# Biot-Savart Law

The Biot-Savart equation expresses the magnetic field created by a current-carrying wire. This conductor or wire is represented as a vector quantity called the current element. Lets take a look at the law and formula of biot-savart law in detail,

### Biot-Savart Law

The magnitude of magnetic induction at a place caused by a tiny element of a current-carrying conductor is stated in the law.

- Directly proportional to the current
- Directly proportional to the length of the element
- Directly proportional to the sine of the angle between the element and the line joining the center of the element to the point and,
- Inversely proportional to the square of the distance of the point from the center of the element

dB α I

dB α dl

dB α sinΘ

dB α 1 / r

^{2}∴ dB α IdlsinΘ / r

^{2}∴ dB = KIdlsinΘ / r

^{2}(K is constant)

**Derivation**

Consider a conductor of any form carrying a current (I) and a tiny length element (dl) (refer to below image). The current flow is depicted vertically upward.

Let (P) be any point a distance (r) from the current-carrying element, and (r) be the position vector of (P) with respect to the current element, and be the angle between (dl) and (r), in the direction of the current.

According to Biot – Savart Law, the magnetic induction at point (P) is given by,

dB = KIdlsinΘ / r^{2}⇢ (1)

Here, K is Constant and its value depends on the system of units and also the medium in which the conductor is situated.

In the SI system, the constant K for vacuum or air is written as [μ_{o} / 4π] where μ_{0} is called the permeability of vacuum or free space.

Substituting K in Eq (1) We get,

dB = μ

_{0}/ 4π [IdlsinΘ / r^{2}] ⇢ (2)SI unit of μ

_{0 }Wb / Am. It`s value is 4π X 10^{-7}Wb / Am(μ

_{0}/ 4π = 10^{-7}Wb / Am)

The direction of magnetic induction is perpendicular to the plane of the figure and directed inside the plane. (as per the right-hand Thumb rule.)

**In vector form,**

Where [r] is the vector drawn from the center of the element to the point and [dl] is the length of the element, in the direction of the current. The total magnetic induction at a point P due to the entire conductor is found by adding the contribution of all such element [dl] and is expressed as,

Summation is replaced by ∫

Now, The magnitude of the magnetic induction due to an infinitely long and straight conductor, carrying a current I, at a point at a distance ‘a’ from the conductor is given by,

**Applications of Biot – savart law**

Following are some importance of the Biot-Savart law:

- We may utilize Biot–Savart law to determine magnetic responses at the atomic or molecule level.
- It may be utilized in aerodynamic theory to determine the velocity promoted by vortex lines.
- This rule may be used to compute the magnetic field produced by a current element.

**Importance of Biot-Savart Law**

Following are some importance of the Biot-Savart law:

- In electrostatics, the Biot-Savart law is analogous to Coulomb’s law.
- The legislation also applies to extremely tiny conductors that convey current.

### Conceptual Problems

**Question 1: State Biot – savart’s law and write it in vector form.**

**Answer:**

Law statement: The Magnitude of magnetic induction at a point due to a small element of a current-carrying conductor is, directly proportional to the current, directly proportional to the length of the element, directly proportional to the sine of the angle between the element and the line joining the center of the element to the point and, inversely proportional to the square of the distance of the point from the center of the element.

Boit savart Law in vector form is given by:

**Question 2: State any two applications of Biot savart’s law.**

**Answer:**

Applications of Biot savart`s law are:

- To calculate magnetic responses at the atomic or molecular level, we can use Biot–Savart law, and
- It may be used in aerodynamic theory to calculate the velocity induced by vortex lines.

**Sample Problems**

**Question 1: A current of 5A is flowing from south to north in a wire kept along the north-south direction. Find the magnetic field due to a 1cm piece of wire at a point 2m northeast from the piece of wire.**

**Solution:**

Given Data: I = 5A

r = 2m

Θ = 45°

dl = 1cm = 1 × 10

^{-2}mTo find: Magnetic field (dB)

Formula: dB = μ

_{0}/ 4π [IdlsinΘ / r^{2}]Calculation: From formula,

dB = 10

^{-7}× 6 × 10^{-2}× sin45° / 2^{2}= 8.84 × 10

^{-10}T

The Magnetic field due to wire at point is 8.84×10^{-10}.

**Question 2: A long straight wire carries a current of 35A. What is the magnetic field B at a point 20cm from the wire?**

**Solution:**

Given Data: I = 35A

a = 30cm = 0.3m

To find: The magnitude field (B)

Formula: B = μ

_{0 }/ 2π (I / a)Calculation: From formula,

B = 4π × 10

^{-7}/ 2π (35 / 0.3)= 2 × 10

^{-7}(116.66)B = 233.32 × 10

^{-7 }T

The magnetic field at point 30cm is 233.32×10^{-7}T.

**Question 3: A long straight wire in the horizontal plane carries a current of 50A in the north-south direction. Give the magnitude and direction of B at a point 3m east of the wire.**

**Solution:**

Given Data: I = 50A

a = 3m

To find: Magnitude and the direction of B

Formula: B = μ

_{0}I / 2πaCalculation: From Formula,

B = 4π × 10

^{-7}× 50 / 2π × 3= 2 × 10

^{-7}× 50 / 3B = 3.333 × 10

^{-6}TIt acts in the vertically upward direction, so direction is downward.

“Note: The direction of the magnetic field is always in a plane perpendicular to the line of element and position vector.”

The magnitude of B is 4×10^{-6}T and the direction is vertically downward.

**Question 4: Two long straight parallel wires in vacuum are 4m apart and carry currents 6A and 8A in the same direction. Find the neutral point, i.e., the point at which the resultant magnetic induction is zero.**

**Solution:**

Given Data: I

_{1}= 4AI

_{2}= 8Ar = 4m

To find: neutral point ( the point at which resultant magnetic induction is zero )

Formula: (i) B

_{1}= μ_{0}/ 4π (2I_{1}/ r_{1})(ii) B

_{2}= μ_{0}/ 4π (2I_{2}/ r_{2})The current flowing through the wire is all going in the same direction. As a result, at any location between the two wires, the two magnetic inductions B

_{1}and B_{2}will have opposing orientations. As a result, the point must be located between the two wires.For the resultant magnetic induction to be zero, we must have B = B

_{2}Let the corresponding point (The neutral point) be at distance r

_{1}from the first wire and r_{2}from the second wire.Calculation:

B = μ

_{0}/ 4π ( 2I_{1}/ r_{1}) and B_{2}= μ_{0 }/ 4π ( 2I_{2}/ r_{2})Now, B

_{1}= B_{2}∴ I

_{1}/ r_{1}= I_{2}/ r_{2}∴ r

_{2}/ r_{1}= I_{2}/ I_{1}∴ r

_{2}/ r_{1}= 8A / 4A= 2

∴ r

_{2}= 2r_{1}But, r

_{1}+ r_{2}= r∴ r

_{1}+ 2r_{1}= 4m∴ r

_{1}= 4/3m and r_{2}= 8/3m

The neutral point lies at a distance of 4/3m from the wire carrying a current of 4A.

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