Sir , please explain the solved example,I did not understand it properly

Asked by modi72879 | 1st Dec, 2017, 11:30: AM

Expert Answer:

As per Amphere law begin mathsize 12px style contour integral B. d l space equals space mu subscript 0 space I end style . Magnetic field summed up in a closed loop equals the current passing through the conductor which is enclosed by the loop.
 
In the problem it is asked to find the magnetic field due to current carrying conductor in the region (1) outside the conductor and (2) inside the conductor.
 
First part is very simple. consider an ampherian circular loop of radius outside the condutor. here the loop radius is grater than condutor radius. the line integral for the closed loop is 2πr if the radius of loop is r ( r>a, a is the conductor radius).
 
Hence B×2πr = μ0 I 
 
begin mathsize 12px style B space equals space fraction numerator mu subscript 0 I over denominator 2 pi r end fraction end style
 
In the second part it is required to get magnetic field at a point inside the conductor. If we draw the circular ampherian loop with radial distance equal to the distance from conductor central axis to the point of intrest, then the enclosed current is proportional to loop area. Current is uniformly distributed across the crosssection of the conductor whose cross section is circular and radius is a. 
 
Hence the current density J is given by :-       J = I / (πa2). 

Hence the enclosed current in the ampherian loop of radius r is

begin mathsize 12px style I subscript r equals space fraction numerator I over denominator pi a squared end fraction cross times pi r squared space equals space r squared over a squared I end style
Hence magnetic field is 
begin mathsize 12px style B subscript r equals space fraction numerator mu subscript 0 I subscript r over denominator 2 pi r end fraction space equals space fraction numerator mu subscript 0 over denominator 2 pi r end fraction r squared over a squared I space equals space fraction numerator mu subscript 0 I space r over denominator 2 pi a squared end fraction end style

Answered by  | 8th Dec, 2017, 03:20: PM

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