the magnetic fields produced by moving charges particularly currents in wires

Chapter 33
The magnetic field of a current

Introduction:
In previous chapter, effect of magnetic field on moving charge was studied.
• In this chapter, we study the magnetic fields produced by moving charges
particularly currents in wires.
• Electric fields produced by charge distributions.
• Magnetic fields produced by current distributions: Ex: Straight wires and
circular loops.
• Relationship between electric and magnetic fields.
33.1. The Magnetic field due to a moving charge:
• In chapter 23, we studied the force experienced by a charged particle moving in
a magnetic field. The moving electric charges are the probes of the magnetic
fields. This was demonstrated by Hans Christian Oersted in 1820.
• When a compass is placed near a straight wire carrying a current, the compass
needle aligns so that it is tangent to a circle drawn around the wire (neglects
influence of Earth’s magnetic field on the compass). Oersted’s discovery
provided the first link between electricity and magnetism (fig. 33-1).
• Another experiment in 1876 by Henry Rowland is shown in fig. 33-2 where by
rotating a disk of charge (by connecting a battery to a layer of gold deposited on
the surface of a disk of insulating material) about its axis, he was able to
produce moving charges and showed their magnetic effect by suspending a
magnetized needle near the disk.

• Goal of this chapter: Magnetic interactions between two moving charges just a Coulomb studied the electric interaction
between charges at rest.
• Similar to the electrostatic measurement, magnetic force between two charges in motion should be measured but the force is
extremely small and very difficult to measure (0.00001 of Earth’s field in Rowland’s experiment).
• It is easy to study of the sources of the magnetic field by studying how a single moving charge produces a magnetic field.
• But this approach is not practical and it is easier to produce magnetic fields in the laboratory by using moving charges in the
form of currents in wires.
• Thought experiment: Project a single charge q with velocity v and detect the field with a suspended magnetic needle that is free
to align in any direction.
• We have a setup an experiment in a region in which Earth’s magnetic field is negligible, this can be done by using current
carrying coils in our laboratories to create fields that cancels Earth’s field.
• Fig. 33-3a shows the outcome of some measurements of B at different locations.
• The moving charge sets up a B and the needle indicates the direction of the field at any
location.
• To determine the magnitude of the field by measuring the force on a second moving
charged particle.
• Here are some properties of magnetic field due to a moving charge:
1. B  v and also to charge q.
2. If v reverses direction or q changes sign, 𝐵 is reversed.
3. B=0 at points along the direction of v (forward and backward). In other directions, B varies
as sin (fig. 33-3 b).
4. B is tangent to circles drawn about v in planes perpendicular to v, with the direction of B
determined by right hand rule (point thumb in the direction of v and fingers will curl in the
direction of B. On any given circle, magnitude of B is same at all points.

5. At points on a line perpendicular to the direction of motion of q (fig. 33-3b) or equivalently on circles of increasing radius
drawn around the line of motion, then magnitude of B decreases like 1/r2, where r is the distance from q to the observation
point.
• One can define B as shown in fig. 33-4. At an arbitrary point P (where magnetic field
can be found), B is perpendicular to the plane determined by v and Ԧ
r (vector that located
P relative to q).
• Since B  q, v and sin and 1/r2.
• In vector form, eqn. 33-1 becomes
• Where K is the proportionality constant to be determined. Here Ƹ
𝑟 is the unit vector in
the direction of Ԧ
r and since Ƹ
𝑟=Ԧ
r/r (similar to Coulomb’s law), we can write eqn. 33-2 as
• To obtain B of a moving charge, K has to be determined as we have inserted the
constant 1/4o into Coulomb’s law.
• The constants in the magnetic and electric field equations are not independent quantities; they are related by the speed of light
(which will be discussed in chapter 38).
• Speed of light is defined by the quantity, our choices are either:
1. To use electric force law (Coulomb’s law) to define the electric constant and then use the value of the speed of light to obtain
the magnetic constant or
2. To use a magnetic force law (force between current-carrying wires, which will be discussed in coming topics) to define
magnetic constant and then use the value of speed of light to obtain electric constant.

• Since the magnetic force between the current-carrying wires can be measured more precisely than the electric
force between the charges, we choose the second method.
• K is defined in SI units to have a exact value 10-7 tesla.meter/ampere (T.m/A).
• Where 0 is known as permeability constant or magnetic constant and has the exact value
• 0 plays a role in calculating magnetic fields similar to o in calculating electric fields. These two constants are
related through the speed of light: c=(o 0)-1/2. Defining 0 and c then determines o exactly.
• Now the complete expression for the magnetic field due to a moving charge is:
• The magnitude of B as
• Where  is the angle between v and Ԧ
r.
• We will use eqn. 33-4 and 33-5 to study steady currents in the next section.

33.2. The Magnetic field of a current:
• In the previous section, we studied that magnetic fields can be produced using current carrying wire rather than the motion of
individual charges.
• In this section, we will find the magnetic field due to a current.
• How to find: Find the magnetic field due to the current in a short element of the wire, and then use integration methods to
find field due to the current in the entire wire. The same method we have used in chapter 26 to find the electric field due to the
continuous charge distribution. (refer to chapter 26.4 for complete details).
• Calculation of magnetic field: Magnetic force exerted on an individual moving charge: 𝐹 B =qvB -----eqn. (32-3)
• Magnetic force exerted on a current element, d𝐹 B =i𝑑𝐿 𝐵 ------eqn. 32-29. 𝑑𝐿 is a vector whose length is equal to the length
of the element of the wire and whose direction is the direction of the current in that element.
• We have gone from an expression describing the force on an individual moving charge to one describing the force on a current
element by replacing qv by i𝑑𝐿 . We can modify eqn. 33-4 in exactly due to a current element, which can be represented by a
charge element q moving with velocity v.
• v=dԦ
s/dt so that dq moves through the displacement dԦ
s in the time interval dt. Now we have
• Substituting eqn. 33-7 into eqn. 33-6, we get
• Eqn. 33-8 is called Biot-Savart law.
• The direction of dB is same of direction of dԦ
sԦ
r. The magnitude of field element dB is
Where  is the angle between dԦ
s and Ԧ
r from the current element to the observation point P.

• Fig. 33-6 shows the vector relationships; compare with fig. 33-4 and note how similar the two
figures are…
• Total field B due to the entire current distribution, integrate over all currents idԦ
s:
• Applications of Biot-Savarat law: Calculate magnetic fields od some current-carrying wires
of different shapes.
1. A straight wire segment: Calculation of B due to current i in a straight wire segment of length
L (fig. 33-7).
• Wire lies along z-axis. To find B at P on the y axis, a distance d from the wire.
• The center of the wire is at origin, so P is on the perpendicular bisector of the wire.
• Step 1 is to choose arbitrary element on idԦ
s, which is located at coordinate z relative to the
origin.
• Using eqn. 33-8 and using right hand rule of fig. 33-7, dԦ
sԦ
r points in the –ve x direction, where
matter where on the wire we choose the current element.
• Adding up all the elements dB, one can find the total field in the –ve x direction.
• ds=dz and z ranges from –L/2 to +L/2.
• To integrate eqn. 33-9, express  and r in terms of integration variable z:

• This problem reminds us of its electrostatic equivalent. In chapter 26, we have derived an expression for E due to a long
charged rod by integration methods, using Coulomb’s law. The same problem was solved using Gauss’ law. Later, we consider
a law of magnetic fields, Ampere’s law, which is similar to Gauss’ law in that it simplifies magnetic field calculations in cases
(such as this case) that have a high degree of symmetry.
• Just as the electric fields, the magnetic field of a current-carrying wire by set of magnetic field lines for a long straight wire is
shown in fig. 33-8.
• The field lines form concentric circles around the wire (By Oersted’s experiment, fig. 33-1) and also indicated by the pattern
of iron fillings near a wire (fig. 33-9).
• At any point, the direction of 𝐵 is tangent to the field line at that point. The field is large where the field lines are close
together (near wire) and small where the field lines are farther apart (far from wire).
• Contrary to electric field lines due to charges ,which begin on positive charges and end on negative charges, the magnetic field
lines due to currents form continuous loops with no beginning or end.
• To find direction of field lines, we use right-hand rule: thumb in the direction of current and fingers would curl around the
wire in the direction of magnetic field.

2. A circular current loop: Circular loop of radius R carrying a current i (fig.33-10).
• Let us calculate 𝐵 at P on axis a distance z from the center of the loop.
•  is the angle between idԦ
s and Ԧ
r is 900.
• Resolve dB into 2 components, one is dBz along the axis of the loop and another, dB⊥ at
right angles to the axis. Only dBz contributes to the total magnetic field at P as they add all
current elements which lie on axis and add directly. The sum of all dB⊥ for complete loop is
zero from symmetry.

• This dependence of the field on the
inverse cube of the distance reminds us of
the electric field of an electric dipole.
• One can consider a loop of wire to be a
magnetic dipole.
• Just as the electric behavior of many
molecules can be characterized in terms of
their electric dipole moment, so also the
magnetic behavior of atoms can be
described in terms of their magnetic
dipole moment.
• In case of atoms, the current loop is due to
the circulation of electrons about the
nucleus.

33.3. Two parallel currents:
• Here, we use long wires carrying parallel (or anti-parallel) currents to illustrate two
properties of magnetic fields: the addition of the fields due to different wires and
the force exerted by one wire on the another.
• First, let us consider the vector addition of the fields due to two different parallel
wires (fig. 33-11).
• 2 wires are perpendicular to the plane of the figure and they carry currents in
opposite directions.
• We will find magnetic field at P due to 2 wires.
• The magnetic field lines due to wire 1 form concentric circles about that wire and
the magnitude of the field at the distance r1 is given by eqn. 33-13, B1= oi1 /2 r1.
• The direction of B1 is tangent to the circular arc passing through P, equivalently B1
is perpendicular to Ԧ
r1, the radial vector from the wire to P.
• Similarly, B2 is perpendicular to Ԧ
r2.
• To find net field: B= B1+B2.
• Magnitude and direction of the total field is obtained by vector addition.
• Fig.33-11 is similar to the method for calculating the total electric field due to two
point charges q1 and q2 i.e., E= E1+E2.
• To observe E at P (fig. 33-11), we measure force on a charged particle moving
through that point or on a third wire carrying a current through P.

• The right-hand rule shows that the direction of B1 at wire 2 is down, as shown in the figure.
• Wire 2 carries current i2 can thus be considered to be immersed in an external magnetic field produces B1
. A length L of this
wire experiences a sideways magnetic force 𝐹 21 =i2 𝐿 𝐵 1 of magnitude
• The vector rule for cross product shows that 𝐹 21 lies in the plane of the wires and points toward wire 1 in fig. 33-15.
• We could equally well have started with wire 2 by first computing B2 produced by wire 2 at the site of wire 1 and then
finding the force 𝐹 21 exerted on a length L of wire 1 by the field of wire 2. This force on wire 1 would, for parallel currents,
point toward wire 2 in fig. 33-15.
• The forces that 2 wires exert on each other are equal in magnitude and opposite in direction; they form an action-
reaction pair according to Newton’s third law.
• Parallel currents attract and antiparallel currents repel. This rule is opposite to the rule for electric charges, in that
like (parallel currents attract, but like (same sign) charges repel.
• The force between long, parallel wires is used to define the ampere.
The interaction between parallel currents:
• Now consider a different calculation which involves two long, straight wires carrying
parallel (or anti-parallel) currents. The result of the magnetic field due to one wire at the
location of the other wire, a magnetic force is exerted on the second wire.
• Similarly, the second wire sets up a magnetic field at the location of the first wire that exerts
a force on that wire.
• In fig. 33-15, wire 1 carrying current i1 produces a magnetic field B1 whose magnitude at the
location of the second wire is, according to eqn. 33-13,

33.4. The magnetic field of a solenoid:
• To obtain a nearly uniform magnetic field is to use a solenoid.
• A solenoid (fig. 33-16a) is a helical winding on a cylindrical core.
• The wires carries a current i and are wound tightly together, so that there are n
windings per unit length along the solenoid.
• To calculate the field along the central axis of the solenoid using previous result for
the magnetic field of a circular loop of wire.
• The calculation of the field off the axis is difficult using Bio-Savart law, but in the
next section, we will discuss a different and much easier way to calculate the field off
axis.
• Fig.33-16 b shows the geometry for calculating the field on the axis.
• The symmetry axis of the solenoid is along z-axis, with the origin at the center of the
solenoid.
• Aim: To find field at P which is at a distance d from the origin along the z axis.
• Assumption: Windings are so narrow that each can be considered as approximately a circular loop of wire which is parallel
to xy plane.
• Solenoid has N turns of turns of wire in a length L, so the number of turns per unit length is n=N/L.
• Consider a long ring of width dz. The number of turns in that ring is ndz and so the total current carried by the ring is ni dz
since each turn carries i.
• The field at P due to ring (by using eqn. 33-19) is
where z-d is the position of the ring relative to P.

• To find total field due to all rings: Integrate eqn. 33-26 from z= -L/2 to z= +L/2.
• Evaluating the integral (see integral 18 of Appendix I), we get
• Eqn. 33-27 gives the field on the axis of the solenoid at a distance d from its center. It is valid for points inside as well as
outside the solenoid. The direction of the field is determined using right-hand rule which implies if current is circulating
counterclockwise as viewed from above, the field is in the positive z direction.
• In an ideal solenoid, the length L is much greater than the radius R. In this case, eqn. 33-27 becomes
• Eqn. 33-28 gives the field of an ideal solenoid at all interior points, off
axis as well as on axis and the field is zero at all points outside the interior
of the solenoid.
• The field calculated from eqn. 33-27 is plotted as a function of location
along the axis in fig. 33-17 for an ideal solenoid and for two different
nonideal solenoids.
• Note: As solenoid becomes longer and narrower thus approaching ideal
behavior, the field along the axis becomes more nearly constant and drops
more rapidly to zero beyond the ends of the solenoid.

• How to understand the field in the interior of a solenoid: Consider the “stretched-out”
solenoid as shown in fig. 33-18.
• Very close to each wire, the magnetic behavior is nearly that of a long, straight wire, with the
field lines forming concentric circles around the wire. This field tends to cancel at points between
adjacent wires.
• Figure says that the fields from the individual loops of wire combine to form field lines that are
roughly parallel to the solenoid axis in its interior.
• In the limiting case of ideal solenoid, the field becomes uniform and parallel to the axis.
• At exterior points (P) in fig. 33-18, the field due to the upper part of the solenoid turns (marked
because the current is out of the page) points to the left and tends to cancel the field due to the
lower parts of the solenoid turns (marked , because the current is into the page), which points to
the right near P. In the limiting case of the ideal solenoid, the field outside the solenoid is zero.
• Taking the external field to be zero is a good approximation for a real solenoid if its length is
much greater than its radius and if we consider only external points, P.
• Figure 33-19: Magnetic field lines for a non ideal solenoid. From the spacing of the field lines
that the field exterior to the solenoid is much weaker than the field in the interior which is very
nearly uniform over the cross section of the solenoid.
• Solenoid is for magnetic fields whereas the parallel plate capacitor is for electric fields: a device
capable of producing a field that is uniform.
• In a parallel-plate capacitor, the electric field is nearly uniform if the plate separation is small
compared with the dimensions of the plate and if we are not too close to the edge of the capacitor.
In the solenoid, the magnetic field is nearly uniform if the radius is small compared with the
length and if we are not too close to the ends. Fig. 33-17, even for a length that is only 10 times
the radius, the magnetic field is with in a few percent of the field of the ideal solenoid over the

• This result is called Ampere’s law and is written as
• Using Gauss’s law, we first constructed an imaginary closed surface (a Gaussian surface)
that enclosed a certain amount of charge.
• Using Ampere’s law, we construct an imaginary closed curve (called Amperian loop , fig.
33-20).
• The LHS of eqn. 33-29 tells us to divide the curve into small segments of length dԦ
s.
• As we travel around the loop (in the direction if dԦ
s), we evaluate 𝐵 .dԦ
s and add
(integrate) all such quantities around the loop.
• Integral on the left of eqn. 33-29 is called the line-integral which has to be integrated
around a closed path.
• Let  represent the angle between the dԦ
s and 𝐵 , we write line integral as
• i represents the total current enclosed by the loop.
33.5. Ampere’s Law:
• Coulombs law is the fundamental law of electrostatics which is used to calculate electric field associated with any distribution
of electric charges.
• Gauss’s law us used to solve problems with high degree of symmetry with ease and also Gauss’s law is more basic than
Coulombs law and Gauss’ law is one of the four fundamental (Maxwell) equations of electromagnetism.
• Similar situation in magnetism. Using Biot-Savart law, one can calculate the magnetic field of any distribution of currents as
we used eqn. 26-6 or eqn.26-13 and 26.14 (which is equivalent to Coulomb’s law) to calculate the electric field of any
distribution of charges.
• A more fundamental approach to magnetic fields uses a law that (Gauss’s law for electric fields) takes advantage of the
symmetry present in certain problems to simplify the calculation of 𝐵 . This law is considered more fundamental law than the
Biot-Savart law and leads to another of the 4 Maxwell equations.

• In analogy with the charges in the case of Gauss’s law,
currents outside the loop are not included.
• Fig. 33-20 a, shows 4 wires carrying current. The 𝐵 at
any point is the net effect of the currents in all the wires.
• In the evaluation of RHS of eqn. 33-29, we include only
i1 and i2 because the wires carrying i3 and i4 do not pass
through the surface enclosed by the loop.
• The two wires that pass through the loop carry currents
in opposite direction.
• A right-hand rule is used to assign signs to currents:
with the fingers of the right hand in the direction in
which the loop is traveled, currents in the direction of
your thumb (such as i1) are taken to be positive whereas
currents in the opposite direction (such as i2) are taken
to be negative. The net current i in the case of fig. 33-20
a is thus i=i1-i2.

• The magnetic field 𝐵 at points on the loop and within the loop certainly depends on the currents i3 and i4;
however, the integral of 𝐵 .dԦ
s around the loop does not depend on currents such as i3 and i4 that do not penetrate
the surface enclosed by the loop., because 𝐵 .dԦ
s for the field established by i1 or i2 always has the same sign as we
travel around the loop; however, 𝐵 .dԦ
s for the fields due to i3 or i4 change sign as we travel around the loop, and in
fact the positive and negative contributions exactly cancel one another.
• Changing the shape of the surface without changing the loop does not change these conclusions.
• In fig. 33-20b the surface has been stretched upward so that now the wire carrying current i4 penetrates the
surface.
• Note: It does twice, one moving downward (-i4) to the total current through the surface, according to right hand
rule and once moving upward (+i4) to the total. Thus total current through the surface does not change; this is
expected because stretching the surface does not change 𝐵 at locations along the fixed loop and therefor line
integral on the left side of Ampere’s law does not change.
• The arbitrary constant 4 in Biot-Savart law reduces the constant that appears in Ampere’s law to simply 0. For
fig. 33-20, Ampere’s law gives
• Eqn. 33-31 is valid for 𝐵 as it varies in both magnitude and direction around the path of the Amperian loop.
• We cannot solve that eqn. for B unless we find a way to remove B from the integral. To do so, we use symmetries
in the geometry to choose an Amperian loop for which B is constant. We used similar trick for calculating electric
fields using Gauss’ law.

• From Oersted’s experiments, 𝐵 has only a tangential component. Thus the angle  is zero and the line integral becomes
• The integral of ds around the path is simply the length of the path or 2d in the case of a circle.
• The right side of Ampere’s law is simply 0i (taken as positive according to right-hand rule).
• Ampere’s law gives
• This to identical to eqn. 33-13, a result that was obtained with Biot-Savart law.
Applications of Ampere’s law:
• Following examples illustrate how Ampere’s law can be used to calculate magnetic
fields in cases with a high degree od symmetry.
1. A long, straight wire (external points): We can use Ampere’s law to find magnetic
field at a distance d from a long straight wire (this problem already solved using
Bio-Savart law).
• In fig. 33-21, choose Amperian loop a circle of radius d centered on the wire with its
plane perpendicular to the wire.
• From the symmetry of the problem, 𝐵 depends only on d (not on angular coordinate
around the circle).
• By choosing a path that is everywhere the same distance from the wire, we know that
B is constant around the path.

2. A long, straight wire (internal points): We can use Ampere’s law to find magnetic
field inside a wire.
• Let us assume a cylindrical wire of radius R in which a total current I is distributed
uniformly over its cross section.
• We have to find the magnetic field at a distance r<R from the center of the wire.
• Fig. 33-22 shows a circular Amperian loop of radius r inside the wire.
• Symmetry suggests that B is constant in magnitude everywhere on the loop and tangent
to the loop , so the left side of the Ampere’s law gives B(2r) exactly as in eqn.33-32.
• The right hand of Ampere’s law involves only the current inside the radius r. If the
current is distributed uniformly over the wire, the fraction of the current inside the
radius r is the same as the fraction of the area inside r, or r2/R2. Ampere’s law then
gives
• where again the right side includes only a fractio0n of the current that passes through
the surface enclosed by the path of integration (Amperian loop).
Solving for B, we obtain
• At the surface of the wire (r=R), eqn. 33-34 reduces to eqn. 33-13 (with d=R) ie., both
the expressions gives the same result for the field at the surface of the wire.
• Fig. 33-23 shows how the field depends on r at points both inside and outside the wire.
• Eqn. 33-34 is valid only for the case in which the current is distributed uniformly over
the wire. If the current density depends on r, a different result is obtained.
• Eqn. 33-13 for the field outside the wire remains valid whether the current density is
constant or a function of r.

3. A solenoid: Consider an ideal solenoid shown in fig. 33-24 and choose an Amperian
loop abcda.
▪ Here, we assume that magnetic field is parallel to the axis of this ideal solenoid and
constant in magnitude along line ab.
▪ We shall prove that field is also uniform in the interior (independent of the distance ab
from the central axis) as suggested by equal spacing of the field lines in fig. 33-24.
▪ The left side of Ampere’s law can be written as the sum of 4 integrals, one for each path
segment:
▪ The first integral on the right is Bh, where B= magnitude of 𝐵 inside the solenoid and h is the arbitrary length of the path
from a to b.
▪ Note: The path ab though parallel to the solenoid axis need not coincide with it. The 2nd and the 4th integrals are zero
because for every element of these paths 𝐵 is either at right angles to the path (for points inside the solenoid) or is zero (for
points outside). In either case, 𝐵 .dԦ
s=0 and the integrals vanish.
▪ The third integral, which includes the part of the rectangle that lies outside the solenoid is zero because we have taken 𝐵 =0
for all external points for an ideal solenoid.
▪ For the entire rectangular path, has the value Bh. The net current I that passes through the rectangular Amperian
loop is not the same ass the current in the solenoid because the windings pass through the loop more than once.
▪ Let n=no. of turns per unit length, then nh=no. of turns inside the loop and the total current passing through the rectangular
Amperian loop of fig. 33-24 is nhi. Ampere’s law then becomes
▪ Above equation agrees with eqn. 33-28, which referred only to points on the central axis of the solenoid because line ab in
fig. 33-244 can be located at any distance from the axis. We can now conclude that the magnetic field inside an ideal
solenoid is uniform over its cross section.

4. A Toroid: fig. 33-2 which may be considered to be a solenoid bent unto the
sshape of a doughnut.
▪ We can use Ampere’s law to find the magnetic field at interior points.
▪ From symmetry, the lines of 𝐵 form concentric circles inside the toroid as shown
in figure. Let us choose a concentric circle of radius r as an Amperian loop and
traverse it in the clockwise direction. Ampere’s law yields
Where I=current in the toroid windings and N is the total number of turns. This gives
• In contrast to the solenoid, B is not constant over the cross section of the toroid.
• We need to show that from Ampere’s law, B=0 for points outside an ideal toroid and in the central cavity.
• Close inspection of above equation (33-36) justifies statement that a toroid is a “solenoid bent into the shape of a
doughnut.”
• The denominator 2r is the central circumference of the toroid and N/ 2r is just n. With this substitution, eqn.
33-36 reduces to B=0in, the equation for the magnetic field in the central region of the solenoid.
• Right-hand rule for direction of B within a toroid: Curl the fingers of right hand in the direction of current
and extended thumb then points in the direction of magnetic field.
• Applications of Toroids: Tokamak, a device showing promise as the basis for a fusion power reactor.

1. https://www.youtube.com/watch?v=VDrnP88WzuA
2. https://www.youtube.com/watch?v=JHNloU9Rfow
3. https://web.iit.edu/sites/web/files/departments/academic-affairs/academic-resource-center/pdfs/Magnetic_Field.pdf
4. https://openstax.org/books/college-physics-ap-courses/pages/22-9-magnetic-fields-produced-by-currents-amperes-
law
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