Like static electric fields are characterized by E or D, static magnetic fields are characterized by H or B.

Magnetostatics

similarities and dissimilarities between electric and magnetic fields

As E and D are related according to D = E for linear material space, H and B are relatedaccording to B = H.

Basic laws

Force law

Source element

Term Electric Magnetic

Field intensity

Flux density

Relationship between fields

Potentials

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Flux

Energy density

Term Electric Magnetic

Poisson's equation

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Two major laws governing magnetostatic fields:

(2) Ampere's circuit law. Just as Gauss's law is a special case of Coulomb's law, Ampere'slaw is a special case of Biot-Savart's law and is easily applied in problems involvingsymmetrical current distribution.

(1) Biot-Savart's law. Like Coulomb's law, it is the general law of magnetostatics.

magnetic field intensity dH produced at a point P, by the differential current clement dl is

Biot-Savart Law

magnetic field intensity dH produced at a point P, by the differential current clement dl isproportional to the product dl and the sine of the angle a between the element and the linejoining P to the element and is inversely proportional to the square of the distance Rbetween P and the element.

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In SI units, So

From the definition of cross product above equation in vector form is

where

direction of dH can be determined by the right hand rule with the right-handthumb pointing in the direction of the current, the right-hand fingers encircling thethumb pointing in the direction of the current, the right-hand fingers encircling thewire in the direction of dH

Customary to represent the direction of the magnetic field intensity H (or current )

current distributions

Line current

Surface currentSurface current

Volume current

field due to a straight current carrying filamentary conductor of

finite length AB

Special case: w

hen the conductor is infinite (with respect to

P) so that point A is now at O

(0, 0, -) w

hile B is at (0, 0,

);

Do it yourself

AMPERE'S CIRCUIT LAWMAXWELL'S EQUATION

Line integral of the tangential component of H around a dosed path is the same as the net current enc enclosed by the path

similar to Gauss's law and it is easily applied to determine H when the current distribution is symmetrical. is symmetrical.

Ampere's law is a special case of Biot-Savart's law; the former may be derived from the latter.

Above equation always holds whether the current distribution is symmetrical or notbut we can only use the equation to determine H when symmetrical currentdistribution exists.

But

Comparing,

By applying Stoke's theorem to the left-hand side

3rd Maxwell's equation

Ampere's law in differential form

Observation: magnetostatic field is not conservativeObservation: magnetostatic field is not conservative

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APPLICATIONS OF AMPERE'S LAW

Infinite Line Current

To determine H at an observation point P, draw a closed path pass through P

Amperian path

Since this path encloses the whole current , according to Ampere's law

Infinite Sheet of CurrentIt cannot have x and z components. Why?

B is perpendicular to K

Applying Ampere's law,

one Bl comes from the top segment, and the other from the bottom

Notice that the field is independent of the distance from the plane, just like the electric field of a uniform surface charge

ToroidA toroid whose dimensions are shown in Figure has N turns and carries current . Determine H inside and outside the toroid.

Ans:

inside: Outside is zero

Find the magnetic field of a very long solenoid, consisting of n closely wound turns per unitlength on a cylinder of radius R and carrying a steady current I

Solenoid

Ans:

insideinside

outside

MAGNETIC FLUX DENSITY BMAXWELL'S EQUATION

similar to the electric flux density D i.e.

magnetic flux through a surface S is given by

Units: Webers/m2 (or) Tesla

total flux through a closed surface in a magnetic field

since an isolated magnetic charge does not exist.

Gauss's law for magnetostatic fields(or) Nonexistence of magnetic monopole

apply the divergence theoremTo get above equation in differential form,

fourth Maxwell's equation

Putting Maxwell equations for static EM fields, we get

Gauss's law

Nonexistence of magnetic monopole

Differential form Integral Form Remarks

Conservativeness of electrostatic field

Ampere's law

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MAGNETIC SCALAR AND VECTOR POTENTIALS

Recall that some electrostatic field problems were simplified by relating the electric potential V to the electric field intensity E (E = V).

Magnetic potential could be scalar Vm or vector A.

Similarly, we can define a potential associated with magnetostatic field B.

Recall two important identities

which must always hold for any scalar field V and vector field A

if J = 0

This satisfies Poissons equation

From Helmholtz theorem

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Like we defined for electrostatic potential V

Can you derive it yourself?

Line current

Surface currentSurface current

Volume current

Illustration of the source point (x', y', z') and the field point (x, y, z)

Proof of Magnetic Vector Potential

where the differentiation is with respect to x, y, and z

Substituting this in we get

Use where

Why?

So,

Now, the equation reduces to

Comparing with we get

So, flux can be found either by using (or)

Given

Problem

Wb/m, calculate the total magnetic flux crossing the surface

Answer: 3.75 Wb

Summary of Magnetostatics

considered the basic laws and techniques commonly used in calculating magneticfield B due to current-carrying elements

Let us study the force a magnetic field exerts on charged particles, current elements, and loops

important to problems on electrical devices such as ammeters, voltmeters, galvanometers, cyclotrons, plasmas, motors, etc.

Forces due to magnetic fields

3 ways in which force due to magnetic fields can be experienced. The force can be3 ways in which force due to magnetic fields can be experienced. The force can be

(a) due to a moving charged particle in a B field,

(b) on a current element in an external B field

(c) between two current elements.

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Electric force Fe on a stationary or moving electric charge Q in an electric field is

A magnetic field can exert force only on a moving charge

magnetic force Fm experienced by a charge Q moving with a velocity u in a magnetic field B is

Fe and E have the same direction

Fm is perpendicular to both u and B.

Comparison:

F independent of the velocity of the charge and can perform work on the charge and Fe independent of the velocity of the charge and can perform work on the charge and change its kinetic energy.

Fm depends on the charge velocity and is normal to it

and does not cause an increase in kinetic energy of the charge.

magnitude of Fm is generally small compared to Fe except at high velocities.

For a moving charge Q in the presence of both fields,

Lorentz force equation

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relates mechanical force to electrical force.

Solution to this equation is important in determining the motion of charged particles inE and B fields.

Remember that in such fields, energy transfer can be only by means of the electric field.

Force on a Current Element

Alternatively, Using and we get

i.e. an elemental charge dQ moving with velocity u (thereby producing convection current element dQ u) is equivalent to a conduction current element dl

B is defined as the force perunit current element

Remember:

Magnetic field produced by the current element dl does not exert force on the element itself just as a point charge does not exert force on itself.

The B field that exerts force on dl must be due to another element.

Force between current elementsforce d(dF1) on element 1 dl1 due to the field dB2 produced by element 2 dl2

From Biot-Savart's lawFrom Biot-Savart's law

So,

total force F, on current loop 1 due to current loop 2

A charged particle of mass 2 kg and charge 3 C starts at point (1, 2 , 0) with velocity4ax + 3az m/s in an electric field 12ax + 10ay, V/m. At time t = 1 s, determine

(a) The acceleration of the particle Ans: 18 ax + 15 ay m/s2b) Its velocity Ans: : 22 ax + 15 ay + 3az m/s(c) Its kinetic energy : 718 J(d) Its position : (14, 5.5, 3)

Problem

A charged particle moves with a uniform velocity 4ax m/s in a region where E = 20 ay V/m and B = B0 az Wb/m2. Determine B0 such that the velocity of the particle remains constant.

MAGNETIC TORQUE AND MOMENTAfter discussing about the force on a current loop in a magnetic field, we can determinethe torque on it.

Important to understand the behavior of orbiting charged particles, d.c. motors, andgenerators.

If the loop is placed parallel to a magnetic field, it experiences a force that tends to rotate it.

The torque T (or mechanical moment of force) on the loop is the vector product ofThe torque T (or mechanical moment of force) on the loop is the vector product ofthe force F and the moment arm r.

Units: Newtons- meters

Apply this to a rectangular loop. From this figure, we notice that dl is parallel toB along sides 12 and 34 of the loop and no force is exerted on those sides. Thus

where because B is uniformwhere because B is uniform

Thus, no force is exerted on the loop as a whole

Fo and Fo act at different points on the loop, thereby creating a couple.

If the normal to the plane of the loop makes an angle with B, as shown in the cross-sectional view of Figure, the torque on the loop is

But the area of the loop

wheremagnetic dipole moment (in A. m2) of the loop.

The magnetic dipole moment is the product of current and area of the loop; its directionThe magnetic dipole moment is the product of current and area of the loop; its directionis normal to the loop.

Direction of torque is axis of rotation (the z-axis in this case)

generally applicable in determining the torque on a planar loop of any arbitrary shape., provided magnetic field must be uniform.

A MAGNETIC DIPOLEA bar magnet (or) a small filamentary current loop

Let us find B at an observation point P due to a circular loop carrying current . The magnetic vector potential A at P is

At far field (r >> a), the loop appears small at P. So A has only -component i.e.

where Is moment of the loop, and

Electric MagneticDoes not existMonopole (point charge)

Dipole (two point charge)Dipole (two point charge)

B lines due to magnetic dipoles

Small current loop with m = S bar magnet with

B lines due to both dipoles are similar

When a short permanent magnetic bar is in a uniform magnetic field B, it experiences a torque given by

Qm i.e. pole strength does not exist without an associated Qm

The torque tends to align the bar with the external magnetic field.

force acting on the magnetic charge is given by

Since both a small current loop and a bar magnet produce magnetic dipoles, they are equivalent if they produce the same torque in a given B field i.e. when

showing that they must have the same dipole moment.

A small current loop L1, with magnetic moment 5 az A . m2 is located at the origin whileanother small loop current L2 with magnetic moment 3 ay A . m2 is located at (4, 3, 10).Determine the torque on L2.

Problem

torque T2 on the loop L2 is due to the field B1 produced by loop L1

Since m1 for loop L1 is along az,

Transform m2 from Cartesian to spherical coordinates i.e.

At (4, 3, 10)

Determine the magnetic moment of an electric circuit formed by the triangular loop

Problem

Ans: 10 (ax + ay + az ) A m2

Similar to polarization of materials in an electric fieldMAGNETIZATION IN MATERIALS

Magnetic dipole moment in a volume v

Internal magnetic field is produced byelectrons orbiting around the nucleus (or)electrons spinning

b is the bound current (bound to the atom).

If there are N atoms in a given volume v and the kth atom has a magnetic moment mk,

A medium for which M is not zero everywhere is said to be magnetized

Magnetization M (amperes/meter) is the magnetic dipole moment per unit volume.

Vector magnetic potential due to m is

Comparing this withand we get

Potential (and hence also the field) of a magnetized object is the same as that produced by a bound volume current density Jb = x M plus a bound surface current density Kb = M x an

In free space, M = 0

Jf is the free current volume density

For M 0

holds for all materials whether they are linear or not

For linear materials, M (in A/m) depends linearly on H such that

Magnetic susceptibility

permeability of the material measured in henry/meter

Problem

Since z = 0 is the lower side of the slab, an = az

CLASSIFICATION OF MAGNETIC MATERIALS

nonmagnetic if m = 0 (or r =1) Ex: Free space, air

Relative Permeability of Some Materials

for most practical purposes we may assumethat r =1 for diamagnetic and paramagneticmaterials. Thus, we may regard diamagneticand paramagnetic materials as linear andnonmagnetic

Ferromagnetic materials are always nonlinearand magnetic except when their temperaturesare above curie temperatureare above curie temperature

Except for superconductors, diamagnetic materials are seldom used in practice

diamagnetism

magnetic fields due to electronic motions of orbiting and spinning completely cancel each other.

permanent (or intrinsic) magnetic moment of each atom is zero and the materials are weakly affected by a magnetic field.

Ex: bismuth, lead, copper, silicon, diamond, sodium chloride

m = 10-5

Paramagnetism occurs in materials where the magnetic fields produced by orbital and spinning electrons do not cancel completely

Paramagnetism Ex: air, platinum, tungsten, potassium

Materials whose atoms have nonzero permanent magnetic moment may be paramagneticor ferromagnetic

spinning electrons do not cancel completely

Unlike diamagnetism, paramagnetism is temperature dependent.

m = 10-5 to 10-3

Ferromagnetism Ex: Iron, cobalt, nickel, and their alloys

occurs in materials whose atoms have relatively large permanent magnetic moment

does not hold for ferromagnetic materials because r depends on B and cannot be represented by a single value

They are nonlinear

typical B-H curve

existence of Br is the cause of having permanent magnets

Hc: coercive field intensity

Problem

MAGNETIC BOUNDARY CONDITIONSconditions that H (or B) field must satisfy at the boundary between two different media.

andUse Gauss law for magnetic fields and Ampere law

Applying to the pillbox (Gaussian surface) and allowing h ! 0

(or)

normal component of B is continuous at theboundary. It also shows that the normalcomponent of H is discontinuous at theboundary; H undergoes some change at theinterface

Apply Ampere law to the closed pathabcda where surface current K on theboundary is assumed normal to the path

As tangential component of H is also discontinuous

If the boundary is free of current (or) the media are not conductors (for K is freecurrent density), K = 0

(or)

tangential component of H is continuous while that of B is discontinuous at the boundary.

If the fields make an angle with the normal to the interface,

gives

gives

Dividing each otherLaw of refraction for magnetic flux lines at a boundary with no surface current

INDUCTORS, MAGNETIC ENERGY

A circuit or part of a circuit that has inductance is called an inductor

If the circuit has N identical turns, flux linkage is

where

if the medium surrounding the circuit is linear, the flux linkage is proportional to the current producing it

L is a constant of proportionality inductance of the circuit

A circuit or part of a circuit that has inductance is called an inductor

Units: Henry

Like capacitances, we may regard inductance as a measure of how much magnetic energy is stored in an inductor

Magnetic energy (in joules) stored in an inductor is

i.e. an inductor is a conductor arranged in a shape appropriate to store magnetic energy.

Ex: toroids, solenoids, coaxial transmission lines, and parallel-wire transmission lines.

Thus the self-inductance of a circuit can becalculated from energy considerations.

If there are two circuits carrying current 1 and 2

self-inductance of circuits 1 and 2 are

and

total energy in the magnetic field is the sum of the energies due to L1, L2, andMI2 (or M21);

positive sign is taken if 1 and 2 flow such that the magnetic fields of the two circuits strengthen each other

For a given inductor, we find the self-inductance L by taking these steps:1. Choose a suitable coordinate system

2. Let the inductor carry current /.

3. Determine B from Biot-Savart's law (or from Ampere's law if symmetry exists)and calculate

4. Finally find L from

Calculate the self-inductance per unit length of an infinitely long solenoid.

Here, number of turns per unit length

If S is the cross-sectional area of the solenoid, the total flux through the cross section is

Since this flux is only for a unit length of the solenoid, the linkage per unit length is

Problem

and thus the inductance per unit length is H/m

Just like we derived,

One can derive similar expression for magnetic energy

MAGNETIC CIRCUITS

Concept is based on solving some magnetic field problems using circuit approach.

Ex: Magnetic devices such as toroids, transformers, motors, generators, and relays may be considered as magnetic circuits

Make it simple by exploiting the analogy between magnetic circuits and electric c.

Magnetomotive force (mmf) (in ampere-turns)

source of mmf in magnetic circuits is usually a coil carrying current

define reluctance as reciprocal of reluctance is permeance

basic relationship for circuit elements is Ohm's law (V = IR):basic relationship for circuit elements is Ohm's law (V = IR):

Based on this, Kirchhoff's current and voltage laws can be applied to nodes and loops of agiven magnetic circuit just as in an electric circuit.

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FORCE ON MAGNETIC MATERIALSpractical interest to determine the force that a magnetic field exerts on a piece ofmagnetic material in the field.useful in electromechanical systems such as electromagnets, relays, rotating machines, and magnetic levitation

The coil has N turns and carries a current .

ignore fringing, the magnetic field in the airgap is the same as that in iron i.e. B1n = B2nTo find the force between the two pieces of iron, we calculate the change in the total iron, we calculate the change in the total energy that would result were the two pieces of the magnetic circuit separated by a differential displacement dl.

Work required to create the displacement is equal to the change in stored energy in the air gap

factor 2 accounts for the two air gaps, and the negative sign indicates that the force acts to reduce the air gap

Remember, force is exerted on the lower piece and not on the current-carrying upper piece giving rise to the field.

The tractive force across a single gap

tractive pressure (in N/m2) in a magnetized surface is

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ProblemThe toroidal core has o = 10 cm and a circular cross section with a = 1 cm. If the core is made of steel ( = 1000 o) and has a coil with 200 turns, calculate the amount of current that will produce a flux of 0.5 mWb in the core.

can be solved in two different ways: using the magnetic field approach (direct), (or) using the electric circuit analog (indirect).

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Method 1: Since o is large compared with a,

Method 2:

ProblemIn the magnetic circuit, calculate the current in the coil that will produce a magnetic fluxdensity of 1.5 Wb/m2 in the air gap assuming that = 5o and that all branches have thesame cross-sectional area of 10 cm2.

Ans: 44.16 A

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