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4. A long coaxial cable, crossection of which is shown in the figure, is
made up of a cylindrical inner wire of radius a and a cylindrical shell
shaped coaxial outer wire of inner radius b and outer radius c. The
r2
current density with magnitude J = K- is spread over the cross section
of the inner wire where the current has out-of-page direction. Outer wire
carries uniformly distributed current i = tka³ having into-the-page
direction.
a) Find the total current of the inner wire iin in terms of i.
Find expressions for the magnitude of magnetic field,
b) inside the inner wire (r < a)
c) between the wires (a <r < b)
d) inside the outer wire (b <r<c)
e) outside the outer wire (r > c)
in terms of µo, i, TI and r. For each case indicate the direction of the magnetic field as well.
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Transcribed Image Text:4. A long coaxial cable, crossection of which is shown in the figure, is made up of a cylindrical inner wire of radius a and a cylindrical shell shaped coaxial outer wire of inner radius b and outer radius c. The r2 current density with magnitude J = K- is spread over the cross section of the inner wire where the current has out-of-page direction. Outer wire carries uniformly distributed current i = tka³ having into-the-page direction. a) Find the total current of the inner wire iin in terms of i. Find expressions for the magnitude of magnetic field, b) inside the inner wire (r < a) c) between the wires (a <r < b) d) inside the outer wire (b <r<c) e) outside the outer wire (r > c) in terms of µo, i, TI and r. For each case indicate the direction of the magnetic field as well.
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Given, 

J=kr2aa. total current of of inner wire, inLet us take an small element ring shape of width dr at a distance r from the centre, dIcurrent flows in this ring. Hence, current density is given byJ=kr2adIdA=kr2adI=kr2adA=kr2a(2πr dr)0IindI=2πka0ar3dr On solving Iin=kπa32=i2out of the page, where i is the current in outside shell.

b.  Since, we know Ampere's Law

B.da=μ0IB.da=B. 2πx0idI=0x2πkar3dri=kπx42aB.da=μ0kπx42aB.2πx=μ0kπx42aB=μ0kx34aThe direction will be along tangent of the circle.

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