Question

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?

Short answer

$\begin{array}{l}a)B=4.8\times {10}^{-3}\text{\hspace{0.17em}T}\\ b)B=9.3\times {10}^{-4}\text{T}\\ c)B=0.0\text{T}\end{array}$

Step-by-step solution

Given

• 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

Understanding the concept

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:

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

(a) Calculate the net magnetic field at a point p and distance r from the center of the hollow cylinder r=1.0mm

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

(b) Calculate the net magnetic field at a point  p and distance r from the center of the hollow cylinder r=3.0mm

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,

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

(c) Calculate the net magnetic field at a point  p and distance r from the center of the hollow cylinder r=5.0mm

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