A cylindrical wire of a linear magnetic material carries a...

A cylindrical wire of a linear magnetic material carries a total current $I$ with the current density proportional to $\rho$. It has a total radius $R$. a) Find $\vec{B}$ for $\rho > R$ and $\rho < R$. b) Find the magnetization currents. c) Confirm $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$ for a circular contour with $\rho_c < R$ and for one with $\rho_c > R$.

Transcript

00:01 Hi in the given problem here this is a cross section of a given co -agial cable.

00:13 This is the inner solid conductor then the outer hollow cylindrical conductor the inner solid conductor the inner solid conductor is having a radius r1 then the inner radius of outer hollow cylindrical conductor is r2 and its outer radius is r3 the current in this conductor in the solid conductor is getting into the plane of paper and the same current is coming out through this outer cylinder and the magnitude of this current is given as i so the only known terms are current i radius of inner solid conductor r1 the inner radius of outer hollow conductor is r2 and its outer radius is r3 now we have to find the magnetic field at various points so for which first of all the observation point is here at any point within the inner solid conductor cylindrical conductor and the observation point is at a distance r so to find the magnetic field at this point first of all we find the current density of the inner solid conductor current density here is given as j is equal to current i divided by this pi r 1 square so when we make an imperial loop passing through this inner point the circumference of that imperial loop will be pi 2 pi into r and its area will be pi r squared so the current passing through the imperial loop will become i -dash is equal to current density j multiplied by this area pi r square so it is given as i by pi r one square multiplied by r square pi r square so this pi got cancelled so the current passing through this ampereal loop of radius smaller is given by i multiplied by r squared divided by r one square.

03:12 So in the first part of the problem, this is the first part of the problem, using amper circuit law, amper circuit law, b .d .l is equal to mu not times i -dash, or we can say this is mu -not times i into r squared divided by r1 square, and for b .d .l it becomes b, multiplied by the circumference of this amperian loop 2 pi r is equal to mu not i r squared divided by r1 squared so finally the expression for this magnetic field comes out to be mu not by 2 pi and this r also got cancelled so mu not by 2 pi into i r r by r one square and it becomes the answer for the first part of the problem now in the second part of the problem the observation point is here which is in the empty region between the inner solid conductor and outer hollow conductor so if we make an amperian loop passing through this point here this ampere in loop will contain whole of the current i capital i so simply applying amper circuit allow here b dot d l this equals to mu not times the total current passing through the inner solid conductor hence it becomes b multiplied by 2 pi r the distance of that observation point is equal to mu not i so the magnetic field here comes out to be mu not by 2 pi into i by r and it becomes the answer for the second part of the problem now for the third part of the problem the observation point is somewhere here within the outer hollow conductor having the same distance as smaller r so the so now the amperial loop is this the outer red circle for which again we will have to find the current passing through this smaller region shaded in red so to find the current passing through this red shaded region first of all we will find the current density of outer cylinder which comes out to be total current divided by the area and this area comes out to be pi r3 square minus r2 square so the current passing through that shaded region i dash is equal to current density into area of that region which is pi r square minus r2 squared so this is i by pi r3 squared minus r2 squared.

06:37 Multiplied by pi r square minus r2 square so this current i dash here becomes canceling this pi here this is i into r square minus r2 squared divided by r3 square minus r2 square and there is one more current as this current is coming out and this i current in the inner solid conductor that is getting in so the net current through this ampereal loop becomes i double dash which is equal to i -dash minus i.

07:20 So the net current using the value of this net current and applying ampir circuit law b .dl becomes mu not times the i double dash...

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