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What is the line integral of a vector? What are the Applications of Ampere’s law?
Line integral or path integral of a vector: Let \(\vec{A}\) be a vector and 81 a very small line segment [Fig.(a)]. This segment can be treated as a vector δ\(\vec{\imath}\), where the magnitude of δ\(\vec{\imath}\) is equal to the length of the segment and its direction is along the tangent to that segment. (In the figure, δ\(\vec{\imath}\) is shown in a magnified form),
The sum of the above dot products along a finite line segment PQ [Fig. (b)] can be expressed as an integral (using the symbol dl instead of δl).
It is called the line integral or path integral of vector \(\vec{A}\) along the path PQ. It is to be noted that the magnitude of vector \(\vec{A}\) may vary between point A to point B and if the direction of \(\vec{A}\) changes, θ will also change. The determination of the magni-tude of a line integral is in general very complicated. However, due to different types of symmetry, the integral can often be determined easily.
Closed line integral: A path that closes on itself is a closed path. To express the line integral of a vector along a closed path the symbol \(\oint\) is used. For example, in the Fig.(c), the line integral of the vector \(\vec{A}\), called the circulation of \(\vec{A}\), is given by
\(\oint \vec{A} \cdot d \vec{l}\) = \(\oint A \cos \theta d l\)
Example:
Line integral of force vector—work done: In Fig., if the force vector \(\vec{F}\) is taken as a special example of vector \(\vec{A}\), according to the definition of work done we can write,
work done by the force \(\vec{F}\) for displacement \(\delta \vec{l}\),
dW = \(\vec{F} \cdot \delta \vec{l}\)= Fδlcosθ
So, the total work done along the segment PQ,
If the force is conservative, the work done is zero.
Naturally, the physical significance of the line integral of force vector is—this integral always indicates the work done along a line.
Similarly, line integrals of different vectors in physics have different physical significances. For example, the line integral of the electrostatic field \(\vec{E}\) around a closed path is zero because elec-trostatic field is a conservation force field.
Statement of Ampere’s circuital law: The line integral of the magnetic field vector along a closed path in any magnetic field is equal to the product of the net current enclosed by the closed path and the permeability of vacuum, i.e.,
\(\oint \vec{B} \cdot d \vec{l}\) = µ0I …… (1)
Here, I = net current enclosed by the closed path.
Explanation:
i) In Fig., if X1 is the closed path, it encloses current I and as a result,
\(\oint_{X_1} \vec{B} \cdot d \vec{l}\) = µ0I
On the other hand, if we consider the X1, closed path X2, it encloses no current and hence I = 0.
So, \(\oint_{X_2} \vec{B} \cdot d \vec{l}\) = 0
ii) If any closed path encloses a number of conductors, carrying currents in different directions, the algebraic sum of the enclosed currents is to be taken. For example, in Fig.,
currents in the parts AB, BC, CD may be taken as I, -I and I, respectively and for the closed path X we can write,
\(\oint_X \vec{B} \cdot d \vec{l}\) = µ0(I – I + I) = µ0I
Again, in the Fig., since an equal current I flows through each of the turns,
\(\oint_X \vec{B} \cdot d \vec{l}\) = µ0NI [N = number of turns]
CGS form Of the law: Substituting B → H, I → i and µ0 → 4π we can write,
for a closed path enclosing a single turn,
\(\oint \vec{H} \cdot d \vec{l}\) = 4πi
and for a closed path enclosing N turns,
\(\oint \vec{H} \cdot d \vec{l}\) = 4πNi
Application of Ampere’s Circuital Law
Magnetic field of a long straight wire: Let a current I flow through a straight long conductor [Fig.]. We have to determine the magnetic field at the point P at a perpendicular distance r from the wire.
Taking the wire as axis, a circular path is drawn through the point P having radius r in such a manner that the path lies in a plane perpendicular to the wire. It is convenient to consider this circular path as Ampere’s closed path. For an element of length \(\delta \vec{l}\) on this closed path, corkscrew rule shows that, the magnetic field \(\vec{B}\) is parallel to \(\delta \vec{l}\) at every place, i.e., the angle between them is 0°. Again, due to symmetry, the magnitude of \(\vec{B}\) (i.e., |\(\vec{B}\)| = B) is the same at every point on the closed path.
So, \(\oint \vec{B} \cdot d \vec{l}\) = \(\oint B d l \cos \theta\)
= \(\oint B d l\) = \(B \oint d l\) = B ᐧ 2πr
Since the current enclosed by the closed path is I, from Ampere’s circuital law,
B ᐧ 2πr = µ0I
or, B = \(\frac{\mu_0 I}{2 \pi r}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\)
this is equation (7) of section 1.4.1.
Solenoid: If a long insulated conducting wire is wound tightly over the surface of a cylinder so that every circular turn is perpendicular to the axis of the cylinder, then this coil is called a solenoid [Fig.], The axis of the cylinder is the axis
of the solenoid. Usually after making a solenoid the inner cylin-der is removed. (A conductor with a coating of insulating mate-rial is called an insulated conductor.)
Magnetic lines of force: The solenoid is placed on a cardboard with its axis lying on the cardboard plane. Now some light iron-filings are scattered over the cardboard and the card-board is slightly tapped. The iron-filings arrange themselves along the magnetic lines of force. The magnetic lines of force are shown in Fig. For identification of N and S-poles of the solenoid see section 2.2. of the chapter ‘Magnetic Properties of Material’s.
Characteristics of the lines of force: The number density of magnetic lines of force inside the solenoid is very high, i.e., the magnetic field in that part is very strong. The magnetic field outside the solenoid can be neglected in comparison. Moreover, the magnetic lines of force inside the solenoid are parallel to its axis. So, a strong and nearly uniform axial magnetic field is generated inside the solenoid.
Magnetic field inside a long straight solenoid: Let the length of a long straight solenoid = L and its number of turns = N. So, the number of turns per unit length of the solenoid, n = \(\frac{N}{L}\). In Fig., a few number of turns of the solenoid is shown. Current through the sole-noid = I.
Here, the rectangle abcd is taken as the Ampere’s closed path whose side ab = L lies along the axis of the solenoid.
Let N be the number of turns enclosed by the rectangular path, \(\overrightarrow{d l}\) is a very small segment on this path and \(\vec{B}\) is the magnetic field produced due to this part of the solenoid. According to Ampere’s circuital law,
On the other hand, magnetic field \(\vec{B}\) is perpendicular to the small segments \(d \vec{l}\) on the parts of bc and da inside the solenoid. As the solenoid is an ideal one, \(\vec{B}\) = 0 for the side cd and also \(\vec{B}\) = 0 for that parts of bc and da which are outside the solenoid. Hence from equations (1) and (2) we get,
\(\oint_{a b c d} \vec{B} \cdot d \vec{l}\) = µ0 NI ….. (3)
or, B = µ0\(\frac{N}{L} \cdot I\) or B = µ0nI ….. (4)
It is to be noted here that the magnitude of the magnetic field depends on the number of turns per unit length n but not on the total number of turns N of the solenoid. Hence, to increase the magnetic field it is not sufficient to increase the number of turns but it is also necessary to make the turns very close to each other so that the value of n increases.
CGS expression: Substituting B → H, I → i and µ0 → 4π in equation (4), we get, H = 4πni.
Toroid: A toroid is a long insulated conducting wire, wound on a donut-shaped core having a circular axis and uniform cross section so that each turn is normal to the axis [Fig.].
A toroid is nothing but a solenoid bent to close on itself. A long straight solenoid has two definite ends but a toroid is an endless solenoid.
Magnetic field of a toroid: Let the radius of the ring of a toroid = r and total number of turns in it = N. So, the circum-ference of the circular axis of the toroid = 2πr, and the number of turns per unit length of it, n = \(\frac{N}{2 \pi r}\). In Fig., a few turns of a toroid is shown. Current through the toroid = I.
Here, the axis of the toroid is considered as the Ampere’s closed path. If any small part \(\delta \vec{l}\) is taken on the axis, according to corkscrew rule, the direction of the magnetic field \(\vec{B}\) is parallel to \(\delta \vec{l}\) at every point, i.e., the angle between them is 0°.
Now, due to symmetry, the magnitude of \(\vec{B}\) is same at all points on the axis. Hence,
\(\oint \vec{B} \cdot d \vec{l}\) = \(\oint B d l \cos 0^{\circ}\) = \(B \oint d l\) = B ᐧ 2πr
Again, the net current enclosed by the axis = current I through each of N turns = NI
So, according to Ampere’s circuital law,
B ᐧ 2πr = µ0NI or, B = \(\mu_0 \cdot \frac{N}{2 \pi r} I\)
or, B = µ0NI ….. (5)
It should be noted that equation (5) is identical with equation (4). From this, it is concluded that if a solenoid is too long, what ever may be the shape, the magnitude of the magnetic field at any point on its axis will be B = µ0nI.
CGS expression: Substituting B → H, I → i and µ0 → 4π inequation (5), we get, H = 4πni.
Core of a solenoid: In the above discussion, air is considered as the core of a solenoid or a toroid. Thus, magnetic permeability is taken as µ0. For any other core (like iron, nickel, etc.), the value of the magnetic permeability changes notably. In that case Ampere’s circuital law takes the form
\(\oint \vec{B} \cdot d \vec{l}\) = µnl, where µ is the permeability of the core.
It is discussed in detail in the chapter: ‘Magnetic Properties of Materials’.
Limitation of Ampere’s circuital law: Maxwell proved that Ampere’s clrcuital law is valid only for steady current. If the enclosed current varies with lime, on the right hand side of
equation (1) [section 1.5], an additional term should be added. By this correction Maxwell arrived at his famous electro magnetic field equations. Elaborate discussion about it has been done in the chapter on Electromagnetic waves. We should remember that, Ampere’s law is not incorrect, it can only be called incomplete. In our discussion, we consider the cases
where electric current remains steady with time and hence, the equations thus obtained from Ampere’s law are accurate.
Numerical Examples
Example 1.
A solenoid with 7 turns per unit length is carrying a current of 2.5 A. What is the magnetic intensity inside the solenoid? [HS 08]
Solution:
Turns per unit length, n = 7 cm-1 = 700 m-1;
if µ be the permeability of the medium inside the solenoid then, magnetic field B = µnI.
∴ Magnetic intensity,
H = \(\frac{B}{\mu}\) = nI = 700 × 2.5 = 1750 A ᐧ m-1
Example 2.
Length of a solenoid is 60 cm and its total number of turns is 1250. If 2 A current is passed through it, what will be the magnetic field at any point on its axis?
Solution:
Number of turns = 1250, length = 60 cm = 0.6 m
∴ Number of turns per unit length, n = \(\frac{1250}{0.6}\) m-1
So, the magnetic field at any point on its axis,
B = µ0nI
= 4π × 10-7 × \(\frac{1250}{0.6}\) × 2
[µ0 = 4π × 10-7Wb ᐧ A-1 ᐧ m-1]
= 5.23 × 10-3 Wb/m2
Example 3.
Two solenoids made of insulated conducting wires and of equal lengths are such that one is wound over another. Resistance of each of them is R and number of turns per unit length is n. The solenoids are now connected in
(i) series and
(ii) parallel and the combi-nation is then connected with a battery of emf E. If current flows through them in the same direction in both cases, determine the magnetic field along the axis of the solenoids in each case.
Solution:
i) In case of series combination, equivalent resistance = 2R
and hence current through each solenoid, Is = \(\frac{E}{2 R}\).
Hence, the resultant magnetic field along the axis,
B = B1 + B2 = µ0nIs + µ0nIs
= 2µ0n ᐧ \(\frac{E}{2 R}\) = \(\frac{\mu_0 n E}{R}\)
ii) In case of parallel combination, terminal potential difference across each solenoid = E.
So, current through each solenoid, Ip = \(\frac{E}{R}\).
The resultant magnetic field along the axis,
B = B1 + B2 = µ0nIp + µ0nIp
= 2µ0n ᐧ \(\frac{E}{R}\) = \(\frac{2 \mu_0 n E}{R}\)
Example 4.
A tong straight solid conductor of radius 5 cm carries a current of 2 A, which is uniformly distributed over its circular cross section. Find the magnetic field at a distance of 3 cm from the axis of the conductor.
Solution:
Let us consider an internal point P at a distance r(= 3 cm) from the axis of the conductor. Imagine a circular path of radius r around the conductor, such that P lies on it. If R is radius of the solid conductor then current enclosed by the circular path,
I1 = \(\frac{I}{\pi R^2} \times \pi r^2\) = \(\frac{I r^2}{R^2}\)
Let B be the magnetic field at point P due to the current carrying conductor. B acts tangentially to the circular path. So according to Ampere’s Circuital law,