Magnetic Fields from Currents Introduction
In the last section we started our study of magnetism by showing how certain materials,
referred to as ferromagnetic, exert magnetic forces on other ferromagnetic material. We also
introduced the concept of a magnetic field which we used to study these forces. More
importantly, we mentioned that fact that there exists an intimate relationship between
electricity and magnetism. We began our study of this relationship by showing that when a
charged particle moves through a magnetic field a magnetic force acts on this particle governed
by the following relationship.
𝑭𝐵 = 𝑞𝒗 𝑥 𝑩
Furthermore, as an electric current is no more than an assemblage of moving charged particles,
(electrons), we went on to define the magnetic force acting on a current carrying wire placed in
a magnetic field as follows:
𝑭𝐵 = 𝑖𝑳 𝑥 𝑩
Although these two observations hint at the intimate relationship between electricity and
magnetism, recall that in both cases the magnetic field was assumed to be generated from
some type of ferromagnetic material. Even more exciting was the discovery, by Hans Christian
Oersted, that magnetic fields are also produced by electric currents. Let’s begin with Oersted’s
observation that led him to this amazing discovery.
In 1820 Oersted noticed that a compass needle is deflected when placed near a current carrying
wire. We could investigate this phenomenon by surrounding a straight current carrying wire
with compasses. As shown below, we would find that the needle aligns itself so that it is
tangent to a circle drawn around the wire. From this observation we can deduce the direction
of the magnetic field. It should come as no surprise that the direction can again be found by
using a right-hand rule. In this case, placing the palm of your right hand on the wire with your
thumb pointing in the direction of the current, when you close your hand your fingers sweep in
the direction of the magnetic field.
𝐼
Further experiments would also reveal that the magnitude of the magnetic field is directly
proportional to the current and inversely proportional to the distance from the wire. After
developing a proper proportionality constant, the relationship can be written as follows:
𝐵 = 𝜇02𝜋∙𝐼
𝑟
Where, 𝐼 is the current measured in amps, 𝑟 is the distance from the wire measured in meters,
and 𝜇0 is called the permeability of free space and is equal to 4𝜋 𝐸−7 𝑇 ∙ 𝑚/𝐴.
As useful as this equation may be, it is valid for straight lengths of wire only, and therefore a
more general relationship is desired. A more general equation was indeed developed and is
known as the Biot-Savart Law. However, before introducing this law, let’s first combine what
we already know about magnetic forces on current carrying wires with the above discovery that
the wire itself generates a magnetic field to find the force between two parallel wires.
Magnetic Forces on Wires
The figure below shows two parallel current carrying wires. Since, as we now know, each wire
generates its own magnetic field, each wire will in effect apply a magnetic force on the other
wire. Note, when the currents are in the same direction the force is attractive, whereas when
the current is in opposite directions the force is repulsive.
𝐼1 𝐼2 𝐵1 𝐵2
𝐵2
𝐵1 𝐼1 𝐼2 𝐹𝐵12 𝐹𝐵21
𝐼1 𝐼2 𝐵1 𝐵2
𝐵2
𝐵1 𝐼1
𝐼2
𝐹𝐵12
𝐹𝐵21
In both cases the magnitude of the force can be found as shown below. Furthermore, note that
𝐹𝐵12 = 𝐹𝐵21, which we should of course expect to be true from Newtons third law.
𝐹𝐵12 = 𝐼1𝐿 𝐵2
𝐹𝐵12 = 𝐼1𝐿 ∙𝜇02𝜋∙𝐼2𝑑
𝐹𝐵12 =𝜇02𝜋∙𝐼1𝐼2𝐿
𝑑
𝐹𝐵21 = 𝐼2𝐿 𝐵1
𝐹𝐵21 = 𝐼2𝐿 ∙𝜇02𝜋∙𝐼1𝑑
𝐹𝐵21 =𝜇02𝜋∙𝐼1𝐼2𝐿
𝑑
Biot-Savart Law
The fundamental relationship, which was developed by Jean-Baptiste Biot and Felix Savart, that
relates the current in a wire to the magnetic field produced is called the Biot-Savart Law. Using
the figure below, we wish to find the magnetic field at a point, 𝑃, due to the current, 𝐼, flowing
in the wire. The Biot-Savart Law is used to find the magnetic field contribution, 𝒅𝑩, due to an
infinitesimal element of the wire, 𝑑𝑙, and is written as below.
𝒅𝑩 = 𝜇04𝜋∙𝐼
𝑟2∙ 𝒅𝒍 𝒙 �̂�
Where, 𝐼 is the current in the wire, 𝒅𝒍 is a vector that points in the direction of the current, 𝒓 is
the displacement vector from 𝑑𝑙 to 𝑃, and �̂� =𝒓
𝑟 is the unit vector in the direction of 𝒓.
𝑑𝑙
𝒅𝒍
𝐼
�̂�
𝒅𝑩 x
𝒓
𝜽
𝑃
Since �̂� is a unit vector we can write the magnitude of the cross product as follows:
|𝒅𝒍 𝒙 �̂�| = 𝑑𝑙 sin(𝜃)
With direction again given by the right-hand rule.
Therefore, the magnitude of 𝒅𝑩 is
𝑑𝐵 = 𝜇04𝜋∙𝐼 sin(𝜃)
𝑟2𝑑𝑙
Naturally, the magnetic field at point 𝑃 is found by integrating over the length of the wire.
𝐵 = ∫𝜇04𝜋∙𝐼 sin(𝜃)
𝑟2𝑑𝑙
To illustrate let’s use this integral to find the magnetic field from a straight wire and see if we
get the same result we stated earlier. Using the figure below we wish to find the magnetic field
at the point 𝑃.
Assuming the wire is in the plane of the page, we know that the direction of the field is into the
page as shown. To determine the magnitude, we can directly use the Biot-Savart integral from
above with 𝑦 varying from −∞ to ∞. However, using symmetry we can integrate instead from
0 to ∞ and multiple the results by 2.
𝐵 = 2 (𝜇0𝐼
4𝜋∫
sin(𝜃)
𝑟2𝑑𝑦
∞
0
)
To solve this integral we can first attempt to write the integrand in terms of the integration
variable, 𝑦. The 𝑟2 can be replaced using Pythagorean theorem as follows:
𝑟2 = 𝑅2 + 𝑦2
And since 𝜃 = 𝜑 + 90 we can replace sin(𝜃) as follows:
sin(𝜃) = sin(𝜑 + 90 ) = cos(𝜑) = 𝑅
𝑟=
𝑅
√𝑅2 + 𝑦2
Re-writing the integral with the above substitutions we have:
𝐵 = 𝜇0𝐼
2𝜋∫
𝑅
√𝑅2 + 𝑦2∙
1
(𝑅2 + 𝑦2) 𝑑𝑦
∞
0
𝐵 = 𝜇0𝐼𝑅
2𝜋∫
1
(𝑅2 + 𝑦2)3 2⁄ 𝑑𝑦
∞
0
This integral can be solved using an advanced technique called trig substitution, however a
table of integrals can also be used to save time. The solution is given below.
𝐵 = 𝜇0𝐼𝑅
2𝜋[
𝑦
𝑅2(𝑅2 + 𝑦2)1 2⁄]0
∞
= 𝜇0𝐼
2𝜋𝑅
Where we used the following from a table of integrals:
∫1
(𝑎2 + 𝑥2)3 2⁄ 𝑑𝑥 =
𝑥
𝑎2(𝑎2 + 𝑥2)1 2⁄
And the following to evaluate 𝑦 = ∞.
lim𝑦→∞
(𝑦
(𝑅2 + 𝑦2)1 2⁄) = lim
𝑦→∞
(
𝑦𝑦⁄
(𝑅2
𝑦2⁄ +𝑦2
𝑦2⁄ )1 2⁄
)
= lim𝑦→∞
(1
(𝑅2
𝑦2⁄ + 1)1 2⁄)
= lim𝑦→∞
(1
(0 + 1)1 2⁄) = 1
The good news is the solution is indeed identical to the one we initially stated above, however
you may have noticed it took quite a bit of work to obtain.
Let’s now step back and recall how we first computed the electric field at a point in space from
a continuous charge distribution. Not surprisingly the equation for the electric field
contribution, 𝒅𝑬, due to an infinitesimal charge element, 𝑑𝑞, is very much like the equation for
the magnetic field contribution, 𝒅𝑩, due to an infinitesimal element of the wire, 𝑑𝑙. Both are
written below.
Coulomb’s Law for the Electric Field Biot-Savart Law for the Magnetic Field
𝒅𝑬 = 𝑘
𝑟2𝑑𝑞 �̂�
𝒅𝑩 = 𝜇04𝜋∙𝐼
𝑟2∙ 𝒅𝒍 𝒙 �̂�
Just as we have seen above for the Biot-Savart law, you may recall computing the electric field
using Coulomb’s law was also not always trivial. Fortunately, we also learned that in cases
where the charge distribution has some form of symmetry Gauss’s Law can be used with
considerably less effort. Is there a similar fundamental law for magnetism that can be used in
place of the Biot-Savart Law? The answer is yes, and the law is called Ampere’s Law. To
highlight the similarities, we state both laws below.
Gauss’s Law for the Electric Field
The distribution of the electric field, (electric flux), through a closed surface is equal to the total net charge enclosed in that surface divided by a constant.
∮𝑬 ∙ 𝑫𝑨 =𝑄𝑒𝑛𝑐𝜀0
Ampere’s Law for the Magnetic Field
The line integral of the magnetic field around a closed curve, (Amperian loop), is equal to the total net current enclosed by the curve multiplied by a constant.
∮𝑩 ∙ 𝒅𝒍 = 𝜇0𝐼𝑒𝑛𝑐
To show the usefulness of Ampere’s law lets once again find the magnetic field from a straight
wire; this time using Ampere’s Law. Based on the Oersted’s experiments mentioned earlier, we
can use the right-hand rule to recognize the direction of the magnetic field shown in the figure
below. We also know that the magnitude of the magnetic field has the same value at all points
that are the same distance, 𝑟, from the wire. In other words, the 𝑩 field has cylindrical
symmetry, and based on this symmetry we will use a circle of radius 𝑟 as our Amperian loop to
simplify the integral on the left-hand side of Ampere’s law.
𝐼
𝑇𝑜𝑝 𝑉𝑖𝑒𝑤
𝑩
𝒅𝒍
𝑩 𝒅𝒍
𝑟
𝐼
𝑩
Ampere’s law is stated as follows:
∮𝑩 ∙ 𝒅𝒍 = 𝜇0𝐼𝑒𝑛𝑐
And since 𝑩 is parallel to 𝒅𝒍 at all points on the circle we can remove the dot product.
𝐵∮ 𝑑𝑙2𝜋𝑟
0
= 𝜇0𝐼
𝐵2𝜋𝑟 = 𝜇0𝐼
𝐵 = 𝜇0𝐼
2𝜋𝑟
Note this is the same answer we computed using the Biot-Savart Law, but with much less
effort!
Another interesting example where Ampere’s Law can greatly simplify the computation of the
magnetic field is that of a coaxial cable. A coaxial cable consists of a single wire carrying a
current surrounded by an insulator, and then another thin cylindrical conducting material that
carries the same current in the opposite direction. Let’s use Ampere’s law to find the magnetic
field inside the center conductor, in the space between the conductors, and outside the cable.
The top figure below shows a front view of a coaxial cable carrying a current, I, directed into the
page for the center conductor with the same current directed out of the page for the outside
conductor. The magnetic field again displays cylindrical symmetry, so we use circles as our
Amperian loops. The bottom figures show the Amperian loop used for each region.
𝑟1
𝑟2
𝐶𝑜𝑎𝑥 𝐶𝑎𝑏𝑙𝑒 𝐹𝑟𝑜𝑛𝑡 𝑉𝑖𝑒𝑤
𝑟 < 𝑟1 𝑟1 < 𝑟 < 𝑟2 𝑟 > 𝑟2
𝐴𝑚𝑝𝑒𝑟𝑖𝑎𝑛 𝐿𝑜𝑜𝑝
𝐼
𝐼
x
Region 1: 𝒓 < 𝒓𝟏
Ampere’s law works out the same as it did for the straight wire above except the enclosed
current is radius dependent.
∮𝑩 ∙ 𝒅𝒍 = 𝜇0𝐼𝑒𝑛𝑐
∮𝐵𝑑𝑙 = 𝜇0𝐼𝑒𝑛𝑐
𝐵 = 𝜇0𝐼𝑒𝑛𝑐2𝜋𝑟
To find the enclosed current we assume the current is evenly distributed so that we can solve
for the enclosed current by comparing ratios as shown below.
𝐼𝑒𝑛𝑐𝐼= 𝜋𝑟2
𝜋𝑟12
𝐼𝑒𝑛𝑐 = 𝐼𝑟2
𝑟12
Substituting we find that the magnetic field varies linearly with distance inside the conductor.
𝐵 = 𝜇02𝜋𝑟
𝐼𝑟2
𝑟12
𝐵 = 𝜇0𝐼
2𝜋𝑟12𝑟
Region 2: 𝒓𝟏 < 𝒓 < 𝒓𝟐
This case is identical to the straight wire example we did above. Note the current outside the
loop does not play a role in the computation of Ampere’s law.
𝐵 = 𝜇0𝐼
2𝜋𝑟
Region 3: 𝒓 > 𝒓𝟐
In this case the magnetic field is zero since the current enclosed in the Amperian loop is zero.
∮𝐵𝑑𝑙 = 𝜇0𝐼𝑒𝑛𝑐
𝐵2𝜋𝑟 = 𝜇0(𝐼 − 𝐼) 𝐵 = 0
Note there are no stray magnetic fields outside a coaxial cable, which makes them ideal for
carrying signals near other sensitive equipment. The figure below illustrates the behavior of the
magnitude of the magnetic field for all regions.
𝝁𝟎𝑰
𝟐𝝅𝒓𝟏
𝒓𝟏 𝒓𝟐
𝝁𝟎𝑰
𝟐𝝅𝒓𝟐
𝑩(𝒓)
𝒓
𝝁𝟎𝑰
𝟐𝝅𝒓𝟏𝟐
𝑩 ∝𝟏
𝒓
𝑩 ∝ 𝒓
Solenoids and Toroid
An electromagnet is a type of magnet in which the magnetic field is produced from an electric
current. Electromagnets are very widely used in devices such as motors, loudspeakers, hard
drives, MRI machines, and even in certain industries to pick up and move heavy iron objects. A
solenoid is one type of electromagnet. A solenoid consists of a long wire that is wound in the
shape of a helix and made to carry a current as illustrated below.
To determine the magnetic field due to the solenoid let’s look first at a single loop of wire
shown in the figure below. Focusing on the top and bottom of the loop where the current
arrows are drawn, we can determine the direction of the field. Place the thumb of your right
hand where the current is shown and close your hand. The field curls in the direction your
hand closes, making circles as shown in the figure. Continuing these ever-larger circles we
notice that the magnetic field lines become almost straight aligning in one direction near the
center of the loop. Alternatively, there is another sort of right-hand rule that can more easily
be used to determine the direction of the magnetic field through the current loop. If you place
the palm of your right hand on the outside of the loop so that when your close your palm your
fingers curl in the direction of the current, your thumb will point in the direction of the
magnetic field.
Placing these circular current loops side by side, as in a solenoid, will serve to increase the
strength of the magnetic field through the solenoid, as shown below.
Let’s now use Ampere’s law to determine the magnetic field inside a solenoid. The figure below
is a cut-view of the solenoid where the current is directed out of the paper at the top and into
the paper at the bottom.
We choose our closed path to be the rectangle with sides 𝑎𝑏, 𝑏𝑐, 𝑐𝑑 and 𝑑𝑎 as shown in the
figure. Therefore, the integral in Ampere’s law can be split into four separate integrals, one for
each side of the rectangle.
∮𝑩 ∙ 𝒅𝒍 = ∫ 𝑩 ∙ 𝒅𝒍𝑏
𝑎
+∫ 𝑩 ∙ 𝒅𝒍𝑐
𝑏
+ ∫ 𝑩 ∙ 𝒅𝒍𝑑
𝑐
+ ∫ 𝑩 ∙ 𝒅𝒍𝑒
𝑑
For the two vertical segments, 𝑏𝑐 and 𝑑𝑎, the vector 𝒅𝒍 is perpendicular to 𝑩, therefore those
terms will be zero. For the two segments, 𝑎𝑏 and 𝑐𝑑, we can see that 𝒅𝒍 and 𝑩 are anti-parallel
and parallel respectively, therefore the above simplifies as shown below.
∮𝑩 ∙ 𝒅𝒍 = ∫ −𝐵𝑂𝑑𝑙𝑏
𝑎
+0 + ∫ 𝐵𝐼𝑑𝑙𝑑
𝑐
+ 0
∮𝑩 ∙ 𝒅𝒍 = − 𝐵𝑂𝐿 + 𝐵𝐼𝐿
∮𝑩 ∙ 𝒅𝒍 = 𝐿(𝐵𝐼 − 𝐵𝑂)
Where, 𝐵𝐼 is the magnetic field inside the solenoid and 𝐵𝑂 is the magnetic field outside the
solenoid.
The field outside a solenoid is very small compared to the field inside, except near the ends. If
we consider an infinitely long solenoid we can let 𝐵𝑂 = 0 and 𝐵𝐼 = 𝐵, therefore
∮𝑩 ∙ 𝒅𝒍 = 𝐵𝐿
and Ampere’s Law becomes
𝐵𝐿 = 𝜇0𝐼𝑒𝑛𝑐
To find 𝐼𝑒𝑛𝑐 we can consider there are 𝑁 loops of wire enclosed in our Amperian loop so that
𝐼𝑒𝑛𝑐 = 𝑁𝐼. Furthermore, if we let 𝑛 = 𝑁/𝐿 be the number of loops per unit length we can
write the magnitude of the magnetic field inside a solenoid as follows.
𝐵 = 𝜇0𝑛𝐼
Which shows that the magnetic field is constant inside an infinitely long solenoid, and the
magnitude is directly proportional to the current, 𝐼, and the number of loops per unit length, 𝑛.
Finally, a toroid is essentially a solenoid which is bent into a shape of a circle. To determine the
magnetic field inside a toroid we would create an Amperian loop in the shape of a circle that
follows the magnetic field lines, as shown below. However, as an alternative we can use the
results of the solenoid from above. The magnetic field of the solenoid can be written as
follows:
𝐵 = 𝜇0𝑁
𝐿𝐼
For the toroid we have 𝐿 = 2𝜋𝑟, therefore the magnitude of the magnetic field for a toroid,
which is not constant as it is in the solenoid, is given as:
𝐵 = 𝜇0𝑁
2𝜋𝑟𝐼
Final Summary for Magnetic Fields from Currents Introduction
Magnetic Field from a Straight Current Carrying Wire
• A straight current carrying wire generates a magnetic field which curls around the wire in the concentric circles with a direction given by the right-hand rule using your thumb pointing in the direction of the current.
• The magnitude of the magnetic field is inversely proportional to the distance from the wire center.
𝐵 = 𝜇02𝜋∙𝐼
𝑟
Magnetic Force Between Two Parallel Current Carrying Wires
• A magnetic field exists between two parallel current carrying wires. o If the currents are flowing in the same direction the force is attractive. o If the currents are flowing in opposite directions the force is repulsive.
The magnitude of the force is given as below.
𝐹𝐵 =𝜇02𝜋∙𝐼1𝐼2𝐿
𝑑
Biot-Savart Law
The magnetic field established by a current carrying conductor can be found using the Biot-Savart Law. The law relates a current carrying element, 𝒅𝒍, to its contribution to the magnetic field, 𝒅𝑩, at a point P, which is a distance 𝑟 from the current element.
𝒅𝑩 = 𝜇04𝜋∙𝐼
𝑟2∙ 𝒅𝒍 𝒙 �̂�
Where, 𝐼 is the current in the wire, 𝒅𝒍 is a vector that points in the direction of the current, 𝒓
is the displacement vector from 𝑑𝑙 to P, �̂� =𝒓
𝑟 is the unit vector in the direction of 𝒓, and 𝜇0
is called the permeability of free space and is equal to 4𝜋 𝐸−7 𝑇 ∙ 𝑚/𝐴 The total magnetic field is found by summing (integrating) all contributions, 𝒅𝑩, on a path along the conductor.
𝑩 = ∫𝒅𝑩
Ampere’s Law
The line integral of the magnetic field around a closed curve, (Amperian loop), is equal to the total net current enclosed by the curve multiplied by a constant.
∮𝑩 ∙ 𝒅𝒍 = 𝜇0𝐼𝑒𝑛𝑐
Solenoid
The magnitude of the magnetic field inside a current carrying solenoid is given as
𝐵 = 𝜇0𝑛𝐼 Where, 𝑛 = 𝑁/𝐿 is the number of turns per unit length. The direction of the field is given by the right-hand rule, where when you close your right hand around the loops in the direction of the current your thumb points in the direction of the magnetic field.
Toroid
The magnitude of the magnetic field inside a current carrying toroid is given as
𝐵 = 𝜇0𝑁
2𝜋𝑟𝐼
Where, 𝑁 is the number of turns and 𝑟 is the distance from the center of the toroid. The direction of the field is given as in the case of the solenoid.
By: ferrantetutoring