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The biot savart law formula can be given as:

Hence, dB ∝ \[\frac{Idlsin\theta}{r^{2}}\] or dB =\[k\frac{Idlsin\theta}{r^{2}}\]

Where, k is constant, depending upon the magnetic properties of medium and system of the units employed. In the SI system of the unit,

k = \[\frac{\mu_{0}\mu_{r}}{4\pi}\]

The final Biot-Savart law derivation is expressed as,

dB = \[\frac{\mu_{0}\mu_{r}}{4\pi}\] × \[\frac{Idlsin\theta}{r^{2}}\]

(image will be uploaded soon)

Consider a long wire carrying current I and at a point p in space. The wire is represented in the picture below by red color. Also, consider an infinitely small length of the wire dl at a distance r from the point P as shown in the diagram. Here, r is the distance vector, which makes the angle θ with the direction of the current in the minute portion of the wire.

Imagining the condition, you can understand that the magnetic field density at point P due to the minute length dl of the wire is proportional to the current carried by this portion of the wire.

As the current through the minute length of the wire is similar to the current carried by the whole wire itself, we can express it as:

dB ∝ I

It is also natural to think that the magnetic field density at point P due to the minute length dl of wire is inversely proportional to the square of the straight distance from Point P to the center of dl. Mathematically it can be expressed as:

dB ∝ \[\frac{1}{r^{2}}\]

Also, the magnetic field density at the point P due to the minute portion of the wire is proportional to the actual length of the minute length dl of wire.

θ is the angle between distance vector r and direction of current through this minute portion of the wire, the component dl of the wire facing perpendicular to the point P is dlsinθ.

Hence, dB ∝ dl sinθ

Now, merging these three statements, we can write,

dB ∝ \[\frac{I.dl.sin\theta}{r^{2}}\]

The formula for Biot-savart law can be stated here as follows:

Now, introducing the value of constant k (which we have already mentioned at the starting of this derivation) in the above formula, we get,

dB ∝ \[k\frac{I.dl.sin\theta}{r^{2}}\]

⇒ dB = \[\frac{\mu_{0}\mu_{r}}{4\pi}\]

Here, μ0 is the absolute permeability of air or vacuum used in the expression of constant k, and its value is 4π10-7 Wb/ A-m in the SI system of units. μr in the expression of constant k is relative permeability of the medium.

Now, flux density (B) at point P due to total length of the current-carrying conductor or wire can be expressed as,

B = \[\int dB\] = dB = \[\int \frac{\mu_{0}\mu_{r}}{4\pi}\times \frac{Idlsin\theta}{r^{2}}\] = \[\frac{\mu_{0}\mu_{r}}{4\pi}\] \[\int \frac{sin\theta}{r^{2}}dl\]

If D is the perpendicular distance of point P from the wire, then

\[rsin\theta\] = D or r = \[\frac{D}{sin\theta}\]

Now, the formula of flux density B at point P can be rewritten as,

B = \[\frac{I\mu_{0}\mu_{r}}{4\pi}\] \[\int \frac{sin\theta}{r^{2}}dl\] = \[\frac{I\mu_{0}\mu_{r}}{4\pi}\] \[\int \frac{sin^{3}\theta}{D^{2}}dl\]

Again, \[\frac{l}{D}\] = \[cot\theta\] ⇒l = D \[cot\theta\]

Therefore, dl = -D\[cosec^{2}\theta d\theta\]

Finally, the expression of B comes as,

B = \[\frac{I\mu_{0}\mu_{r}}{4\pi}\] \[\int \frac{sin^{3}\theta}{D^{2}}[-Dcosec^{2}\theta d\theta\]

= - \[\frac{I\mu_{0}\mu_{r}}{4\pi D}\] \[\int sin^{3}\theta cosec^{2}\theta d\theta\]

= - \[\frac{I\mu_{0}\mu_{r}}{4\pi D}\] \[\int sin\theta d\theta\]

This angle θ is dependent upon the length of the wire and position of point P. Say for a certain partial length of the wire, angle θ, as indicated in the figure above, varies from θ1 to θ2. Hence, magnetic flux density at the point P due to the total length of the conductor is,

B = - \[\frac{I\mu_{0}\mu_{r}}{4\pi D}\] \[\int_{\theta 1}^{\theta 2} sin\theta d\theta\]

= - \[\frac{I\mu_{0}\mu_{r}}{4\pi D}\] \[[-cos\theta]_{\theta 1}^{\theta 2}\]

= \[\frac{I\mu_{0}\mu_{r}}{4\pi D}\] \[[\cos\theta_{1}-cos\theta_{2}]\]

Let's consider the wire is infinitely long, then θ will change from 0 to π that is θ_{1} = 0 to θ_{2} = π.

Placing these two values in the above final expression of Biot Savart law, we get,

B = \[\frac{I\mu_{0}\mu_{r}}{4\pi D}\] \[[\cos0-cos\pi]\] = \[\frac{I\mu_{0}\mu_{r}}{4\pi D}\] [1 - (-1)]

FAQ (Frequently Asked Questions)

1. Give the Applications of Biot-Savart’s Law.

Ans- Some of the application of Biot Savart law is given below.

Used to calculate the magnetic response at atomic or molecular level.

Used to calculate the velocity induced by the vortex line in aerodynamic theory.

2. What is the Difference Between Ampere's Law and Biot-Savart's Law?

Ans- Their difference is similar to that of the difference between Coulomb's law and Gauss's law.

Both Coulomb's law and Gauss's law help us in establishing charge distribution. Still, Gauss's law is simpler and more convenient because it utilizes symmetry, e.g., spherical, cylindrical, cuboidal, etc.

Also, both BiotSavart law and Ampere's law support getting magnetic field distributions, however, Ampere's law takes symmetry into account as it’s a closed line integral (Amperian loop). Both laws can be utilized to calculate the net magnetic field generated at a point by various distributions of current.

Sometimes biot-savart law can be very complicated and may require computation to resolve the equations. In case of asymmetric distribution methods, Ampere’s law is highly recommended to use.

3. Do Maxwell's Equations Imply the Biot-Savart Law?

Ans- Yes, Maxwell’s equation implies Biot Savart law. Considering parallel with the Coulomb’s and Gauss law, a slight increase in the complexity we can get the Biot Savart law from Ampere's law.

As we're studying with magnetostatics here, so Ampere's Law simply reads

∇×B=μ_{0}J∇×B=μ_{0}J

We know that the vector capability is associated with the magnetic field by B=∇×AB=∇×A. Replacing this and using a famous vector calculus identity, we get

∇(∇⋅A)−∇2A=μ_{0}J∇(∇⋅A)−∇2A=μ_{0}J

Now, using the gauge freedom coupled with potentials, and choose the Coulomb gauge, i.e., ∇⋅A=0∇⋅A=0. We end up with ∇2A=−μ_{0}J∇2A=−μ_{0}J.

This is similar to Laplace's equation in electrostatics. It indicates that the solution for AA is similar to the electrostatic potential solution given some charge density distribution. This means that

A(r)=−μ_{0}4π∭J(r′)|r−r′|dr′A(r)=−μ_{0}4π∭J(r′)|r−r′|dr′

To obtain the magnetic field, we consider the curl on both sides. Consider the curl is with respect to r, not r'! the final result of all this (you'll need another identity here) is

B=μ_{0}4π∭J(r′)×r−r′|r−r′|3dr′B=μ_{0}4π∭J(r′)×r−r′|r−r′|3dr′

That is the Biot-Savart law.