From the study of subatomic particles to the laws of motion, Physics Topics offer insights into the workings of the world around us.
Is Biot-Savart Law and Laplace Law same? What is Magnetic Intensity or Magnetising Field?
Diagrammatic representation of a magnetic field is its representation using magnetic lines of force:
1) the direction of the tangent drawn at any point on a magnetic line of force is the direction of the magnetic field at the point;
2) comparing the number density of magnetic lines of forces at different points in a magnetic field, the field-strengths at those points can be compared. But to define magnetic field precisely as a definite physical quantity, at every point its magnitude should be represented by a number and an associated unit. This is not possible from the concept of magnetic lines of force only.
As a physical quantity, the usual symbol of magnetic field is \(\vec{B}\) (as it is a vector). This vector \(\vec{B}\) is named magnetic field or magnetic induction or magnetic flux density. In the region surrounding a current carrying conductor
- the direction of \(\vec{B}\) is determined by Maxwell’s corkscrew rule and
- the magnitude of \(\vec{B}\) (i.e., B) is determined by Biot-Savart law.
Statement of Biot-Savart Law:
Let δl = a small elementary part of a conducting wire [Fig.]
I = current through the wire
r = distance of any external point of from the element δl
θ = angle between the element \(\delta \vec{l}\) and the position vector \(\vec{r}\) of the external point
δB = magnitude of the magnetic field at that external point.
Then the Biot-Savart law states that,
- δB ∝ δl,
- δB ∝ I,
- δB ∝ \(\frac{1}{r^2}\) and
- δB ∝ sinθ.
This means, δB ∝ \(\frac{I \delta l \sin \theta}{r^2}\) or, δB = \(k \frac{I \delta l \sin \theta}{r^2}\) …… (1)
This law is also called Laplace’s law. The value of the constant k in equation (1) depends on
- the nature of the medium between the conductor and the point under consideration and
- the system of units used for different physical quantities. In this chapter we shall consider only vacuum as the medium.
SI unit: SI unfts of different physical quantities used in equation (1) are : δI and r : metre (m); I : ampere (A); δB: weber/metre2 (Wb ᐧ m-2) or tesla (T).
For the definition of the unit of magnetic field (Wb ᐧ m-2), see Section 1.6.
If these units are used in equation (1) then another constant is traditionally used instead of Á. This constant is µ0 = 4πk,
i.e., k = \(\frac{\mu_0}{4 \pi}\)
So, the usual form of Biot-Savart law in vacuum,
δB = \(\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}\) …… (2)
The constant µ0 is called the permeability of free space.
Unit of µ0: From equation (2),
µ0 = \(\frac{4 \pi r^2 \delta B}{I \delta l \sin \theta}\)
∴ The unit of µ0 = \(\frac{\mathrm{m}^2 \times \mathrm{Wb} \cdot \mathrm{m}^{-2}}{\mathrm{~A} \times \mathrm{m}}\) = wb ᐧ A-1 ᐧ m-1
Here, Wb ᐧ A-1 is also called henry (H). [see the chapter ‘Electromagnetic Induction’]
Hence, the unit of µ0 is henry/metre (H ᐧ m-1).
Value of µ0: µ0 = 4π × 10-7H ᐧ m-1
It is to be noted that, \(\frac{\mu_0}{4 \pi}\) = 10-7H ᐧ m-1
Biot-Sovort low for extended conductors: An extended conductor is assumed to be composed of a number of smaller parts δl1, δl2, … etc. and the Biot-Savart law is applied for each part. At any point, the magnetic field due to the whole conductor will be,
B = ΣδB = \(\frac{\mu_0}{4 \pi} \sum \frac{I \delta l \sin \theta}{r^2}\)
The Σ (summation) sign indicates the sum of a number of terms. \(\vec{B}\) is a vector; so to determine the resultant, the rule of vector addition is applied. Usually this vector addition is very complicated. But in case of symmetrical conductors like straight conducting wire, circular coil, etc., determination of the resultant magnetic field is not so troublesome.
Vector form of Biot-Savort law: In Fig., if the unit vector towards the point P with respect to δl be \(\hat{r}\), then \(\vec{r}\) = r\(\hat{r}\);. Hence, the magnitude of the vector product \((\delta \vec{l} \times \hat{r})\) is,
\(|\vec{l} \times \hat{r}|\) = δl ᐧ 1 ᐧ sinθ [∵ |\(\hat{r}\)| = 1]
Again, the direction of \((\delta \vec{l} \times \hat{r})\) is downward with respect to the plane of the figure, which is actually the direction of the magnetic field as per corkscrew rule. So, the vector form of the equation (2) is,
\(\delta \vec{B}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{I \delta \vec{l} \times \hat{r}}{r^2}\) …… (3)
From equation (3) we get,
\(\vec{B}\) = \(\frac{\mu_0}{4 \pi} \sum \frac{I \delta \vec{l} \times \hat{r}}{r^2}\) = \(\frac{\mu_0}{4 \pi} \sum \frac{I \delta \vec{l} \times \vec{r}}{r^3}\) …… (4)
Usually with respect to the plane of the paper, an upward vector is denoted by sign to mean the tip of an arrow and a downward vector by sign to mean the tail of an arrow. Thus, the direction of the magnetic field at the point P [Fig.] is denoted by sign, and at the point Q, sign is used.
Magnetic intensity or magnetic field strength or magnetising field:
In SI: Magnetic permeability of vacuum is µ0; this value of µ0 is also used for air because in presence of air no remarkable change in the magnetic field is observed. But for any other medium, permeability of that medium is denoted by µ. For different media the magnitude of p is different. In case of any medium, the general form of Biot-Savart law is,
δB = \(\frac{\mu}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}\) ….. (5)
From equations (2) and (5) we see that, the quantity \(\frac{1}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}\) can be treated as the cause of the magnetic field in a medium. In absence of the multiplier µ0 or µ in this expression, this quantity does not depend on the medium. To determine the magnetic field in a medium, this quantity is just multiplied by the magnetic permeability of that medium. This quantity is called magnetic Intensity or magnetising field. It is expressed by H.
From the equation (5) we can write,
Definition: At any point in a medium the ratio of the acting magnetic field and the magnetic permeability of the medium is called the magnetic intensity or magnetising field at that point.
Unit: Unit of H \(=\frac{\text { unit of } B}{\text { unit of } \mu}\) = \(\frac{\mathrm{Wb} \cdot \mathrm{m}^{-2}}{\mathrm{~Wb} \cdot \mathrm{A}^{-1} \cdot \mathrm{m}^{-1}}\)
= A ᐧ m-1 (ampere/metre)
CGS or Gaussian system: In this system \(\vec{H}\), i.e., the magnetic intensity (instead of \(\vec{B}\)) is considered as the primary vector in magnetism. In this case, the unit of magnetic intensity H is oersted (0e) and the unit of magnetic field B is gauss (G). It may be of interest to note that 1 oersted is the same as 1 dyn per unit pole, and 1 gauss is the same as 1 maxwell/cm2 according to the old definition of magnetic intensity.
Relation between SI and CGS units:
1A ᐧ m-1 = 4π × 10-3Oe, 1 Wb ᐧ m-2 = 104G
In CGS system, electromagnetic unit (abbreviated as emu) of current is used as the unit of current and is expressed by the symbol i. This electromagnetic unit is so defined that if the
other quantities in equation (6) be expressed in CGS units, the constant \(\frac{1}{4 \pi}\) can be replaced by 1. In that case, the CGS form of the equation be,
δH = \(\frac{i \delta l \sin \theta}{r^2}\) …….. (8)
So, the different units used in this equation are—
δl and r: cm; i : emu of current; δH : Oe.
This equation (8) indicates the Biot-Savart law or Laplace’s law in the CGS or Gaussian system.
emu of current: The current which, when flowing through a conducting wire of length 1 cm , bent in the form of an arc of a circle of radius 1 cm, produces a magnetic intensity of 1 0e at the centre of the arc, is called 1 electromagnetic unit(emu) of current [Fig.].
The two parts of the wire except the circular arc shown in the figure are kept along the radius of the circle; thus no magnetic field is produced at the centre of the circle due to the current flowing through these two parts.
Definition of the electromagnetic unit of current can be explained from the discussion of magnetic field produced due to current in a circular conductor [Article 1.4.2].
Relation between ampere and emu of current:
1 emu of current = 10 A
Rules of conversion from SI to CGS: The following replacements convert SI expressions into the corresponding CGS expressions:
- Magnetic intensity \(\vec{H}\), in place of magnetic field \(\vec{B}\) ;
- Electric current i in emu, in place of I in amperes;
- The constant 1 in place of \(\frac{\mu_0}{4 \pi}\), i.e., 4π in place of µ0.
Application of Biot-Savart Law (Long Straight Conductor)
Let I = strength of current in an elecrical circuit. AB is a straight conductor and OP = r = perpendicular distance of the point P from the current carrying wire [Fig.(a)].
The two end points of the conducting wire make angles θ1 and θ2 at the point P with respect to OP. In the direction of current, if the angle θ2 is taken as positive then in the opposite direction, θ1 will be negative, i.e., it will be -θ1.
The conducting wire and its adjacent region is shown in Fig.(b) in a magnified form. From a very small part dl of the wire, the distance of the point P is x and the angle between the direction of current and x is α.
So, according to Biot-Savart law, the magnitude of the magnetic field at the point P due to dl is,
dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{I d l \sin \alpha}{x^2}\) …… (1)
From the diagram,
α = 180° – β = 180° – (90° – θ) = 90° + θ
So, sin α = sin(90° + θ) = cosθ …… (2)
Again, l = rtan θ
Since, r = constant, differentiating the equation we get,
dl = rsec2θdθ ….. (3)
Again, cosθ = \(\frac{r}{x}\) or, x2 = \(\frac{r^2}{\cos ^2 \theta}\) = r2sec2θ ….. (4)
Now putting the values of sin a, dl and x2 from equations (2), (3) and (4) into equation (1) we get,
dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot r \sec ^2 \theta d \theta \cdot \cos \theta}{r^2 \sec ^2 \theta}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \cdot \cos \theta d \theta\) ………. (5)
So, for the entire wire AB, magnetic field at the point P is,
This is the expression for the magnitude of magnetic field at the point P due to the current carrying wire.
Special cases:
i) For infinitely long conducting wire: Let the length of the wire below the point O be L1 and above this point, L2. If these two lengths are much greater than r, i.e., r\(\ll\)L1 and r\(\ll\)L2, the wire can be treated as of infinite length with
respect to the point P. In that case, for the lowermost end of the wire, θ1 ≈ 90° and for the uppermost end of the wire θ2 ≈ 90°. So from the equation (6), we can write,
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left(\sin 90^{\circ}+\sin 90^{\circ}\right)\)
∴ B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\) …….. (7)
ii) For semi-Infinitely long conducting wire: Let the lower most end of the wire be O and the length of the wire = L. If r \(\ll\)L, the wire can be called semi-infinitely long with respect to the point P. In that case, θ1 = 0° and θ2 = 90°.
So, from the equation (6), we can write,
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\) (sin 0° + sin 90°)
∴ B = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\) …. (8)
iii) For a point on the extended part of the wire: If point P1 be located according to Fig.(a), both θ1 and θ2 will be negative with respect to r which is the perpendicular distance between point P1 and extension of the wire AB.
So, the magnetic field at P1,
B1 = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left[\sin \theta_1+\sin \left(-\theta_2\right)\right]\)
[taking θ2 negative in equation (6)]
= \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left(\sin \theta_1-\sin \theta_2\right)\) …. (9)
If we consider the point P2 to lie on the extension of the wire AB [Fig.(b)], then θ1 = 90° and θ2 = 90°
Putting these in equation (9) we get, the magnetic field at the point P2,
B2 = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\)[sin 90° – sin 9o°] = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}[1-1]\) = 0
So, at any point lying along the length of a current carrying conductor, the magnetic field due to that conductor will be zero.
Expressions in CGS or Gaussian system: In the above mentioned equations (6), (7) and (8), if we substitute B → H, I → i and µ0 → 4π then we get:
Magnetic intensity due to a straight conductor,
H = \(\frac{i}{r}\)(sin θ1 + sin θ2)
Magnetic intensity due to an infinitely long conductor,
H = \(\frac{2 i}{r}\)
Magnetic intensity due to a semi-infinitely long conductor,
H = \(\frac{i}{r}\)
In each of the following examples to determine the direction of the magnetic field Maxwell’s corkscrew rule has been used.
Numerical Examples
Example 1.
The distance between two long straight conductors is 5 m. Currents 2.5 A and 5 A are flowing through them in the same direction. What will be the magnetic field at the mid-point between them? [CBSE 2000]
Solution:
According to the corkscrew rule, the magnetic fields at the mid-point due to the two conductors will be opposite in directions.
So, the relation, B = \(\frac{\mu_0}{4 \pi}, \frac{2 I}{r}\), gives magnetic field at the mid-point due to the first conductor,
B1 = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{\frac{r}{2}}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{4 I_1}{r}\)
and magnetic field at the mid-point due to the second conductor,
Example 2.
5 A current is flowing in mutually opposite directions through each of two parallel straight conducting wires kept 0.2 m apart [Fig.]. Determine the magnitudes and directions of magnetic field at the points P, Q and R lying on the plane containing the two wires.
Solution:
Magnetic field at P due to the first wire,
B1 = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \times 5}{0.1}\)(upward)
Magnetic field at P due to the second wire,
B2 = \(\frac{\mu_0}{4 \pi}, \frac{2 \times 5}{0.3}\) (downward)
Since, B1 > B2, the resultant magnetic field will be upward from the plane.
∴ Bp = B1 – B2
= \(\frac{\mu_0}{4 \pi}(2 \times 5)\left(\frac{1}{0.1}-\frac{1}{0.3}\right)\)
= 10-7 × 10 10 × \(\frac{2}{3}\)
= 6.67 × 10-6Wb ᐧ m-2
Similarly, upward magnetic field at the point Q,
BQ = 6.67 × 10-6 Wb ᐧ m-2
At the point R, magnetic field due to the first wire as well as for the second wire is downward. Hence, the resultant magnetic field will be the sum of these two fields.
So, BR = 2B1 = 2 × \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \times 5}{0.1}\)
= 2 × 10-7 × 10 × 10
= 2 × 10-5 Wb ᐧ m-2
Example 3.
An infinitely long conducting wire POQ is bent through right angles at O [Fig.]. If a current I is sent through this bent wire, what will be the magnitude of the mag-netic field at the point A at a distance r from each p part of the wire?
Solution:
From the Fig., AM = AN = r So, ∠OAM = ∠OAN = 45°
Now, the magnetic field at A due to PO is equal to that due to OQ both in magnitude and direction (downwards).
Again, the points P and Q make angles at A relative to AM and AN. For infinitely long wires, each of these angles is close to 90°.
So the resultant magnetic field at A is
B = 2 × \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\) (sin90° + sin45°)
= \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}(2+\sqrt{2})\) (directed downwards)
Example 4.
Through each of two wires POQ and P’O’Q’ an electric current I is passing [Fig.]. The points Q, O, O’ and Q’ are collinear. Determine the magnetic field at the midpoint A of OO’.
Solution:
The point A lies along the length of the two parts OQ and O’ Q’. Hence, no magnetic field exists at A due to these two parts.
The two parts PO and P’O’ are semi-infinite conducting wires with respect to the point A. Hence, for each part, the magnetic field at A = \(\frac{\mu_0}{4 \pi}, \frac{I}{r}\); and these two fields are downwards.
Hence, the downward magnetic field at the point A,
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}+\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\)
Example 5.
5 A current is flowing through a long straight conducting wire. What is the magnitude of magnetic field at a distance 10 cm from the wire?
Solution:
We know, B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\) [for a straight infinite wire]
Here, I = 5 A, r = 10 cm = 0.1m, µ0 = 4π × 10-7 H ᐧ m-1
∴ B = \(\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \times 5}{0.1}\) = 10-5Wb ᐧ m-2