Question

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$\mathrm{OC}=\mathrm{c}$:

• A. $\frac{{\mathrm{\mu }}_0\mathrm{Ic}}{\mathrm{\pi }\left({\mathrm{b}}^2-{\mathrm{c}}^2\right)}$
• B. $\frac{{\mathrm{\mu }}_0\mathrm{Ic}}{2\mathrm{\pi }\left({\mathrm{b}}^2-{\mathrm{c}}^2\right)}$
• C. $\frac{{\mathrm{\mu }}_0\mathrm{I}\left({\mathrm{b}}^2-{\mathrm{c}}^2\right)}{2\mathrm{\pi c}}$
• D. $\frac{{\mathrm{\mu }}_0\mathrm{Ic}}{2{\mathrm{\pi a}}^2{\mathrm{b}}^2}$

Answer 1

The correct option is B

$\frac{{\mathrm{\mu }}_0\mathrm{Ic}}{2\mathrm{\pi }\left({\mathrm{b}}^2-{\mathrm{c}}^2\right)}$

Step 1: Given data

In the question's statement, it is given that:

Radius of the hole $=\mathrm{a}$

Current carried by the rod $=\mathrm{I}$

Step 2: Formula used

To solve this question, we will be using the Ampere circuital law.

The integral value of the magnetic field density along an imaginary line is equal to the product of the permeability of open space and the current encompassed by that path, according to the law.

$\oint \mathrm{B}\cdot \mathrm{dl}={\mathrm{\mu }}_0\mathrm{I}$

Step 3: Apply Ampere's circuital law

Apply Ampere’s circuital law, assuming a circular Gaussian surface of radius $\mathrm{c}$, we get:
$\oint \mathrm{B}\cdot \mathrm{dl}={\mathrm{\mu }}_0\mathrm{I}$ ……………(1)
Here, $\mathrm{I}$ is the enclosed circuit in the circular Gaussian surface of radius $\mathrm{c}$ and its value will be:
$\mathrm{I}=\mathrm{J}\cdot {\mathrm{\pi c}}^2$
Here, $\mathrm{J}$ is the current flowing per unit cross-sectional area.

The current density of the rod if it did not have a cavity, will be:
$\mathrm{J}=\frac{\mathrm{I}}{{\mathrm{\pi b}}^2}$
Since the metal rod has a cavity of radius $\mathrm{a}$, hence the resulting cross-sectional area will be the cross-sectional area of the rod subtracted by the cross-sectional area of the cavity. Thus, its value will be:
$\mathrm{J}=\frac{\mathrm{I}}{{\mathrm{\pi b}}^2-{\mathrm{\pi a}}^2}$

And the value of $\mathrm{I}$ will be:

$\mathrm{I}=\frac{\mathrm{J}}{{\mathrm{\pi b}}^2-{\mathrm{\pi a}}^2}$ …………….(2)

Step 4: Putting the value of "$\mathrm{I}$" in Ampere's circuital law
Substituting these value of $\mathrm{I}$ in equation (1) we get:

$\oint \mathrm{B}·\mathrm{dl}={\mathrm{\mu }}_0\frac{\mathrm{I}·{\mathrm{\pi c}}^2}{{\mathrm{\pi b}}^2-{\mathrm{\pi a}}^2}$

as, B is constant and parallel to dl so, taking B out of the line integral.

$B.\oint \mathrm{dl}={\mathrm{\mu }}_0\frac{\mathrm{I}·{\mathrm{\pi c}}^2}{{\mathrm{\pi b}}^2-{\mathrm{\pi a}}^2}$

Step 5: Find the value of magnetic induction
As, the radius of the circular Gaussian surface is c, thus, the circumference or dl will be $2\mathrm{\pi c}$.

So, now we will put the value of dl in the above equation.

On calculating the line integral, we will get the following equation:
$\mathrm{B}\left(2\mathrm{\pi c}\right)={\mathrm{\mu }}_0\frac{\mathrm{I}·{\mathrm{\pi c}}^2}{{\mathrm{\pi b}}^2-{\mathrm{\pi a}}^2}$

$\Rightarrow \mathrm{B}={\mathrm{\mu }}_0\frac{\mathrm{I}·{\mathrm{\pi c}}^2}{\left(2\mathrm{\pi c}\right).{\mathrm{\pi b}}^2-{\mathrm{\pi a}}^2}$

$\Rightarrow \mathrm{B}=\frac{{\mathrm{\mu }}_0.\mathrm{I}.{\mathrm{c}}^2}{2\mathrm{c}.({\mathrm{b}}^2-{\mathrm{a}}^2)}$

Thus, the value of magnetic induction on the axis of the hole, where $\mathrm{OC}=\mathrm{c}$ is $\mathrm{B}=\frac{{\mathrm{\mu }}_0.\mathrm{I}.{\mathrm{c}}^2}{2\mathrm{c}.({\mathrm{b}}^2-{\mathrm{a}}^2)}$