Answer:
Explanation:
we can consider an element of radius r < a and thickness dr. and Area of this element is
[tex]dA=2\pi r dr[/tex]
since current density is given
[tex]J=kr[/tex]
then , current through this element will be,
[tex]di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr[/tex]
integrating on both sides between the appropriate limits,
[tex]\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr \\\\ I=\frac{2\pi\,ka^3}{3} -------------------------------(1)[/tex]
Magnetic field can be found by using Ampere's law
[tex]\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}[/tex]
for points inside the wire ( r<a)
now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.
by applying the Ampere's law, we can write
[tex]\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc} [/tex]
by symmetry [tex]\vec{B}[/tex] will be of uniform magnitude on this loop and it's direction will be tangential to the loop.
Hence,
[tex]B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)= \\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3} \\\\B_{in}=\frac{\mu_0kl^2}{3} [/tex]
now using equation 1, putting the value of k,
[tex]B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3} \\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3} [/tex]
B)
now, for points outside the wire ( r>a)
consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.
applying the Ampere's law
[tex]\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc} [/tex]
by symmetry [tex]\vec{B}[/tex] will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence
[tex]B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr) \\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3} \\\\B_{out}=\frac{\mu_0ka^3}{3r} [/tex]
again using,equaiton 1,
[tex]B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3} \\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}[/tex]
