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**Magnetic Force on Current **

The magnetic field at any point due to steady current is called a magnetostatic field. The magnetic force on a charge Q , moving with velocity v in a magnetic field is, This is known as Lorentz force law.

In the presence of both electric and magnetic fields, the net force on Q would be:

**Current in a Wire**

A line charge λ traveling down a wire at a speed constitutes a current

Magnetic force on a segment of current-carrying wire is,

since points in the same direction**Surface Current Density**

When charge flows over a surface, we describe it by the surface current

is the current per unit width-perpendicular to flow.

Also where is σ surface charge density and is its velocity.

Magnetic force on surface current

**Volume Current Density**

When the flow of charge is distributed throughout a three-dimensional region, we describe it by the volume current density .

is the current per unit area-perpendicular to flow.

Also where ρ is volume charge density and is its velocity.

Magnetic force on volume current

Current crossing a surface S is **Example 1: A wire ABCDEF (with each of side of length L ) bent as shown in figure and carrying a current I is placed in a uniform magnetic induction B parallel to the positive y-direction. Find the force experienced by the wire.**

FE and BA are parallel to magnetic induction . Magnetic force on each of them will be zero. DE and CB are perpendicular to . They carry currents in opposite directions forces on them will be equal in magnitude and opposite in direction. The net force due to these portions of wire will be zero. Now force on side DC is

**Example 2: A semi–circular wire of radius R carries a current I and is placed in a uniform field B acting perpendicular to the plane of the semi–circle. Calculate force acting on the wire.**

Consider an element of length dl of the wire. The force on this element is obtained by

(Horizontal component cancels only perpendicular component add up).

**Continuity Equation**

The total charge per unit time leaving a volume V is

Because charge is conserved, whatever flows out through the surface must come at the expense of that remaining inside:

(The minus sign reflects the fact that an outward flow decreases the charge left in V.) Since this applies to any volume, we conclude that

This is the precise mathematical statements of **local charge conservation.**

**Note:** When a steady current flows in a wire, its magnitude I must be the same all along the line; otherwise, charge would be piling up somewhere, and it wouldn't be a steady current. Thus for magnetostatic fields and hence the continuity equation becomes:

**Biot-Savart Law**

The magnetic field of a steady line current is given by

where

For surface and volume current Biot-Savart law becomes: **Magnetic Field due to Wire**

Let us find the magnetic field a distance d from a long straight wire carrying a steady current I.

In the diagram, points out of the page and has magnitude dl′ sinα = dl′cos θ

Since,

From Biot–Savart law:**For Infinite wire:****Note:**

- Magnetic field a distance r from a long straight wire carrying a steady current I is

- Magnetic field a distance r from a infinite wire carrying a steady current I is:
- Force (per unit length) of attraction between two long, parallel wires a distance d apart, carrying currents I
_{1}and I_{2}in same direction are: - If currents are in opposite direction they will repel with same magnitude.

**Magnetic Field due to Solenoid and Toroid**

The magnetic field of a very long solenoid, consisting of n closely wound turns per unit length of a cylinder of radius R and carrying a steady current I is:

Magnetic field due to Toroid is

where N is the total number of turns.

**Example 3: Find the force of attraction between two long, parallel wires a distance d apart, carrying current I _{1} and I_{2} in the same direction.**

The field at (2) due to (1) is

Force on (2) is

Force per unit length is towards (1) and net force is attractive.

**Example 4: Find the magnetic field a distance d above the center of a circular loop of radius R, which carries a steady current I.**

The field attributable to the segment as shown. As we integrate

around the loop, sweeps out a cone. The horizontal components cancel, and the vertical components combine to give.

**Example 5: Find the force on a square loop placed as shown in figure, near an infinite straight wire. Both the loop and the wire carry a steady current I.**

The force on the two sides cancels.

At the bottom,

At the top,

Thus

**Ampere's Law**

The magnetic field of an infinite wire is shown in the figure (the current is coming out of the page). Let us find the integral of around a circular path of radius r, centered at the wire, is

Notice that the answer is independent of r; that is because B decreases at the same rate as the circumference increases. If we use cylindrical coordinates ( r ,φ,z ) , with the current flowing along the z axis,

In general we can write

where I_{enc} is the total current enclosed by the amperian loop.

since

**Right hand Rule**

If the fingers of your right hand indicate the direction of integration around the boundary, then your thumb defines the direction of a positive current.

**Example 6: A steady current I flow down a long cylindrical wire of radius a. Find the magnetic field, both inside and outside the wire, if**

**The current is uniformly distributed over the outside surface of the wire.****The current is distributed in such a way that J is proportion to r, the distance from the axis.**

(1)

(2)

**Example 7: Find the magnetic field of an infinite uniform surface current , flowing over the x–y plane.**

Since have no x-component because B is ⊥

^{r}_{ }to x-component i.e. in the direction of

Also, have no z-component: For y > 0 , B is along and for y < 0 , B is along − thus field cancels each other.⇒ has only y-component:

For z >0 , points left () and for z <0 , point’s right ( ).

Draw a rectangular amperian loop parallel to the yz plane and extending an equal distance above and below the surface. Now apply ampere’s law, we find

{One Bl from top segment, and the other from bottom}

Note: The field is independent of the distance from the plane, just like the electric field of a uniform surface charge.

**Magnetic Vector Potential **

Since

For magnetostatic fields,

if goes to zero at infinity, for volume current.

For line and surface currents,

**Example 8: What current density would produce the vector potential (where K is a constant), in cylindrical coordinates ?**

**Magnetostatic Boundary Condition (Boundary is sheet of current, )**

Just as the electric field suffers a discontinuity at a surface charge, so the magnetic field is discontinuous at a surface current. Only this time it is the tangential component that changes.

Since

For tangential components

Thus the component of that is parallel to the surface but perpendicular to the current is discontinuous in the amount μ_{0}K . A similar amperian loop running parallel to the current reveals that the parallel component is continuous. The result can be summarized in a single formula:

where is a unit vector perpendicular to the surface, pointing “upward”. Like the scalar potential in electrostatics, the vector potential is continuous across, a boundary:

For guarantees that the normal component is continuous, and

in the form

But the derivative of inherits the discontinuity of

**Example 9: (a) Find the magnetic vector potential at a distance r from an infinite straight wire carrying a current I. ****(b) Find the magnetic potential inside the wire, if it has radius R and the current is uniformly distributed.**

(a)point in the same direction as I and is a function of r (the distance from the wire). In cylindrical coordinates(b)

where b is arbitrary constant.

must be continuous at R,which means that we must pick a and b such that

**Example 10: Find the vector potential of an infinite solenoid with n turns per unit length, radius R, and current I.**

Since where φ is the flux of through the loop in question.

Inside solenoid:

Outside solenoid:

**Multiple Expansion of Vector Potential **

We can derive approximate formula for the vector potential of a localized current distribution, valid at distant points. We can always write the potential in the form of a power series in 1/r, where r is the distance to the point in question. Thus we can always write

First term, monopole (no magnetic monopole)

Second term, dipole

where is the magnetic dipole moment:

where is area vector

Thus

Hence**Note:** (a) When a magnetic dipole is placed in a uniform magnetic field

net force on the dipole is zero and it experiences a torque

(b) In non-uniform field, dipoles have net force and torque

(c) Energy of an ideal dipole in an magnetic field

(d) Interaction energy of two dipoles separated by a distance is

**Example 11: A phonograph record of radius R, carrying a uniform surface charge σ is rotating at constant angular velocity ω. Find its magnetic dipole moment.**

Magnetic moment of a ring of radius r and thickness dr is, dm = Iπr

^{2}where

I = σvdr= σrωdr

**Example 12: A spherical shell of radius R, carrying a uniform surface charge σ, is set spinning at angular velocity ω. Find its Magnetic dipole moment.**

The total charge on the shaded ring is

dq = σ (2π R sinθ ) Rdθ

Time for one revolution is

⇒Current in the ring

Area of the ring =π(R sin θ)^{2}, so the magnetic moment of the ring is

**Magnetisation **

If a piece of magnetic material is examined on an atomic scale we will find tiny currents: electrons orbiting around nuclei and electrons spinning about their axes. For macroscopic purpose, these current loops are so small that we may treat them as magnetic dipoles. Ordinarily they cancel each other out because of the random orientation of the atoms. But when a magnetic field is applied, a net alignment of these magnetic dipoles occurs, and medium becomes magnetically polarized, or magnetized.

Magnetization is magnetic dipole moment per unit volume.

**The Field of a Magnetized Object (Bound Currents) **

Consider a piece of magnetized material with magnetization

Then the vector potential of a single dipole is given by

In the magnetized object, each volume element dτ ′ carries a dipole moment so the total vector potential is

The equation can be written as

The first term is like potential of a volume current

while the second term is like potential of a surface current,

where is the normal to the unit vector. With these definitions, the field of a magnetized object is

This means the potential(and hence also the field) of a magnetized object is the same as would be produced by a volume current throughout the material, plus a surface current on the boundary. We first determine these **bound currents**, and then find the field they produce.

**Example 13: An infinitely long circular cylinder carries a uniform magnetization **

**parallel to its axis. Find the magnetic field (due to ) inside and outside the cylinder.**

The field is that of a surface current that is the case of a

solenoid,So the field outside is zero.

Field inside is:

**Example 14: A long circular cylinder of radius R carries a magnetization **

**where K is a constant; r is the distance from the axis. Find the magnetic field due to **

** for points inside and outside the cylinder.**

So the bound current flows up the cylinder, and returns down the surface.

Outside point: I

_{enc}= 0⇒B =0

Inside point:

**The Auxiliary field **

Ampere’s Law in in presence of Magnetic Materials In a magnetized material the total current can be written as where is bound current and

is free current.

The quantity in parentheses is designated by the letter

In integral form where is the total free current passing through the amperian loop.

plays a role in magnetostatic analogous to in electrostatic: Just as

allowed us to write Gauss's law in terms of the free charge alone, permits us to express Ampere's law in terms of the free current alone- and free current is what we control directly. Note: When we have to find or in a problem involving magnetic materials, first look for symmetry. If the problem exhibits cylindrical, plane, solenoid, or toroidal symmetry, then we can get directly from the equation

Magnetic Susceptibility and Permeability

For most substances magnetization is proportional to the field where χ_{m} is magnetic susceptibility of the material.

where(μ = μ_{0}μ_{r}= μ_{0}(1 + χ_{m}) is permeability

of material.

**Boundary Condition **

The boundary between two medium is a thin sheet of free surface current K_{f} . The Ampere’s law states that

Since

Thus

**Example 15: A current I flows down a long straight wire of radius a. If the wire is made of linear material with susceptibility χ _{m} , and the current is distributed uniformly, what is the magnetic field a distance r from the axis? Find all the bound currents. What is the net bound current following down the wire?**

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