- Introduction
- |
- Biot Savart Law
- |
- Comparison between Coulomb’s laws and Biot Savart laws
- |
- Applications of Biot Savart law
- |
- Magnetic Field due to steady current in an infinitely long straight wire
- |
- Force between two long and parallel current carrying conductor
- |
- Magnetic Field along axis of a circular current carrying coil
- |
- Magnetic Field at the center of a current carying arc
- |
- Ampere's circuital law
- |
- Magnetic field of a solenoid
- |
- Magnetic Field of a toriod
- |

- It is experimentally established fact that two current carrying conductors attract each other when the current is in same direction and repel each other when the current are in opposite direction

- Figure below shows two long parallel wires separated by distance d and carrying currents I
_{1}and I_{2}

- Consider fig 5(a) wire A will produce a field B
_{1}at all near by points .The magnitude of B_{1}due to current I_{1}at a distance d i.e. on wire b is

B_{1}=μ_{0}I_{1}/2πd ----(8)

- According to the right hand rule the direction of
**B**is in downward as shown in figure (5a)_{1}

- Consider length l of wire B and the force experienced by it will be (I
_{2}**l**X**B**) whose magnitude is

- Direction of F
_{2}can be determined using vector rule .F_{2}Lies in the plane of the wires and points to the left

- From figure (5) we see that direction of force is towards A if I
_{2}is in same direction as I_{1}fig( 5a) and is away from A if I_{2}is flowing opposite to I_{1}(fig 5b) - Force per unit length of wire B is

- Similarly force per unit length of A due to current in B is

and is directed opposite to the force on B due to A. Thus the force on either conductor is proportional to the product of the current

- We can now make a conclusion that the conductors attract each other if the currents are in the same direction and repel each other if currents are in opposite direction

- Let there be a circular coil of radius R and carrying current I. Let P be any point on the axis of a coil at a distance x from the center and which we have to find the field

- To calculate the field consider a current element Id
**l**at the top of the coil pointing perpendicular towards the reader

- Current element Id
**l**and**r**is the vector joining current element and point P as shown below in the figure

- From Biot Savart law, the magnitude of the magnetic field due to this current element at P is

where Φ is the angle between the length element dl and r

- Since Id
**l**and**r**are perpendicular to each other so Φ=90.Therefore

- Resolving dB into two components we have dBsinθ along the axis of the loop and another one is dBcosθ at right angles to the x-axis

- Since coil is symmetrical about x-axis the contribution d
**B**due to the element on opposite side ( along -y axis ) will be equal in magnitude but opposite in direction and cancel out. Thus we only have dBsinθ component - The resultant B for the complete loop is given by,

B=∫dB

Now from figure 6

sinθ=R/r =R/√(R^{2}+ x^{2}) So

eq

- If the coil has N number of turns then

- Direction of magnetic field at a point on the axis of circular coil is along the axis and its orientation can be obtained by using right hand thumb rule .If the fingers are curled along the current, the stretched thumb will point towards the magnetic field

- Magnetic field will be out of the page for anti-clockwise current and into the page for clockwise direction

- At the center of the coil x=0

so

- In this case R in the denominator can be neglected hence

- For coil having N number of turns

- If the area of the coil is πR
^{2}then

- m=NIA represents the magnetic moment of the current coil. Thus from equation (17) we have

- Consider an arc of radius R carrying current I as shown below in the figure

- According to the Biot Savart law the magnetic field at any point P is given by

Here dl=RdΦ

So

- If l is the length of the arc then

l=RΦ so that

- Equation 19 and 20 gives us magnetic field only at the center of curvature of a circular arc of current

- For semi circular loop put Φ=π in equation 19 and for full circle Φ=2π in equation 19 and calculate to find the result

- If the circular current loop lies on the plane of the paper then magnetic field will be out of the page for anticlockwise current and into the page for clockwise current as shown below in the figure

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
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- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

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