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Moving Charges And Magnetism

Question
CBSEENPH12039120

State Biot-Savart law, giving the mathematical expression for it.

Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis.

How does a circular loop carrying current behave as a magnet? 

Solution

Biot-Savart law states that the magnetic field strength (dB) produced due to a current element of current I and length dl at a point having position vector  to current element is given by,
                                       stack d B with rightwards harpoon with barb upwards on top space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space stack d l with rightwards harpoon with barb upwards on top space cross times space r with rightwards harpoon with barb upwards on top over denominator r cubed end fraction 

where, mu subscript o  is permeability of free space.

The magnitude of magnetic field is given by, 
d B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space d l space sin straight space theta over denominator r squared end fractiontheta spaceis the angle between the current element and position vector.

Magnetic field at the axis of a circular loop:

Consider a circular loop of radius R carrying current I. Let, P be a point on the axis of the circular loop at a distance x from its centre O.
Let,  be a small current element at point A. 
                            

Magnitude of magnetic induction dB at point P due to this current element is given by,
 delta B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space delta l space sin alpha over denominator r squared end fraction italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space... space left parenthesis 1 right parenthesis

The direction of stack delta B with rightwards harpoon with barb upwards on top is perpendicular to the plane containing stack delta l with rightwards harpoon with barb upwards on top space a n d space r with rightwards harpoon with barb upwards on top
Angle between stack delta l with rightwards harpoon with barb upwards on top space a n d space r with rightwards harpoon with barb upwards on top space i s space 90 to the power of o

Therefore, 
stack delta B with rightwards harpoon with barb upwards on top space equals space fraction numerator mu subscript o I over denominator 4 pi end fraction fraction numerator delta l space sin 90 to the power of 0 over denominator r squared end fraction equals fraction numerator mu subscript o I delta l space over denominator 4 pi r squared end fraction                           ... (2)
The magnetic induction stack delta B with rightwards harpoon with barb upwards on top can also be resolved into two components, PM and PN’ along the axis and perpendicular to the axis respectively. Thus if we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to the axis will be added up. 
Thus, resultant magnetic induction B with rightwards harpoon with barb upwards on top at axial point P is given by, 
T h e space c o m p o n e n t space o f space stack delta B with rightwards harpoon with barb upwards on top space a l o n g space t h e space a x i s comma space delta B subscript x space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space delta l space space s i n alpha over denominator r squared end fraction
Error converting from MathML to accessible text. 

Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by, 

B = contour integral fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction. d l space equals space fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction contour integral d l space
contour integral d l space equals l e n g t h space o f space t h e space l o o p space equals space 2 pi R

italic space T h e r e f o r e italic comma italic space
B space equals space fraction numerator mu subscript o I R over denominator italic 4 pi italic left parenthesis R to the power of italic 2 italic plus x to the power of italic 2 italic right parenthesis to the power of italic 3 italic divided by italic 2 end exponent end fraction left parenthesis 2 pi R right parenthesis
space space space equals space fraction numerator mu subscript o I R squared over denominator 2 left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction space T e s l a
If the coil contains N turns, then B equals space fraction numerator mu subscript o N I R squared over denominator 2 left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction space T e s l a is the required magnetic field.
A magnetic needle placed at the center and axis of a circular coil shows deflection. This implies that a circular coil behaves as a magnet.