A Little Bit About Magnetics

Magnetostatics
Electric charges are source of electric fields. An electric field exerts force on an
electric charge, whether the charge happens to be moving or at rest.
One could similarly think of a magnetic charge as being the source of a magnetic
field. However, isolated magnetic charge ( or magnetic monopoles) have never
been found to exist. Magnetic poles always occur in pairs ( dipoles) – a north
pole and a south pole. Thus, the region around a bar magnet is a magnetic field.
What characterizes a magnetic field is the qualitative nature of the force that it
exerts on an electric charge. The field does not exert any force on a static charge.
However, if the charge happens to be moving (excepting in a direction parallel to
the direction of the field) it experiences a force in the magnetic field.
It is not necessary to invoke the presence of magnetic poles to discuss the source
of magnetic field. Experiments by Oersted showed that a magnetic needle gets
deflected in the region around a current carrying conductor. The direction of deflection
is shown in the figure below.
Current into the page Current out of the page
Thus a current carrying conductor is the source of a magnetic field. In fact, a
magnetic dipole can be considered as a closed current loop.
Lorentz Force :
We know that an electric field ~E exerts a force q~E on a charge q. In the presence
of a magnetic field ~B , a charge q experiences an additional force
~F
m = q~v × ~B
where ~v is the velocity of the charge. Note that
1
• There is no force on a charge at rest.
• A force is exerted on the charge only if there is a component of the magnetic
field perpendicular to the direction of the velocity, i.e. the component of the
magnetic field parallel to ~v does not contribute to ~Fm.
• ~v · ~Fm = 0, which shows that the magnetic force does not do any work.
In the case where both ~E and ~B are present, the force on the charge q is given by
~F = q(~E +~v × ~B)
This is called Lorentz force after H.E. Lorentz who postulated the relationship.
It may be noted that the force expression is valid even when ~E and ~B are time
dependent.
Unit of Magnetic Field
From the Lorentz equation, it may be seen that the unit of magnetic field is
Newton-second/coulomb-meter, which is known as a Tesla (T). (The unit is occasionally
written as Weber/m2 as the unit of magnetic flux is known as Weber).
However, Tesla is a very large unit and it is common to measure ~B in terms of a
smaller unit called Gauss,
1T = 104 G
It may be noted that ~B is also referred to as magnetic field of induction or simply
as the induction field. However, we will use the term “magnetic field”.
Motion of a Charged Particle in a Uniform Magnetic Field
Let the direction of the magnetic field be taken to be z- direction,
~B
= Bˆk
we can write the force on the particle to be
~F
m = m
d~v
dt
= q~v × ~B
The problem can be looked at qualitatively as follows. We can resolve the motion
of the charged particle into two components, one parallel to the magnetic field and
the other perpendicular to it. Since the motion parallel to the magnetic field is not
affected, the velocity component in the z-direction remains constant.
vz(t) = vz(t = 0) = uz
2
where ~u is the initial velocity of the particle. Let us denote the velocity component
perpendicular to the direction of the magnetic field by v⊥. Since the force (and
hence the acceleration) is perpendicular to the direction of velocity, the motion in
a plane perpendicular to ~B is a circle. The centripetal force necessary to sustain
the circular motion is provided by the Lorentz force
mv2

R
=| q | v⊥B
where the radius of the circle R is called the Larmor radius, and is given by
R =
mv⊥
| q | B
The time taken by the particle to
complete one revolution is
T =
2πR
v⊥
The cyclotron frequency ωc is given
by
ωc =

T
=
| q | B
m
F V
B
B Magnetic field into the page.
Figure shows directions of force and velocity
for a positive charge.
Motion in a Magnetic Field – Quantitative
Let the initial velocity of the particle be ~u. we may take the direction of the
component of ~u perpendicular to ~B as the x− direction, so that
~u = (ux, 0, uz)
Let the velocity at time t be denoted by ~v
~v = vxˆı + vyˆ + vzˆk
we can express the force equation in terms of its cartesian component
m
dvx
dt
= q(vyBz − vzBy) = qvyB
m
dvy
dt
= q(vzBx − vxBz) = −qvxB
m
dvz
dt
= q(vxBy − vyBx) = 0
3
where we have used Bx = By = 0 and Bz = B.
The last equation tells us that no force acts on the particle in the direction in which
~B
acts, so that
vz = constant = uz
The first two equationsmay be solved by converting them into second order differential
equatons. This is done by differentiating one of the equations with respect
to time and substituting the other equation in the resulting second order equation.
For instance, the equation for vx is given by
d2vx
dt2 =
qB
m
dvy
dt
= −
q2B2
m2 vx
The equation is familiar in the study of simple harmonic motion. The solutions
are combination of sine and cosine functions.
vx = Asin ωct + B cos ωct
where
ωc =
qB
m
is called the cyclotron frequency and A and B are constants. These constants have
to be determined from initial conditions. By our choice of x and y axes, we have
vx(t = 0) = ux
so that B = ux. Differentiating the above equation for ux,
dvx
dt
= Aωc cos ωct − Bωc sin ωct

qB
m
vy = ωcvy
Since vy = 0 at t = 0, we have A = 0. Thus, the velocity components at time t
are given by
vx = ux cos ωct
vy = −ux sin ωct = ux sin(ωct +
π
2
)
which shows that vx and vy vary harmonically with time with the same amplitude
but with a phase difference of π/2. Equation of the trajectory may be obtained by
4
integrating the equations for velocity components
x(t) =
ux
ωc
sin ωct + x0
y(t) =
ux
ωc
cos ωct + y0
z(t) = uzt + z0
where x0, y0 and z0 are constants of integration representing the initial position
of the particle. The equation to the projection of the trajectory in the x-y plane is
given by
(x − x0)2 + (y − y0)2 =
u2
x
ω2
c
which represents a circle of radius ux/ωc, centered about (xo, y0). As the z- component
of the velocity is constant, the trajectory is a helix.
Helical motion of a charged particle
-1
0
1
x
-1
0
1
y
-10
-5
0
5
10
z
A plot of the motion of a charged particle in a constant magnetic field.
Motion in a crossed electric and magnetic fields
The force on the charged particle in the presence of both electric and magnetic
fields is given by
~F = q(~E +~v × ~B)
5
Let the electric and magnetic fields be at right angle to each other, so that,
~E
· ~B = 0
If the particle is initially at rest no magnetic force acts on the particle. As the
electric field exerts a force on the particle, it acquires a velocity in the direction of
~E
. The magnetic force now acts sidewise on the particle.
For a quantitative analysis of the motion, let ~E be taken along the x-direction and
~B
along z-direction. As there is no component of the force along the z-direction,
the velocity of the particle remains zero in this direction. The motion, therefore,
takes place in x-y plane.
The equations of motion are
m
dvx
dt
= q(~E +~v × ~B)x = qE + qBvy (1)
m
dvy
dt
= q(~E +~v × ~B)y = −qBvx (2)
As in the earlier case, we can solve the equations by differentiating one of the
equations and substituting the other,
m
d2vx
dt2 = qB
dvy
dt
= −
q2B2
m
vx
which, as before, has the solution
vx = Asin ωct
with ωc = qB/m. Substituting this solution into the equation for vy, we get, using
Eqn. (1)
vy = −
E
B
+
m
qB
dvx
dt
= −
E
B
+
m
qB
Aωc cos ωct
Since vy = 0 at t = 0, the constant A = Eq/mωc = E/B, so that Thus we have
vx =
E
B
sin ωct
vy =
E
B
(cos ωct − 1)
6
The equation to the trajectory is obtained by integrating the above equation and
determining the constant of integration from the initial position (taken to be at the
origin),
x =
E
Bωc
(1 − cos ωct)
y =
E
Bωc
(sin ωct − ωct)
The equation to the trajectory is
x −
E
Bωc2
+ y −
Et
B 2
=
E2
B2ω2
c
which represents a circle of radius
R =
E
Bωc
whose centre travels along the negative y direction with a constant speed
v0 =
E
B
The trajectory resembles that of a point on the circumference of a wheel of radius
R, rolling down the y-axis without slipping with a speed v0. The trajectory is
known as a cycloid.
7
-30
-25
-20
-15
-10
-5
0
0 1 2
y
x
Cycloidal motion of a charged particle
Exercise : Find the maximum value of x attained by the particle during the cycloidal
motion and determine the speed of the particle at such points. (Ans.
xmax = 2E/Bωc speed 2E/B.)
Current Density :
The current in a wire is a measure of the amount of charge flowing through any
point of the wire in unit time. If the charge density in the wire is λ, the current is
given by I = λv where v is the drift velocity of the charge carriers. The carriers
travel a distance dl = vdt in time dt so that the amount of charge that flows in
time dt through any point of the wire is dQ = Idt = λvdt. The unit of current is
Coulomb per second, known as Ampere.
When we consider motion of charges on a surface or in a volume, the charges
move in all directions and we have to consider average velocities. We need to
introduce the concept of a charge sensity which is then a vector quantiy at a point
in the material. For instance, consider charges flowing two dimension.
8
If we consider a ribbon along the
direction of flow in the surface, we
could talk of the amount of charge
flowing past a length dl⊥ normal to
the direction of flow at a point. The
surface current density that we talk
of at this point is the current per
unit length oriented perpendicular
to the flow at this point.
dl
Flow Direction
Thus the current density is
K =
dI
dl⊥
The unit obviously is Ampere per meter. If σ is the surface charge density and v
is the average velocity, then the amount of charge flowing past the length element
in time dt is clearly dQ = σdl⊥vdt. The current flowing past the length element
is dQ/dt = σdl⊥v so that the current density is K = σv.
The Lorentz force on charges in the surface is
~Fm = Z (σda)~v × ~B = Z (~K × ~B)da
.
In three dimensions, the concept is
very similar. One has to consider the
current flowing through a surface element
oriented perpendicular to the
direction of flow.
dS
flow
The volume current density ~ J is the defined as
~ J =
dI
dS⊥
which has a direction perpendicular to the surface element but going in the direction
of flow. The Lorentz force acting on the charges in the volume is then given
by
~F
m = Z (ρdτ )~v × ~B = Z ( ~ J × ~B )dτ
9
Equation of Continuity :
The current through any closed volume is thus given by
I = ZS
JdS⊥ = ZS
~ J · d~S
where S is the surface bounding the volume. Using divergence theorem, we may
conver the surface integral to a volume integral over the divergence, so that
I = Z ∇ · ~ Jdτ
We know that ∇ · ~ J gives the outward flux through the surface. The net outward
flux must result in a net decrease of charges within the volume. In other words,
we must have,
Z ∇ · ~ Jdτ = −
d
dt Z ρdτ = −Z ∂ρ
∂t

As the relation is true for arbitrary volume, we have the equation of continuity
∇ · ~ J +
∂ρ
∂t
= 0
In magnetostatics, we study effects of what are known as steady current. The
word steady in thius context does not mean stationary because current itself implies
that charges are moving in time. Steady implies that the rate of flow is
constant so that ∂ρ/∂t = 0. For steady currents, the equation of continuity becomes
∇ · ~ J = 0
Biot- Savarts’ Law
We have seen that electrostatics was formulated on the basis of empirical law of
Coulomb. The corresponding law for magnetostatics is Biot-Savart’s law. which
provides an expression for the magnetic field due to a current segment. The field
d~B at a position ~r due to a current segment Id~l is experimentally found to be
perpendicular to ~dl and ~r. The magnitude of the NEODYMIUM MAGNETS field is
• proportional to the length | dl | and to the current I and to the sine of the
angle between ~r and ~dl.
• inversely proportional to the square of the distance r of the point P from the
current element.
10
Mathematically,
d~B ∝ I
~dl × ˆr
r2
In SI units the constant of proportionality
is μ0/4π, where μ0 is the
permeability of the free space. The
value of μ0 is
μ0 = 4π × 10−7 N/amp2
The expression for field at a point P
having a position vector ~r with respect
to the current element is
d~B =
μ0

I
~dl × ˆr
r2
r
B
I
q
dl
For a conducting wire of arbitrary shape, the field is obtained by vectorially adding
the contributions due to such current elements as per superposition principle,
~B
(P) =
μ0

I Z ~dl × ˆr
r2
where the integration is along the path of the current flow.
Example 1 : Field due to a straight wire carrying current
The direction of the field at P due to a current element ~dl is along ~dl ×~r, which is
a vector normal to the page (figure on the left) and coming out of it.
r
a
l
dl q
P
O
11
We have,
~dl × ˆr
r2 =
| dl sin θ |
r2
ˆk
where the plane of the figure is taken as the x-y plane and the direction of outward
normal is parallel to z-axis. If a be the distance of the oint P from the wire, we
have
r = a/ sin θ
l = a cot θ
dl = −(a/ sin2 θ)dθ
Thus
~dl × ˆr
r2 =
sin θ
a
dθˆk
The direction of the magnetic field at a distance a from the wire is tangential to a
circle of radius a,as shown.
I
a
B
Since the magnetic field due to all current elements at P are parallel to the zdirection,
the field at P due to a wire, the ends of which make angles φ1 and φ2 at
P is given by a straightforward integration
12
~B
=
μ0I
4πa Z 2
1
sin θdθˆk
=
μ0I
4πa
[−cos θ]2
1
ˆk
=
μ0I
4πa
[−cos θ1 − cos θ2] ˆk
=
μ0I
4πa
[sin φ1 + sin φ2] ˆk
P
O
q
f
q
f 1
a
2
1 2
Note that both the angles φ1 and φ2 are acute angles.
If we consider an infinite wire (also called long straight wire), we have φ1 = φ2 =
π/2, so that the field due to such a wire is
~B
=
μ0I
2πa
ˆk
where the direction of the field is given by the Right hand rule.
Exercise :
A conductor in the shape of an n-sided polygon of side a carries current I. Calculate
the magnitude of the magnetic field at the centre of the polygon. [ Ans.
(μ0In/πa) sin(π/n).]
Example 2 : Field due to a circular coil on its axis
Consider the current loop to be in the x-y plane, which is taken perpendicular to
the plane of the paper in which the axis to the loop (z-axis) lies. Since all length
elements on the circumference of the ring are perpendicular to ~r, the magnitude
of the field at a point P is given by
dB =
μ0I

dl
r2
13
The direction of the field due to every
element is in the plane of the paper
and perpendicular to ~r, as shown.
Corresponding to every element ~dl
on the circumference of the circle,
there is a diametrically opposite element
which gives a magnetic field
d~B in a direction such that the component
of d~B perpendicular to the
axis cancel out in pairs.
dl
r dB
I q
z
a
The resultant field is parallel to the axis, its direction being along the positive
z-axis for the current direction shown in the figure. The net field is
~B
=
μ0I
4π Z dl
r2 cos θ
=
μ0I

cos θ
r2 Z dl
=
μ0I cos θ
4πr2 2πa =
μ0Ia cos θ
2r2
In terms of the distance z of the point P and the radius a, we have
B =
μ0I
2
a2
(a2 + z2)3/2
The direction of  Neodymium magnet is determined by the following Right Hand
Rule.
I
B
If the palm of the right hand is curled
in the direction of the current, the
direction in which the thumb points
gives the direction of the magnetic
field at the centre of the loop. The
field is, therefore, outward in the figure
shown.
Note that for z ≫ a, i.e. the field due to circular loop at large distances is given
by
B =
μ0Ia2
2z3 =
μ0μ
2πz3
where μ = Iπa2 is the magnetic moment of the loop. The formula is very similar
to the field of an electric dipole. Thus a current loop behaves like a magnetic
dipole.
14
Example 3 :
A thin plastic disk of radius R has a uniform surface charge density σ. The disk is
rotating about its own axis with an angular velocity ω. Find the field at a distance
z along the axis from the centre of the disk.
z−axis
w
s
R
r
dr
z
P
Solution :
The current on the disk can be calculated by assuming the rotating disk to be
equivalent to a collection of concentric current loops. Consider a ring of radius
r and of width dr. As the disk is rotating with an angular speed ω, the rotating
charge on the ring essentially behaves like a current loop carrying current σ ·
2πrdr · ω/2π = σωrdr.
The field at a distance z due to this ring is
dB =
μ0(σωrdr)
2
r2
(r2 + z2)3/2
The net field is obtained by integrating the above from r = 0 to r = R
B =
μ0σω
2 Z R
0
r3
(r2 + z2)3/2 dr
=
μ0σω
2 Z R
0
r2 + z2 − z2
(r2 + z2)3/2 rdr
The integral above may easily be evaluated by a substitution x = r2 + z2. The
result is
B = μ0σω  R2 + 2z2
(R2 + z2)3/2 − 2z
The field at the centre of the disk (z = 0) is
B(z = 0) = μ0σωR
15
Exercise : Find the magnetic moment of the rotating disk of Example 7. [Ans.
πωR4/4]
Example 4 : Field of a solenoid on its axis
Consider a solenoid of N turns. The solenoid can be considered as stacked up
circular coils. The field on the axis of the solenoid can be found by superposition
of fields due to all circular coils. Consider the field at P due to the circular turns
between z and z +dz from the origin, which is taken at the centre of the solenoid.
The point P is at z = d. If L is the length of the solenoid, the number of turns
within z and z + dz is Ndz/L = ndz, where n is the number of turns per unit
length.
The magnitude of the field at P due
to these turns is given by
dB =
μ0NIdz
2L
a2
[a2 + (z − d)2]3/2
The field due to each turn is along
ˆk; hence the fields due to all turns
simply add up. The net field is
~B
=
μ0NIa2
2L Z L/2
−L/2
dz
[a2 + (z − d)2]3/2
ˆk
z=−L z=0 z z=L
dz
P
q
d
a
b
The integral above is easily evaluated by substituting
z − d = a cot θ
dz = −acosec2θdθ
The limits of integration on θ are α and β as shown in the figure. With the above
substitution
~B
= −
μ0nI
2 Z

sin θdθ =
μ0nI
2
(cos α − cos β)ˆk
For a long solenoid, the field on the axis at points far removed from the ends of
the solenoid may be obtained by substituting α = 0◦ and β = 180◦, so that,
~B
= μ0nIˆk
The field is very nearly constant. For points on the axis far removed from the ends
but outside the solenoid, α ≈ β so that the field is nearly zero.
16
Example 5 :
Determine the field at the point located
at the centre P of the semicircular
section of the hairpin bend
shown in the figure.
a
I
Solution :
The field at P may be determined by superposition of fields due to the two straight
line sections and the semicircular arc. The contribution due to all three sections
add up as the field due to each is into the plane of the paper.
The field due to each straight line section is obtained by putting φ1 = 90◦ and
φ2 = 0◦ in the expression obtained in Example 5 above. The field due to each
wire is μ0I/4πa.
For the semi-circular arc, each length element on the circumference is perpendicular
to ~r, the vector from the length element to the point P. Thus
Barc =
μ0I
4π Z ~dl × ˆr
r2
=
μ0I
4πa2 Z dl =
μ0I
4πa2 · πa =
μ0I
4a
The net field due to the current in the hairpin bend at P is
B =
μ0I
2πa
+
μ0I
4a
Example 6 : Force between two long and parallel wires.
Force due to the first wire at the position of the second wire is given by
~B
= −
μ0I1
2πd
ˆk
where ˆk is a unit vecor out of the page.
17
The force experienced by the second
wire in this field is
~F
= Z (~I2 × ~B )dl
= −
μ0I1I2
2πd
(ˆ × ˆk) Z dl
= −
μ0I1I2
2πd
ˆı Z dl
I
1 I2
F
B
d
x
y
Thus the force between the wires carrying current in the same direction is attractive
and is μ0I1I2/2πd per unit length.
Exercise :
Determine the magnetic field at the point P for the two geometries shown in the
figures below. [Ans . (a) μ0I/4R (b)
μ0I(R1 − R2)θ
4πR1R2
]
1
2
I
I
q
P
R
R
I
(b)
(a)
P
R
18
Example 7 :
Find the field at the centre of a conductor
shaped as shown in the figure.
The curved sections are quadrants of
circles of radii a and b and the conductor
carries a current I.
A
B
D C
P
dl
r
f
Solution :
For straight segments AB and CD d~l × ~r = 0 and there is no contribution to the
magnetic field. For the arcs BC and DA use cylindrical coordinates. For instance,
for the arc of radius b, we have d~l = −bdlφˆφ, the minus sign is because the current
is in the clockwise direction, and ~r = −bˆρ so that
d~l × ~r = b2dφˆφ × ˆρ = −b2dφˆk
Similarly, for the smaller arc, since the current is counterclockwise,
d~l × ~r = a2dφˆk
where ˆk is in the z-direction, i.e. perpendicular to the plane of the arcs. Using
Biot-Savart’s law, we get
~B
=
μ0I
4π Z /2
0
dφ1
a

1
b
=
μ0I
8
ˆk 1
a

1
b
Ampere’s Law
Biot-Savart’s law for magnetic field due to a current element is difficult to visualize
physically as such elements cannot be isolated from the circuit which they
are part of. Andre Ampere formulated a law based on Oersted’s as well as his
own experimental studies. Ampere’s law states that “the line integral of magnetic
field around any closed path equals μ0 times the current which threads the surface
bounded by such closed path.. Mathematically,
I ~B · ~dl = μ0Ienclosed (1)
19
In spite of its apparent simplicity, Ampere’s law can be used to calculate magnetic
field of a current distribution in cases where a lot of information exists on the
behaviour of ~B. The field must have enough symmetry in space so as to enable us
to express the left hand side of (1) in a functional form. The simplest application
of Ampere’ s law consists of applying the law to the case of an infinitely long
straight and thin wire.
Example : Magnetic Field of a long wire
By symmetry of the problem we know that the magnitude of the field at a point
can depend only on the distance of the point from the wire. Further, the field is
tangential to the circle of radius r, its direction being given by the right hand rule.
Thus the integral around the circle is
I ~B · ~dl = B I dl = B · 2πr
Equating this to μ0I, we get
B =
μ0I
2πr
which is consistent with the result
obtained from Biot-Savart’s law.
I
B
r
Let us see if the result above is consistent with a path which is not circular, as
shown in the figure. The field at every element ~dl of the path is perpendicular to
~r. From geometry, it can be seen that
~B
· ~dl = Bdl cos φ = Brdθ
20
Thus
I ~B · ~dl = I μ0I
2πr
rdθ
=
μ0I
2π I dθ = μ0I
We need to specify the direction
along which the path is traversed.
This is done by Right Hand Rule. If
we curl the fingers of our right hand
along the path of integration, the direction
along which the thumb points
is the direction of current flow.
I
B
dq
r
rdq
df d l
For the case where the path of integration
lies totally outside the path of
the current, for every element ~dl at P,
there exists another element at P’ for
which ~B · ~dl has opposite sign. Thus
when complete line integral is taken,
the contributions from such pairs add
to zero
I ~B · ~dl = 0
Combining these, we get Ampere’s
law in the form of Eqn. (1)
B
r
rdq
df
dq
d l
I
d l
B
P
P’
Example 8 : Calculate the field due to a uniform current distribution in an infinite
wire of cross sectional radius R.
Solution :
Let the cross section of the wire be circular with a radius R. Take the current
direction to be perpendicular to the page and coming out of it. Symmetry of the
problem demands that the magnitude of the field at a point is dependent only on
the distance of the point from the axis of the wire. Consider an amperian loop of
21
radius r. As before we have
I ~B · ~dl = 2πrB
R
r 2
r
1
0
R 2R 3R 4R 5R
1
0.5
B(r) −−−>
~ r ~ 1/r
r −−>
If r > R (as in loop 1), the entire current is enclosed by the loop. Hence Ienclosed =
I so that
B =
μ0I
2πr
If r < R (loop 2), the current enclosed is proportional to the area, i.e.
Ienclosed = I
πr2
πR2 = I
r2
R2
so that
B =
μoI
2πR2 r
The field distribution with distance is as shown.
Exercise :
A long wire of cross sectional radius R carries a current I. The current density
varies as the square of the distance from the axis of the wire. Find the magnetic
field for r < R and for r > R. ( Hint : First show that the current density
J = 2Ir2/πR4 and obtain an expression for current enclosed for r < R. Answer
: B = μ0I/2πr for r > R and B = μ0Ir3/2πR4 for r < R.)
Exercise :
22
A hollow cylindrical conductor of
infinite length carries uniformly distributed
current I from a < r < b.
Determine magnetic field for all r.
(Answer : Field is zero for r < a,
B =
μ0I
2πr
r2 − a2
b2 − a2 for a < r < b and
B = μ0I/2πr for r > b.)
a
I b
Exercise :
A coaxial cable consists of a solid conductor of radius a with a concentric shell of
inner radius b and outer radius c. The space between the solid conductor and the
shell is supported by an insulating material.
b
c
a
I
A current I goes into the inner conductor and is returned by the outer shell. Assume
the cuurent densities to be uniform both in the shell and in the inner conductor.
Calculate magnetic field everywhere. (Ans. B = μ0Ir/2πb2 inside
the inner conductor, B = μ0I/2πr between the shell and the inner conductor,
B =
μ0I
2πr
c2 − r2
c2 − b2 )
Exercise :Determine the magnetic field in a cylindrical hole of radius a inside a
cylindrical conductor of radius b. The cylinders are of infinite length and their
axes are parallel, being separated by a distance d. The conductor carries a current
I of uniform density. (Hint : The problem is conveniently solved by imagining
currents of equal and opposite densities flowing in the hole and using superposition
principle to calculate the field. Answer : The field inside the hole is constant
B = μ0Id/2π(b2 − a2))
Example 9 : Field of a long solenoid
We take the solenoid to be closely wound so that each turn can be considered to
be circular. We can prove that the field due to such a solenoid is entirely confined
to its interior, i.e. the field outside is zero, To see this consider a rectangular
amperian loop parallel to the axis of the solenoid.
23
z
B
D C
A
E F
H G
b a
Field everywhere on AB is constant and is B(a). Likewise the field everywhere
on CD is B(b). By Right hand rule, the field on AB is directed along the loop
while that on CD is oppositely directed. On the sides AD and BC, the magnetic
field direction is perpendicular to the length element and hence ~B · ~dl is zero
everywhere on these two sides. Thus
I ~B · ~dl = L(B(a) − B(b)
By Ampere’s law, the value of the integral is zero as no current is enclosed by the
loop. Thus B(a) = B(b). The field outside the solenoid is, therefore, independent
of the distance from the axis of the solenoid. However, from physical point of
view, we expect the field to vanish at large distances. Thus B(a) = B(b) = 0.
To find the field inside, take an amperian loop EFGH with its length parallel to
the axis as before, but with one of the sides inside the solenoid while the other is
outside. The only contribution to H ~B · ~dl comes from the side GH. Thus,
I ~B · ~dl = BL = μ0Ienclosed = μ0nLI
where I is the current through each turn and n is number of turns per unit length.
Ienclosed = nLI because the number of turns threading the loop is nL. Hence,
B = μ0nI
is independent of the distance from the axis.
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Exercise : Field of a toroid
A toroid is essentially a hollow tube bent in the form of a circle. Current carrying
coils are wound over it. Use an amperian loop shown in the figure to show that
the field within the toroid is μ0NI/L, where N is the number of turns and L the
circumference of the circular path.
B
Note that as the circumference of the circular path varies with the distance of the
amperian loop from the toroid axis, the magnetic field in the toroid varies over its
cross section.
Take the inner radius of the toroid to be 20cm and the outer radius as 21cm. Find
the percentage variation of the field over the cross section of the toroid. (Ans.
2.9%)
Example 10 : Field of an infinite current sheet
An infinite conducting sheet carries a current such that the current density is λ per
unit length. Take an amperian loop as shown.
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b
a
P
Q
R
S
The contribution to the line integral of ~B from the sides QR and SP are zero as ~B
is perpendicular to ~dl. For PQ and RS the direction of ~B is parallel to the path.
Hence
I ~B · ~dl = 2bB = μ0λb
giving B = μ0λ/2.
Exercise :
Calculate the force per unit area between two parallel infinite current sheets with
current densities λ1 and λ2 in the same direction. ( Ans.μ0λ1λ2/2)
Ampere’s Law in Differential Form
We may express Ampere’s law in a differential form by use of Stoke’s theorem,
according to which the line integral of a vector field is equal to the surface integral
of the curl of the field,
I ~B · ~dl = ZS
curl ~B · d~S
The surface S is any surface whose boundary is the closed path of integration of
the line integral.
26
J
C
S
In terms of the current density ~ J, we
have,
ZS
J~ · d~S = Iencl
where Iencl is the total current
through the surface S. Thus, Ampere’s
law H ~B · ~dl = Iencl is equivalent
to
Z curl ~B · d~S = μ0 Z J~ · d~S
which gives
curl ~B = μ0 ~ J
You may recall that in the case of electric field, we had shown that the divergence
of the field to be given by ∇~E = ρ/epsilon0. In the case of magnetic field there
are no free sources (monopoles). As a result the divergence of the magnetic field
is zero
∇ · ~B = 0
The integral form of above is obtained by application of the divergence theorem
ZS
~B
· d~S = ZV
∇ · ~B dV = 0
Thus the flux of the magnetic field through a closed surface is zero.
Vector Potential
For the electric field case, we had seen that it is possible to define a scalar function
φ called the “potential” whose negative gradient is equal to the electric field ∇φ =
~E
. The existence of such a scalar function is a consequence of the conservative
nature of the electric force. It also followed that the electric field is irrotational,
i.e. curl ~E = 0.
For the magnetic field, Ampere’s law gives a non-zero curl
curl ~B = μ0 ~ J
Since the curl of a gradient is always zero, we cannot express ~B as a gradient of a
scalar function as it would then violate Ampere’s law.
27
However, we may introduce a vector function ~A(~r) such that
~B
= ∇ × ~A
This would automatically satisfy∇·~B = 0 since divergence of a curl is zero. vecA
is known as vector potential. Recall that a vector field is uniquely determined by
specifying its divergence and curl. As ~B is a physical quantity, curl of ~A is also
so. However, the divergence of the vector potential has no physical meaning and
consequently we are at liberty to specify its divergence as per our wish. This
freedom to choose a vector potential whose curl is ~B and whose divergence can
be conveniently chosen is called by mathematicians as a choice of a gauge. If ψ
is a scalar function any transformation of the type
~A
−→ ~A + ∇ψ
gives the same magnetic field as curl of a gradient is identically zero. The transformation
above is known as gauge invariance. (we have a similar freedom for
the scalar potential φ of the electric field in the sense that it is determined up to
an additive constant. Our most common choice of φ is one for which φ → 0 at
infinite distances.)
A popular gauge choice for ~A is one in which
∇ · ~A = 0
which is known as the “Coulomb gauge”. It can be shown that such a choice can
always be made.
Exercise :
Show that a possible choice of the vector potential for a constant magnetic field ~B
is given by ~A = (1/2)~B × ~r. Can you construct any other ~A ? (Hint : Take ~B in
z-direction, express ~A in component form and take its curl.)
Biot-Savart’s Law for Vector Potential
Biot-Savart’s law for magnetic field due to a current element ~dl
d~B =
μ0I

~dl × ˆr
r2 = −
μ0I

~dl × ∇(
1
r
)
may be used to obtain an expression for the vector potential. Since the element ~dl
does not depend on the position vector of the point at which the magnetic field is
calculated, we can write
d~B =
μ0I

∇ × (
~dl
r
)
28
the change in sign is because ∇(
~dl
r
) = ∇(1/r) × ~dl.
Thus the contribution to the vector potential from the element ~dl is
d~A =
μ0I
4πr
~dl
The expression is to be integrated over the path of the current to get the vector
potential for the system
~A
=
μ0I
4π Z ~dl
r
Example 11 : Obtain an expression for the vector potential at a point due to a
long current carrying wire.
Solution :
Take the wire to be along the z-direction, perpendicular to the plane of the page
with current flowing in a direction out of the page. The magnitude of the field at a
point P is μ0I/2πr with its direction being along the tangential unit vector ˆθ at P,
~B
=
μ0I
2πr
ˆθ
The direction of ~B makes an angle (π/2) +
θ with the x direction, where tan θ = y/x.
Thus
ˆθ = ˆı cos π
2
+ θ + ˆ sin π
2
+ θ
= −ˆı sin θ + ˆ cos θ
= −ˆı
y
r
+ ˆ
x
r
x
y
q
q
^r
^
P
q
Hence we have
~B= μ0I

(−yˆı + xˆ)
x2 + y2
We wish to find a vector function ~A whose curl is given by the above. One can
see that the following function fits the requirement
~A
= −
μ0I

ln(x2 + y2)ˆk (1)
29
In the following, we will derive this directly from the expression for Biot-Savart’s
law.
If ρ is the distance of P from an element
of length dz at z of the wire, we have,
r2 = x2 + y2 + z2 = ρ2 + z2
Thus
~A
=
μ0I
4π Z dz
(ρ2 + z2)1/2
If the above integral is evaluated from
z = −∞ to z = +∞, it diverges. However,
we can eliminate the infinity in the
following manner. Let us take the wire
to be of length 2L so that the neodymium magnets can recover.
~A
=
μ0I

ˆk Z L
−L
dz
(ρ2 + z2)1/2
z
dz
r
P
r
q
I
The integral is evaluated by substituting z = ρ tan θ, so that dz = ρ sec2 θdθ. We
get
~A
=
μ0I

ˆk Z

sec θdθ
=
μ0I

ˆk ln(sec α + tan α)
where tan α = L/ρ.
In terms of L and ρ, we have
sec α =
(L2 + ρ2)1/2
ρ
=
L
ρ 1 +
ρ2
L21/2

L
ρ 1 +
ρ2
2L2
Thus to leading order in L,
A =
μ0I

ˆk ln(2L/ρ) =
μ0I

(ln 2L − ln ρ)ˆk
As expected, for L → ∞, the expression diverges. However, since ~A itself is
not physical while curl of ~A is, the constant term (which diverges in the limit of
30
L → ∞) is of no consequence and ~A is given by
A = −
μ0I

ˆk ln ρ = −
μ0I

ln(x2 + y2)
which is the same as Eqn. (1)
Example 12 : Obtain an expression for the vector potential of a solenoid.
Solution :
We had seen that for a solenoid, the field is parallel to the axis for points inside
the solenoid and is zero outside.
B = μ0nIˆk insidesolenoid
= 0outside
Take a circle of radius r perpendicular to the axis of the solenoid. The flux of the
magnetic field is
R
r
r
1 2
z
q^
q^
Z ~B · d~S = μ0nI.πr2 forr ≤ R
= μ0nI.πR2 forr ≥ R
Since ~B is axial, ~A is directed tangentially to the circle. Further, from symmetry,
the magnitude of ~A is constant on the circumference of the circle.
Use of Stoke’s theorem gives
Z ~B · ~ dS = Z curl ~A · d~S
= I ~A · ~dl
= | A | 2πr
31
Thus
~A
=
μ0nIπr2
2πr
ˆθ =
μ0nIr
2
ˆθ for ≤ R
=
μ0nIπR2
2πr
ˆθ =
μ0nIR2
2r
ˆθ for ≥ R
where ˆθ is the unit vector along the tangential direction.
Exercise : Obtain an expression for the vector potential inside a cylindrical wire
of radius R carrying a current I. (Ans. −μ0Ir2/4πR2)
The existence of a vector potential whose curl gives the magnetic field directly
gives
div B = 0
as the divergence of a curl is zero. The vector identity
∇ × ~B = ∇ × ∇ × ~A = ∇(∇ · ~A) − ∇2~A
can be used to express Ampere’s law in terms of vector potential. Using a Coulomb
gauge in which ∇ · ~A = 0, the Ampere’s law ∇ × ~B = −μ0 ~ J is equivalent to
∇2~A = −μ0 ~ J
which is actually a set of three equations for the components of ~A, viz.,
∇2Ax = −μ0Jx
∇2Ay = −μ0Jy
∇2Az = −μ0Jz
which are Poisson’s equations.

 

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