Hi the given problem here, this is cross sectional view. So far screen, solid cylindrical current caring conductor was radius. It's given us capital R is equal to 10.0 centimetre. Then in order to find magnetic field at any point within this conductor, supposed that observation point at which we have to find magnetic will suppose that point lies actually stands smaller from the center of this conductor. So first of all, we will find current density. That current density J is given as current passing through the conductor or particular to its area of cross section or unit area of cross section. No, if this is the absolution point, we will consider an empty eerie in lieu passing through this observation point and having a radius R. So three years of imperial loop will be equal to dispense of observation point from the center conductor is here in this general case that dispenses are so area of imperial loop. That will be buy into our square. So finally, but I didn't passing through imperial luke will be given by let it be I dash and that I dish will be equal to current density multiplied by area of the emperor liu. There this is I buy bye arty square into by small artist square canceling this pie. Finally the current linked through the emperor liu will be given by I. R. Squared by capital artist square. Well no using um peer circuit A law to find the magnetic field at that observation point line integral of magnetic field at that point will be given by mu note times the current wearing the imperial blue. So as A B means magnetic field at that point will be constant at all. The points of imperial loop. So it can be taken as a constant out. Then remaining means the integration of deal will be equal to circumference of imperial loop needs to buy our so it will be new not times the I dash which is I into our square by ari square canceling this small. Are finally expression for magnetic field at any point within this solid conductor will be given by um You not by two pi into I. R. By Are square. Or if we multiply it by two x 2. It may also be given as new. Not by four by Into two i. R. Invited by are the square Now we can find a magnetic fields at all. The given points in this problem. Most of all we have to find a magnetic field At our .8. Here the distance of point is zero means we have to find magnetic bill at the center. So this be a means magnetic field at this point will come out with zero. Then in the second case this is Observation Point B whose distance is 4.00 centimetres. So this be its magnetic field at this point B will be given by Yeah, the values are Our vic was to 4.00 cm Radius of the conductor. We have already seen it to be 10 centimetre and current passing through the conductor wars 1.35 M. P. R. So this BB will be given by For me, not upon four by the suspended for -7 into Two times of high into armies between 2 1.35 multiplied by Three years of the imperial which is four centim or foreign to. And it's part -2 meter divided by the radius of the solid conductor square of the radius means this is turned into and it's about minus the square. So finally this be at the observation point here comes out to be 1.08 Into 10. To the power -6 Tesla. Then in the third case, this time the observation point R. C. This is Equal to and zero Centim is equal to the radius of conductor. Is the observation point is at the surface of the conductor. So this magnetic will be C will come out to be Again using the same formula. Um you're not upon four pi into this time, this is capital, are in place of smaller, divided by our square. So canceling this, Finally putting the known values. This is standard for -7 into two into 1.35, invited by our which is 10 cm or turn into 10 for -2 m. So finally this magnetic will now at the surface of the conductor comes out to be 2.7 Into Standish Power -6 Islam. one more answer for this given problem. Until now we have obtained three answers. No. In the 4th case. Now the observation point is outside the street conductor. That distance is given as 4.2 d. This is 16.0 centimetre and we do at any point outside the conductor of the magnetic field is simply given by You, not upon 45 into two, I by our only. So this is tending bar -7 into into 1.35, invited by Our which is 16 cm or 16 to 10 days per -2 m. So finally, this magnetic field here comes out to be 1.72 And it's AR -6 islam. The 4th answer for the same problem, and the direction of all these magnetic fields is counter clockwise using right and and rule using right hand rule, which is a rule to find the direction of magnetic field around the street current carrying conductor directions of all the fields. His counter clockwise around this even conduct. Thank you.