91Ó°ÊÓ

• 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:

$âˆ\circledR \stackrel{→}{B}·\stackrel{→}{ds}={\mathrm{μ}}_0{I}_{enclosed}$

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

At

$$\begin{gathered} âˆ\circledR {\stackrel{→}{B}}_w·\stackrel{→}{ds}={\mathrm{μ}}_0{I}_{enclosed}={\mathrm{μ}}_{0}i \\ {B}_w·2\mathrm{Ï€}r={\mathrm{μ}}_{0}i \end{gathered}$$

This gives

$B_w=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

and

$\begin{array}{c}âˆ\circledR {\stackrel{→}{B}}_câ‹\dots \stackrel{→}{ds}={\mathrm{μ}}_0{I}_{enclosed}={\mathrm{μ}}_{0}I=0.0\\ B=B_c+B_w=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€}r}=\frac{1.26×{10}^{−6}×24}{2×\mathrm{Ï€}×1.0×{10}^{−3}}\\ B=4.8×{10}^{−3}\text{â€Í¿}\end{array}$

Hence,$B=4.8×{10}^{−3}\text{â€Í¿}$

At$r=3.0×{10}^{−3}\text{″¾}=3.0\text{ }\text{mm}$

$âˆ\circledR {\stackrel{→}{B}}_w·\stackrel{→}{ds}={\mathrm{μ}}_0{I}_{enclosed}={\mathrm{μ}}_{0}i$

This gives,

$B_w=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

And

$âˆ\circledR {\stackrel{→}{B}}_c·\stackrel{→}{ds}={\mathrm{μ}}_0{I}_{enclosed}={\mathrm{μ}}_0{I}_{enclosed}$

Now,

$I_{enclosed}=I·\left(\frac{\mathrm{π}{r}^2-\mathrm{π}{r}_i^2}{\mathrm{π}{r}_0^2-\mathrm{π}{r}_i^2}\right)$

Hence,

$B_c=\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}r}â‹\dots Iâ‹\dots \left(\frac{\mathrm{Ï€}{r}^2−\mathrm{Ï€}{r}_i^2}{\mathrm{Ï€}{r}_0^2−\mathrm{Ï€}{r}_i^2}\right)$

Since i = - I

$\begin{array}{c}B=\frac{2{\mathrm{μ}}_{0}i}{4\mathrm{Ï€}r}â‹\dots \frac{(r_0^2−r^2)}{(r_0^2−r_i^2)}=\frac{(2×{10}^{−7}×24×(4^2−3^2))}{3.0×{10}^{−3}×(4^2−2^2)}\\ B=9.3×{10}^{−4}\text{T}\end{array}$

Hence,

$B=9.3×{10}^{−4}\text{ }\text{T}$

At $r=5.0×{10}^{−3}\text{″¾}=5.0\text{″¾m}$

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

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

Hence, B=0.0T