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Moving Charges And Magnetism

Question
CBSEENPH12038987

(a) Using Biot-Savart’s law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop.

(b) What does a toroid consist of? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside and exterior to the toroid is zero.

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 fraction ; theta 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                                       ... (1) 
The direction of delta B with italic 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 space equals space italic 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 italic space equals space 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 italic space
Error converting from MathML to accessible text.
therefore space delta B subscript x equals space fraction numerator mu subscript o I space delta l space over denominator 4 pi r squared end fraction. R over r equals fraction numerator mu subscript o I R over denominator 4 pi r cubed end fraction delta l space
therefore space delta B subscript x 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 delta l 

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
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

Therefore, 
B 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 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

b) A long solenoid on bending in the form of closed ring is called a toroidal solenoid.

 

i) For points inside the core of a toroid,

 As per Ampere’s circuital law, 
Error converting from MathML to accessible text. 
where, I is the current in the solenoid.

 So, resultant net current = NI
therefore space contour integral space straight B with rightwards harpoon with barb upwards on top space. space dl with rightwards harpoon with barb upwards on top space equals space straight mu subscript straight o space NI space

rightwards double arrow space open vertical bar straight B close vertical bar 2 straight pi space straight r space equals space straight mu subscript straight o space NI space

rightwards double arrow space space open vertical bar straight B close vertical bar space equals space fraction numerator straight mu subscript straight o space NI space over denominator 2 πr end fraction

space space space space space space space space space space space space space space equals straight mu subscript straight o space straight n space straight I space space space space space space space open square brackets because space straight n space equals space fraction numerator straight N over denominator 2 πr end fraction close square brackets 

Since no current is flowing through the points in the open space inside the toroid.

Therefore, I = 0.
So, 
text ∮ end text B with rightwards harpoon with barb upwards on top space. space stack d l with rightwards arrow on top space equals space mu subscript o space I space equals space 0 space

rightwards double arrow space open vertical bar B close vertical bar subscript i n s i d e end subscript space equals space 0 space