Question

The figure shows the cross section of a long cylindrical wire of

radius aa, which carries a uniform current ii0. Determine the module of the field

magnetic produced by the current at a distance from the axis of the wire equal to (a)

0, (b) a/2, (c) a (wire surface) and (d) 2a.

Answer 1

Hi there!

Recall Ampere's Law:
$\oint B \cdot dl = \mu_0 i_{encl}$

B = Magnetic Field Strength (T)
μ₀ = Permeability of Free Space (Tm/A)

i =  Enclosed Current (A)

dL = differential path length

To begin, we must derive an expression for the magnetic field strength inside a wire that contains a uniformly-distributed current.

Using the expression:

$i = \int J \cdot dA$

Where 'J' is the density of current, and A is the cross-sectional area:
$A = \pi r^2\\\\dA = 2\pi r dr$

We know that:
$J = \frac{i_0 }{A}\\\\J = \frac{i_0}{\pi a^2}$

This is the current density. We can now integrate:

$i = \int\limits^r_0 {\frac{i_0}{\pi a^2} \cdot 2\pi r} \, dr\\ \\i =\frac{i_0}{a^2}\int\limits^r_0 {2r } \, dr\\\\i = \frac{i_0}{a^2} \cdot r^2 = \frac{i_0 r^2}{a^2}$

Now, substitute this expression back into the above equation for the magnetic field:
$\oint B \cdot dl = \mu _0 \frac{i_0r^2}{a^2}$

The path of integration is a closed loop of length 2πr, so:
$B \cdot 2\pi r = \mu_0 \frac{i_0r^2}{a^2}\\\\B = \frac{\mu_0 i_0r}{2\pi a^2}$

We can now use this equation for the first 2 parts.

a)
If 'r' equals 0:
$B = \frac{\mu _0 i_0 (0)}{2\pi a^2} = \boxed{0 T}$

b)

If 'r' equals a/2:
$B = \frac{\mu _0 i_0 (\frac{a}{2})}{2\pi a^2} =B =\boxed{ \frac{\mu _0 i_0 }{4\pi a} T}$

c)
At the wire's surface, 'r' = a:
$B = \frac{\mu _0 i_0 (a)}{2\pi a^2} =B =\boxed{ \frac{\mu _0 i_0 }{2\pi a} T}$

d)

At 'r' = 2a, since this is outside of the wire, the relationship between magnetic field and distance from the wire becomes a 1/r (inverse) relationship. This is found using Ampere's Law:
$B = \frac{\mu_0 i_0}{2\pi r}$

$B = \frac{\mu_0 i_0}{2\pi (2a)} \\\\\boxed{B = \frac{\mu_0 i_0}{4\pi a}\\}$