Chapter 20, Problem 42P

(a)

To determine

Number of turns on the solenoid.

Answer to Problem 42P

Number of turns on the solenoid is $1.8\times {10}^4$.

Explanation of Solution

Write an expression for the number of turns on the solenoid.

  $N=\frac{BL}{{\mu }_{0}I}$                                                                                                             (I)

Here, $N$ is the number of turns in the solenoid, $B$ is the magnetic field, $L$ is the length of the solenoid ${\mu }_0$ is the permeability of free space and $I$ is the current.

Conclusion:

Substitute $1.5\text{T}$ for $B$, $1.8\text{m}$ for $L$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$ and $120\text{A}$ for $I$  in equation (I) to find $N$.

  $\begin{array}{c}N=\frac{\left(1.5\text{T}\right)\left(1.8\text{m}\right)}{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(120\text{A}\right)}\\ =\frac{2.7\text{T}\cdot \text{m}}{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(120\text{A}\right)}\\ =1.8\times {10}^4\end{array}$

Thus, the number of turns on the solenoid is $1.8\times {10}^4$.

(b)

To determine

The energy stored in the magnetic field during normal operation.

Answer to Problem 42P

The energy stored in the magnetic field during normal operation is $620\text{kJ}$.

Explanation of Solution

Write an expression for energy stored in the magnetic field during normal operation.

  $U_B=\frac{B^2}{2{\mu }_0}\pi {\left(\frac{d}{2}\right)}^{2}L$                                                                                                 (II)

Here, $U_B$ is energy stored in the magnetic field during normal operation and $d$ is the diameter.

Conclusion:

Substitute $1.5\text{T}$ for $B$, $1.8\text{m}$ for $L$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$ and $75\text{cm}$ for $d$  in equation (II) to find $U_B$.

  $\begin{array}{c}{U}_B=\frac{{\left(1.5\text{T}\right)}^2}{2\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)}\pi {\left(\frac{\left(75\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}{2}\right)}^2\left(1.8\text{m}\right)\\ =\frac{{\left(1.5\text{T}\right)}^2}{2\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)}\pi {\left(0.35\text{m}\right)}^2\left(1.8\text{m}\right)\\ =\left(620\times {10}^3\text{J}\right)\left(\frac{1\text{kJ}}{{10}^3\text{J}}\right)\\ =620\text{kJ}\end{array}$

Thus, the energy stored in the magnetic field during normal operation is $620\text{kJ}$.

(c)

To determine

The energy stored.

Answer to Problem 42P

The energy stored is $110\text{kJ}$.

Explanation of Solution

The energy stored is proportional to the square of the magnitude of magnetic field. The magnitude of magnetic field is proportional to the current. Thus, the energy stored is proportional to the square of the current.

Write an expression for the energy stored.

  ${U^{\prime }}_B=U_B\frac{{I^{\prime }}^2}{I^2}$                                                                                               (III)

Here, ${U^{\prime }}_B$ is the energy stored and $I^{\prime }$ is the current.

Conclusion:

Substitute $620\text{kJ}$ for $U_B$, $120\text{A}$ for $I$  and $50\text{A}$  for $I$ in equation (III) to find ${U^{\prime }}_B$.

  $\begin{array}{c}{U^{\prime }}_B=\left(620\text{kJ}\right)\frac{{\left(50\text{A}\right)}^2}{{\left(120\text{A}\right)}^2}\\ =\left(\frac{25}{144}\right)\left(620\text{kJ}\right)\\ =110\text{kJ}\end{array}$

Thus, the energy stored is $110\text{kJ}$.