arham_jain_hsr
- 25
- 9
- 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?