s Law states: If the component of B along a path (that is, Bcos , where  is the angle
between B and dL ) is multiplied by a teensy weensy hunk of the path ( dL ) 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.
VII-8
Unit VII – CALCULATING MAGNETIC FIELDS
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 B -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 B and dL
along these sections is 90º. Hence,
{recall that cd and bea is “arc cd” and “arc bea”}
Now
B2 =
and a little algebraic messing around gives
B1r = B2R.
From the sketch we can see that cd = Rf and ab = rf so
.
Therefore, the line integral becomes
.
But
.
Thus
Ü AMPERE'S LAW
So you see, we can fiddle around with the path all we want and still get the same result!
VII-9
Unit VII – CALCULATING MAGNETIC FIELDS
Now lets evaluate
Evaluating
ò B × dL along a path that doesn’t contain any current.
ò B × dL 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 B -field is
given by
Check this formula for the limiting cases of a = 0 and r = a. What is the direction of B ?
b) Make a rough plot of B from r = 0 to r = ∞.
VII-10
Unit VII – CALCULATING MAGNETIC FIELDS
Solution:
a) Applying Ampere’s law
ò B × dL =
let’s integrate dL around a path within the body
of the cylinder where B is constant in magnitude and always in the direction of dL .
That is, integrate dL 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 B : point right thumb in the direction of the current I and wrap your hand
around the cylinder. Your right fingers indicate B is counter-clockwise as viewed from
the end of the cylinder (the direction of dL in the sketch).
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 2a.
VII-11
Unit VII – CALCULATING MAGNETIC FIELDS
VII-12
Unit VII – CALCULATING MAGNETIC FIELDS
b) For r < a, B = 0.
For r > b, the B -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,
B µr
Plotting these three ranges of r give the graph shown above.
At this point try problems 18 through 20 at the back of this unit.
B -field Inside a Solenoid by Ampere’s Law
A solenoid is a tightly wound helix of wire. When a current flows in the
wire a B -field is set up roughly the shape shown. The direction of B
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 B -field inside the solenoid a cinch. These
are (1) the B -field outside the solenoid close to its surface is zero and (2) the B -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 B -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 Ä 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 B is in the correct direction.
Using the assumptions above, evaluate
b to c to d and back to a.
ò B × dL over the path a to
1)
2)
(VII-11 #1)
VII-13
Unit VII – CALCULATING MAGNETIC FIELDS
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 B inside the solenoid:
B = __________________
(VII-12 #5)
B -Field Inside a Toroid
A 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, ( Ä ). Consider the dotted paths shown. If
we evaluate Ampere’s Law along path 1 we find that B 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 B 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 B within the coils of the toroid itself. Applying the righthand rule and noting the symmetry of the toroid, you should be
able to convince yourself that, for a given R, the B -field is
constant in magnitude and tangent to the circle labeled path 2.
Evaluating
ò B × dL around path 2:
(VII-12 #10)
VII-14
Unit VII – CALCULATING MAGNETIC FIELDS
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 B -field we find that B inside a toroid is given by
B = _________________.
Now try problems 21 through 25 at the back of this unit.
– End Unit VII –
(VII-13 #3)
VII-15
Unit VII – CALCULATING MAGNETIC FIELDS
In-Text Fill-In Answers
Page
Fill-In #
Vii-3
1
Fill-In Answer
10-7 T, 10-3 gauss, Out of page
2
3
VII-4
VII-5
ò dL , where the limits of integration are L = 0 to L = 2R
1
Bat P =
2
Bat P =
3
out of the page
4
10-4 T or 1.0 G
5
100 C, . . . Bat P is now 10-2 T, out of the page.
6
dB =
7
zero
1
sin = R/r
2
BP =
3
integrate dL from L = 0 to L = 2R
4
BP =
5
BP =
6
7
dB =
VII-16
Unit VII – CALCULATING MAGNETIC FIELDS
VII-5 (Cont’d)
8
x = D tan

9
dx = D sec2d

10
R = D/cos  = Dsec 
11
cos  = sin 

VII-6
simplify Þ dB =
1
dB =
2
limits: from  = - /2 to  = /2
3
BP =
4
Ä , into the page
5
VII-9
6
Fon #2 =
, up toward the top of the page
7
Fon #2 =
, up toward the top of the page
8
Fon #1 = -Fon #2 =
1
1)
2)
3)
4)
, down toward the bottom of the page
VII-17
Unit VII – CALCULATING MAGNETIC FIELDS

VII-11
1
1)
2)
VII-12
1
3)
4)
2
ò B × dL = BL =
3
4
ò B × dL = BL =
5
B=
6
7
8
9
10
VII-13
1
2
3
Btoroid =
VII-18
Unit VII – CALCULATING MAGNETIC FIELDS
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;
, toward the top of the page]
;
VII-19
Unit VII – CALCULATING MAGNETIC FIELDS
End of Unit Problems
6. 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-20
Unit VII – CALCULATING MAGNETIC FIELDS
End of Unit Problems
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-21
Unit VII – CALCULATING MAGNETIC FIELDS
End of Unit Problems
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-22
Unit VII – CALCULATING MAGNETIC FIELDS
End of Unit Problems
20. 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