A long, hollow, cylindrical conductor (with inner radius 2.0mm and outer radius 4.0mm) carries a current of 24A distributed uniformly across its cross section. A long thin wire that is coaxial with the cylinder carries a current of 24A in the opposite direction. What is the magnitude of the magnetic field (a) 1.0mm,(b) 3.0mm, and (c) 5.0mm from the central axis of the wire and cylinder?

• Current flowing through hollow cylinder l=24A
• Current flowing through long thin wire placed coaxial with hollow cylinder i=24Aopposite to the direction of l
• Inner radius of cylinder r_i=2mm
• Outer radius of cylinder r_o=4mm

We can find the magnitude of magnetic field by using Ampere’s law separately for a long thin wire and hollow cylinder. As the direction of currents is opposite, we can add both fields by considering the direction of the fields produced by them.

Formula:

$\oint \overrightarrow{B}·\overrightarrow{ds}={\mu }_0{I}_{enclosed}$

Consider B_c andB_w as the magnitude of magnetic field at distancerfrom the center due to the cylinder and thin wire then

At

$$\begin{gathered} \oint {\overrightarrow{B}}_w·\overrightarrow{ds}={\mu }_0{I}_{enclosed}={\mu }_{0}i \\ {B}_w·2\pi r={\mu }_{0}i \end{gathered}$$

This gives

$B_w=\frac{{\mu }_{0}i}{2\pi r}$

and

$\begin{array}{c}\oint {\overrightarrow{B}}_c\cdot \overrightarrow{ds}={\mu }_0{I}_{enclosed}={\mu }_{0}I=0.0\\ B=B_c+B_w=\frac{{\mu }_{0}i}{2\pi r}=\frac{1.26\times {10}^{-6}\times 24}{2\times \pi \times 1.0\times {10}^{-3}}\\ B=4.8\times {10}^{-3}\text{\hspace{0.17em}T}\end{array}$

Hence,$B=4.8\times {10}^{-3}\text{\hspace{0.17em}T}$

At$r=3.0\times {10}^{-3}\text{\hspace{0.17em}m}=3.0\text{mm}$

$\oint {\overrightarrow{B}}_w·\overrightarrow{ds}={\mu }_0{I}_{enclosed}={\mu }_{0}i$

This gives,

$B_w=\frac{{\mu }_{0}i}{2\pi r}$

And

$\oint {\overrightarrow{B}}_c·\overrightarrow{ds}={\mu }_0{I}_{enclosed}={\mu }_0{I}_{enclosed}$

Now,

$I_{enclosed}=I·\left(\frac{\pi r^2-\pi r_i^2}{\pi r_0^2-\pi r_i^2}\right)$

Hence,

$B_c=\frac{{\mu }_0}{2\pi r}\cdot I\cdot \left(\frac{\pi r^2-\pi r_i^2}{\pi r_0^2-\pi r_i^2}\right)$

Since i = - I

$\begin{array}{c}B=\frac{2{\mu }_{0}i}{4\pi r}\cdot \frac{(r_0^2-r^2)}{(r_0^2-r_i^2)}=\frac{(2\times {10}^{-7}\times 24\times (4^2-3^2))}{3.0\times {10}^{-3}\times (4^2-2^2)}\\ B=9.3\times {10}^{-4}\text{T}\end{array}$

Hence,

$B=9.3\times {10}^{-4}\text{T}$

At $r=5.0\times {10}^{-3}\text{\hspace{0.17em}m}=5.0\text{\hspace{0.17em}mm}$

$I_{enclosed}=i-i=0$in the ampere loop is zero

Hence$B=B_c+B_w=0.0\text{T}$

Hence, B=0.0T