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Question 3 (option): 6) A very long line of charge with charge density = 20 nC/em lies along the s-axis in free space. Draw (schematically) the E-field due to this line of charge. Explain what principles you used to sketch the E-field and comment on the praph. Calculate the electric field strength and displacement field at a distance d 50 cm from the line Comment on any assumptions and approximations used. A very long conducting cylinder cames a curent that is uniformly distributed across its circular cross-section. The radius of the cylinder is R = 1 cm and the current density it carries is J = 0.5 mA/cm². The relative permeability of the cylinder is = 1 and the relative permeability of the medium outside the cylinder is = 2. The cylinder is shown in Figure 4. a. Calculate and sketch the magnitude of the magnetic field (B and H) inside and outside this cylinder as a function of distance r counted from the centre of the cylinder (see Figure 4). b. What field, B or H, experiences a discontinuity on the surface of the cylinder and why? What is the magnitude of this discontinuity? J-0mAcm

Answer 1

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(a) is Electric field due to infinetely long linearly Charge wire. =) linear Charge density (X) = q elm l We use gauss's law to denieve the electric field 1 z 1 - Gaussian swe face (cylinder) de l Electric field 1(E) has direction • radially outward from the charge at point a 2 VA2 L Total flux associated with the gaussian surface his summation of all 3 Sloefaces fluxes line 1,2 L & 3. 10 = a + 2 + 13 S8, 2027 Thas perpendicular = 0 + 0 + $ Ž.cs Toinection with 10 = 7 8 d3 Electric field E ds Cosgo =0 0 = Z X 2T 8 -0 s care ved surface - Area = 27Tre

= 0.5 mA/cm2 Radius R=1cm Top view (9) Drawing the top view of the cylinder field - for magnetic inside the cylinder be op view - circui tal Applying ampere. F - ( Is is flowing law along circular loop off e into into the paper) radius. _6 H dị - Lenclosed Hi gode = J. IT ² ST is the wafaith | Is is Plowng. h 2778 = F Saga and that is enclosed by loop. a is tangential direction. 2 With the of R points Curling help of right hand thumb rule the direction field is determined that is the thumb in the direction of current and fingers towards the direction of magnetic

and we know the relation I B = ur so, B. = My Mahi Mp=1 . For magnetic field outside the cylinder (ko) again considering a circular loop of radius 8 which is outside the cylinder and applying ampere circuital law along the circular loop. 3 hodľ s. IS. ITR² L Is is flowing with TR² area Conly 1 ) ho 2F = Js HR² Fe = Js RP a direction vermering at the same. and Bo= uho 80 Bo = 2Mo J. R² 20 T a Mo J. R2

- because LB field will experiences on the surface of the contain the term a discontinuity Cylinder u in its expression le 1 to 2 Lis 2. and at the sweface My Changes from therefore the magnitude of discontinuity