Question

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a constant and r is the distance from the axis of the wire. Find expressions for the magnitudes of the magnetic field inside and outside the wire as a function of r. (Hint: Find the current through an Ampèrian loop of radius r using Ithru = J · dA. Use the following as necessary: μ0, I, a and r. Do not substitute numerical values; use variables only.)a. B inside=?b. B outside=?

Answer 1

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr.  and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr \\\\ I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

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

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)= \\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3} \\\\B_{in}=\frac{\mu_0kl^2}{3}$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3} \\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}$

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

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr) \\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3} \\\\B_{out}=\frac{\mu_0ka^3}{3r}$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3} \\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$