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A long cylindrical shell, with inner radius a and
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6). Current Density for the given cylinder, J= \frac{I}{\pi(b^2-a^2)} ................................(1)

Now for the Magetic Field inside the cylinder(0r<a), Using Ampere's Circuital Law,

\oint \vec{B}.\vec{dl}= \mu_o I_{encl}=0  (Since no current in enclosed in this region)

thus   B_1= 0 for region 0<r<a.........................................(A)

Now for the region of the cylinder, a<r<b. Using Ampere's Circuital Law,

\oint \vec{B_2}.\vec{dl}= \mu_o I_{encl}

B_2( 2 \pi r)= \mu_o (J \pi (r^2-a^2))

B_2= \mu_o\frac{J (r^2-a^2)}{2r}

Using equation 1 in above,

  B_2=\frac{I \mu_o (r^2-a^2)}{2r\pi (b^2-a^2)}

thus we can say that B_2\propto rfor region a<r<b..................................(B)

Now for the region of the cylinder, r>b. Using Ampere's Circuital Law,

\oint \vec{B_3}.\vec{dl}= \mu_o I_{encl}

B_3( 2 \pi r)= \mu_o I (SInce then whole of the current lies in the region)

B_3=\frac{\mu _0 I}{2 \pi r}

thus we can say that B_3\propto \frac{1}{r} for region r>b..................................(C)

Then following equation A, B and C we can say that graph (D). depicts correctly the variation of Magnetic Field with the distance.

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