Chapter 28, Problem 80P

(a)

To determine

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r,r_1$ , $r_1<r<r_2$ and $r>r_2$ .

Answer to Problem 80P

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r,r_1$ is $0$ , for $r_1<r<r_2$ is $\frac{{\mu }_{0}I}{2\pi r}$ and $r>r_2$ is $0$ .

Explanation of Solution

Formula Used:

The relation for the magnetic is given by,

  $u_{\text{m}}=\frac{B^2}{2{\mu }_0}$

Calculation:

The magnetic field inside the cylinder $\left(r<r_1\right)$ and outside the cylinder $\left(r>r_2\right)$ is zero because the net enclosed current is zero.

  $B_{r<r_1}=0$

And,

  $B_{r>r_2}=0$

The magnetic field between the cylinder is calculated as,

  $\begin{array}{c}\left(2\pi r\right)B={\mu }_{0}I\\ B=\frac{{\mu }_{0}I}{2\pi r}\end{array}$

Conclusion:

Therefore, the magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r,r_1$ is $0$ , for $r_1<r<r_2$ is $\frac{{\mu }_{0}I}{2\pi r}$ and $r>r_2$ is $0$ .

(b)

To determine

The proof that the magnetic energy density in the region between the cylinder is by $u_{\text{m}}=\frac{1}{2}\left({\mu }_0/4\pi \right)I^2/\left(\pi r^2\right)$ .

Answer to Problem 80P

The proof that the magnetic energy density in the region between the cylinder is by $u_{\text{m}}=\frac{1}{2}\left({\mu }_0/4\pi \right)I^2/\left(\pi r^2\right)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the magnetic energy density in the region between the cylinder is given by,

  $u_m=\frac{B^2}{2{\mu }_0}$

The magnetic field between the cylinder is given as,

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Calculation:

The magnetic energy density in the region between the cylinder is calculated as,

  $\begin{array}{c}{u}_m=\frac{{\left(\frac{{\mu }_{0}I}{2\pi r}\right)}^2}{2{\mu }_0}\\ =\frac{{\mu }_0{I}^2}{8{\pi }^2{r}^2}\\ =\frac{1}{2}\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{I^2}{\pi r^2}\right)\\ u_{\text{m}}=\frac{1}{2}\left({\mu }_0/4\pi \right)I^2/\left(\pi r^2\right)\end{array}$

Conclusion:

Therefore, the proof that the magnetic energy density in the region between the cylinder is by $u_{\text{m}}=\frac{1}{2}\left({\mu }_0/4\pi \right)I^2/\left(\pi r^2\right)$ is stated above.

(c)

To determine

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $U_{\text{m}}=\left({\mu }_0/4\pi \right)I^{2}l\ln \left(r_2/r_1\right)$

Answer to Problem 80P

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $U_{\text{m}}=\left({\mu }_0/4\pi \right)I^{2}l\ln \left(r_2/r_1\right)$ is stated below.

Explanation of Solution

Formula Used:

The magnetic energy density in the region between the cylinder is given by,

  $u_{\text{m}}=\frac{1}{2}\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{I^2}{\pi r^2}\right)$

The magnetic energy $d{U}_{\text{m}}$ in the cylindrical element of volume $dV$ is given by,

  $\begin{array}{c}d{U}_m=u_{\text{m}}dV\\ =\left(\frac{1}{2}\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{I^2}{\pi r^2}\right)\right)\left(l2\pi rdr\right)\\ =\frac{{\mu }_0{I}^{2}l}{4\pi }\cdot \frac{dr}{r}\end{array}$ ……. (I)

Calculation:

Integrate equation (I) over the limits $r_1\text{to}{r}_2$ .

  $\begin{array}{c}{\displaystyle \int d{U}_m}={\displaystyle \underset{r_1}{\overset{r_2}{\int }}\frac{{\mu }_0{I}^{2}l}{4\pi }\cdot \frac{dr}{r}}\\ U_{\text{m}}=\frac{{\mu }_0{I}^{2}l}{4\pi }{\displaystyle {\int }_{r_1}^{r_2}\frac{1}{r}dr}\\ =\frac{{\mu }_0{I}^{2}l}{4\pi }{\left[\ln r\right]}_{r_1}^{r_2}\\ =\frac{{\mu }_0{I}^{2}l\ln \left(r_2/r_1\right)}{4\pi }\end{array}$

Conclusion:

Therefore, the proof that the total magnetic energy in a cable of volume of length $l$ is given by $U_{\text{m}}=\left({\mu }_0/4\pi \right)I^{2}l\ln \left(r_2/r_1\right)$ is stated above.

(d)

To determine

The proof that self inductance per unit length of the cab arrangement is given by $L/l=\left({\mu }_0/2\pi \right)\ln \left(r_2/r_1\right)$ .

Answer to Problem 80P

The proof that self inductance per unit length of the cab arrangement is given by $L/l=\left({\mu }_0/2\pi \right)\ln \left(r_2/r_1\right)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the total magnetic energy is given by,

  $U_{\text{m}}=\frac{{\mu }_0{I}^{2}l\ln \left(r_2/r_1\right)}{4\pi }$ …… (II)

The energy in the magnetic in terms of $L$ and $I$ is given by,

  $U_{\text{m}}=\frac{1}{2}L{I}^2$ …….. (III)

Calculation:

The self inductance per unit length is calculated as,

  $\begin{array}{c}{U}_{\text{m}}=\frac{1}{2}L{I}^2\\ L=\frac{2\left(U_{\text{m}}\right)}{I^2}\\ =\frac{2\left(\frac{{\mu }_0{I}^{2}l\ln \left(r_2/r_1\right)}{4\pi }\right)}{I^2}\\ \frac{L}{l}=\frac{{\mu }_0\ln \left(r_2/r_1\right)}{2\pi }\end{array}$

Conclusion:

Therefore, the proof that self inductance per unit length of the cable arrangement is given by $L/l=\left({\mu }_0/2\pi \right)\ln \left(r_2/r_1\right)$ is stated above.