- Introduction
- |
- Biot Savart Law
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- Comparison between Coulomb’s laws and Biot Savart laws
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- Applications of Biot Savart law
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- Magnetic Field due to steady current in an infinitely long straight wire
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- Force between two long and parallel current carrying conductor
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- Magnetic Field along axis of a circular current carrying coil
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- Magnetic Field at the center of a current carying arc
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- Ampere's circuital law
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- Magnetic field of a solenoid
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- Magnetic Field of a toriod
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- In the previous chapter we have defined concept of magnetic field represented by vector
**B**

- We defined magnetic field B in terms of force it exerts on moving charges and on current carrying conductors

- We also know that magnetic field is produced by the motion of the electric charges or electric current

- In this chapter we would study the magnetic field produced by the steady current

- we would study about how various factors affect the magnitude and direction of the magnetic field

- We will also learn to calculate the equation for magnetic field B if the current configuration is known using Biot-savart's law and ampere circuital law

- We know that electric current or moving charges are source of magnetic field

- A Small current carrying conductor of length
**dl**(length element ) carrying current I is a elementary source of magnetic field .The force on another similar conductor can be expressed conveniently in terms of magnetic field**dB**due to the first

- The dependence of magnetic field
**dB**on current I ,on size and orientation of the length element**dl**and on distance**r**was first guessed by Biot and savart

- The magnitude of the magnetic field
**dB**at a distance**r**from a current element**dl**carrying current I is found to be proportional to I ,to the length dl and inversely proportional to the square of the distance |**r**|

- The direction of the magnetic Field is perpendicular to the line element
**dl**as well as radius**r**

- Mathematically, Field
**dB**is written as

Here (μ_{0}/4π) is the proportionality constant such that

μ_{0}/4π=10^{-7}Tesla Meter/Ampere(Tm/A)

- Figure below illustrates the relation between magnetic field and current element

- if in figure, Consider that line element
**dl**and radius vector**r**connecting line element mid point to the field point P at which field is to be found are in the plane of the paper

- From equation (1) ,we expect magnetic field to be perpendicular to both
**dl**and**r**.Thus direction of**dB**is the direction of advance of right hand screw whose axis is perpendicular to the plane formed by**dl**and**r**and which is rotated from**dl**to**r**( right hand screw rule of vector product)

- Thus in figure ,
**dB**at point P is perpendicular directed downwards represented by the symbol (x) and point Q field is directed in upward direction represented by the symbol (•)

- The magnitude of magnetic field is

where θ is the angle between the line element dl and radius vector**r**

- The resultant field at point P due to whole conductor can be found by integrating equation (1) over the length of the conductor i.e.

**B**=∫d**B**

**Relation between permeability (μ**_{0}and permittivity (ε_{0}) of the free space

- We know that

μ_{0}/4π=10^{-7}N/A^{2}----(a)

and

1/4πε_{0}=9*10^{9}N-m^{2}/C^{2}----(b)

Dividing equation a by b we get

μ_{0}ε_{0}=1/(9*10^{16}) (C/Am)^{2}

we know that

1C=1A-s So μ_{0}ε_{0}=1/(3*10^{8}m/s)^{2}

And 3*10^{8}m/s is the speed of the light in free space

So μ_{0}ε_{0}=1/c^{2}

or c=1/√(μ_{0}ε_{0})

Class 12 Maths Class 12 Physics

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