Figure 29-80 shows a cross-section of a long cylindrical conductor of radius $\mathit{a}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{\text{cm}}$containing a long cylindrical hole of radius$\mathit{b}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{cm}}$. The central axes of the cylinder and hole are parallel and are distance$\mathit{d}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{\text{cm}}$apart; currentis uniformly distributed over the tinted area. (a) What is the magnitude of the magnetic field at the center of the hole? (b) Discuss the two special cases$\mathit{b}\mathbf{=}\mathbf{0}$and$\mathit{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$.

1. Radius of a long cylinder conductor $a=4\text{cm}$
2. Radius of a cylindrical hole$b=1.50\text{cm}$
3. Distance between centers$d=2\text{cm}$
4. Current$i=5.25\text{A}$

An equation known as the Biot-Savart Law describes the magnetic field produced by a steady electric current. It connects the electric current's strength, direction, length, and proximity to the magnetic field.

We use Biot-Savart law to find the magnetic field inside the wire.Since the current is uniform we use the current density formula to find the current in a solid cylinder of different radii. Using these different values of current we can find the magnetic field inside the two different regions. Adding these two fields at the required distance, we get the net magnetic field at that point.

Formulae:

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}\mathbf{r}}{\mathbf{2}\mathbf{\pi }{\mathbf{R}}^{\mathbf{2}}}$

$\mathit{J}\mathbf{=}\frac{\mathbf{i}}{\mathbf{A}}$

Magnitude of magnetic field at the center of the hole.

The magnetic field at any point inside the wire of which the current is uniformly distributed over its cross section is given by

$B=\frac{{\mu }_{0}ir}{2\pi R^2}$ …(i)

Where $r$is the distance from the axis of the cylinder and $R$is the radius of the cylinder.

The current density is given by

$J=\frac{i}{A}$

The cross-sectional area of the cylinder with the hole is

$A=\pi \left(a^2-b^2\right)$

Hence, the current density with the hole is given by

$J=\frac{i}{\pi \left(a^2-b^2\right)}$ …(ii)

Now the current without the hole is given by

$I_1=JA=\pi a^{2}J$

Substituting the value of $J$from equation (ii) we get

$$\begin{gathered} I_1=\pi a^2\frac{i}{\pi \left(a^2-b^2\right)} \\ =\frac{i{a}^2}{\left(a^2-b^2\right)} \end{gathered}$$ …(iii)

This current produces the magnetic field inside the cylinder at distance $r_1$; therefore, from equation (i)

$B_1=\frac{{\mu }_0{I}_1{r}_1}{2\pi a^2}$ …(iv)

Substituting the value of (iii) in (iv) we get

$$\begin{gathered} B_1=\frac{{\mu }_{0}i{a}^2{r}_1}{2\pi a^2\left(a^2-b^2\right)} \\ =\frac{{\mu }_{0}i{r}_1}{2\pi \left(a^2-b^2\right)} \end{gathered}$$

Now the current inside the cylinder that fills the hole is

$$\begin{gathered} I_2=\pi J{b}^2 \\ =\frac{i{b}^2}{\left(a^2-b^2\right)} \end{gathered}$$ …(v)

This current produces the magnetic field inside a distance $r_2$from the axis, and has magnitude

$$\begin{gathered} I_2=\pi J{b}^2 \\ =\frac{i{b}^2}{\left(a^2-b^2\right)} \end{gathered}$$ …(vi)

Substituting the value of from (v) in (vi) we get

$$\begin{gathered} B_2=\frac{{\mu }_{0}i{b}^2{r}_2}{2\pi b^2\left(a^2-b^2\right)} \\ =\frac{{\mu }_{0}i{r}_2}{2\pi \left(a^2-b^2\right)} \end{gathered}$$

At the center of the hole $r_2=0$, hence, this field is zero at the center. Hence, the net field at the center of the hole is only $B_1$

And the distance $r_1$from the center of cylinder to center of hole is $d$

$$\begin{gathered} B_1=\frac{{\mu }_{0}i{r}_1}{2\pi \left(a^2-b^2\right)} \\ B=\frac{{\mu }_{0}id}{2\pi \left(a^2-b^2\right)} \end{gathered}$$

${\mu }_o=4\pi \times {10}^{-7}\raisebox{1ex}{$Tm$}\!\left/ \!\raisebox{-1ex}{$A$}\right.,i=5.25\mathrm{A},d=2\times {10}^{-2}m,a=4\times {10}^{-2}\text{m}$ …(vii)

$b=1.50\times {10}^{-2}\text{m}$

$$\begin{gathered} B=\frac{\left(4\pi \times {10}^{-7}\frac{\text{T.m}}{\text{A}}\right)\left(5.25\text{A}\right)\left(2\times {10}^{-2}\text{m}\right)}{2\pi \left({\left(4\times {10}^{-2}\text{m}\right)}^2-{\left(1.50\times {10}^{-2}\text{m}\right)}^2\right)} \\ =1.53\times {10}^{-5}\text{T} \end{gathered}$$

Special cases : $b=0,d=0$

If $b=0$then from equation (vii) the net field is

$B=\frac{{\mu }_{0}id}{2\pi \left(a^2-b^2\right)}$

$B=\frac{{\mu }_{0}id}{2\pi a^2}$

This is the field due to a solid cylinder carrying a uniform current $i$inside the cylinder at a distance from the axis of the cylinder.

If $d=0$

Then B = 0

This is the field on the axis of the cylindrical shell carrying uniform current because the current enclosed in this case is zero.