Finding magnetic field using Ampere's Circuital Law

[arham_jain_hsr]

Homework Statement

A long straight metal rod has a very long hole of radius ′a′ drilled parallel to the rod axis as shown in the figure. If the rod carries a current ′I′. Find the magnetic induction on the axis of the hole, where OC=c.

Relevant Equations

Ampere's Circuital Law

I followed the following approach which is also the listed solution:
First of all, from Ampere’s circuital law, we get:
$∮B⋅dl=μ_0I$
Here, $I$ is the enclosed circuit in the circular Gaussian surface of radius $c$ and its value will be:
$I=J⋅πc^2$
Here, $J$ is the current flowing per unit cross-sectional area. Current density of the rod if it did not have a cavity, will be as follows:
$J=\frac{I}{πb^2}$
Since the metal rod has a cavity of radius $a$, the resulting cross-sectional area will be the cross-sectional area of the rod subtracted by the cross-sectional area of the cavity, and its value will be:
$J=\frac{I}{πb^2−πa^2}$
Substituting these values in Ampere’s circuital law we get:
$∮B⋅dl=μ_0\frac{I⋅πc^2}{πb^2−πa^2}$
or $B∮dl=μ_0\frac{I⋅πc^2}{πb^2−πa^2}$
On calculating the line integral, we will get the following equation:
$B(2πc)=μ_0\frac{I⋅πc^2}{πb^2−πa^2}$
$\implies B=μ_0\frac{I⋅πc^2}{(2πc)(πb^2−πa^2)}$
$\therefore B=\frac{μ_0Ic^2}{2π(b^2−a^2)}$
Thus, the value of magnetic induction on the axis of the hole, where $OC=c$
is $\frac{μ_0Ic^2}{2π(b^2−a^2)}$

However, this just doesn't seem appropriate to me to find the magnetic field strength from magnetic flux by simply multiplying the latter with $2\pi c$ because I don't think the magnetic field strength in the cavity is the same as the magnetic field strength in the rod at a distance $c$. A similar point has also been raised here: "https://www.physicsforums.com/insights/a-physics-misconception-with-gauss-law/"
So, is the listed solution/answer incorrect? Or, am I missing something here?

 

[PhDeezNutz]

The answer is correct but I can see why you find it unconvincing

Try finding the magnetic field using ampere's law at a random distance from the center of the cavity (Do it along the line $OC$ (say a distance $d$ from $C$) and then set $d=0$ at the end. You will recover your answer.

Edit: at a random distance from the center of the cavity but still in the cavity.

 

[kuruman]

There is faulty algebra here, probably a typo
If $B(2πc)=μ_0\frac{I⋅πc^2}{πb^2−πa^2}$,
it doesn't follow that
$\therefore B=\frac{μ_0Ic^2}{2π(b^2−a^2)}$.

More importantly, there is faulty application of Ampere's circuital law.
This is correct $∮B⋅dl=μ_0\frac{I⋅πc^2}{πb^2−πa^2}$ assuming uniform current density $J$.

However, it doesn't necessarily follow that $∮B⋅dl=B\oint dl$. The B-field on the Amperian loop must be shown to (a) be tangent to the loop at all points and (b) have the same magnitude. For the equality to hold, one needs to show that the field is not different on the part of the loop that's in the hole from the field on the part of the loop that is in the conductor.

. . . because I don't think the magnetic field strength in the cavity is the same as the magnetic field strength in the rod at a distance c.

You are right. To see what's going on, you need to find the field at an arbitrary point in the cavity. You can do that by superposition. Consider the superposition of two fields, one from a conductor of radius $b$ carrying current density $\mathbf J_1$ and a second conductor of radius $a$ carrying current density $-\mathbf J_1$ parallel to the first with axis separation $c$. Of course, you will have to relate the magnitude $J_1$ to the given current $I$.

 

[Steve4Physics]

Treat the current (I) as consisting of two separate superposed currents $I_1$ and $I_2$:

With $\vec {J_1} = - \vec {J_2}$ the hole corresponds to a region with zero net current.

In terms of magnitudes $I = I_1 - I_2$.

You now have two cylindrically symmetric current distributions so it's easy to (correctly) apply Ampere’s law and find the field-contribution due to each.

 

[arham_jain_hsr]

Thanks for your response!

There is faulty algebra here, probably a typo

Yes, sorry, the final answer should've been $B=\frac{μ_0Ic}{2 \pi (b^2-a^2)}$.

You are right. To see what's going on, you need to find the field at an arbitrary point in the cavity. You can do that by superposition. Consider the superposition of two fields, one from a conductor of radius b carrying current density J1 and a second conductor of radius a carrying current density −J1 parallel to the first with axis separation c. Of course, you will have to relate the magnitude J1 to the given current I.

With $\vec{J_1}=-\vec{J_2}$ the hole corresponds to a region with zero net current.

So, the magnetic field strength at a radial distance $c$ in a solid cylindrical rod with $+I$ current should be $B=\frac{μ_0Ic}{2 \pi (b^2-a^2)}$. For a solid cylindrical rod with current $-I$, the magnetic field strength at the centre is $0$. Therefore, the magnitude of net magnetic field strength at point $C$ would simply become $B=\frac{μ_0Ic}{2 \pi (b^2-a^2)}+0=\frac{μ_0Ic}{2 \pi (b^2-a^2)}.$

 

[kuruman]

Yes.