Chapter 28, Problem 21P

(a)

To determine

The magnetic flux through the coil when the axis of the coil is vertical.

Answer to Problem 21P

The magnetic flux through the coil when the axis of the coil is vertical is $0$ .

Explanation of Solution

Given:

The turns of the coilis $n=25$ .

The radius of the coil is $r=5.0\text{cm}$ .

The axis of coil is vertical.

Formula used:

The expression for the magnetic flux through the coil is given by,

  ${\varphi }_m=NBA\cos \theta$

Calculation:

The expression for the magnetic flux through the coil is given by,

  $\begin{array}{c}{\varphi }_m=NBA\cos \theta \\ =NB\pi r^2\cos \theta \\ =25\left(0.70\text{G}\times \frac{1\text{T}}{{10}^4\text{G}}\right)\pi {\left(5.0\times {10}^{-2}\text{m}\right)}^2\cos \theta \\ =\left(\text{13}\text{.7}\text{μWb}\right)\cos \theta \end{array}$

If the plane of the coil is vertical, $\theta =90°$ ,

  $\begin{array}{c}{\varphi }_m=\left(\text{13}\text{.7}\text{μWb}\right)\cos 90°\\ =0\end{array}$

Conclusion:

Therefore, the magnetic flux through the coil when the axis of the coil is vertical is $0$ .

(b)

To determine

The magnetic flux through the coil when the axis of the coil is horizontal.

Answer to Problem 21P

The magnetic flux through the coil when the axis of the coil is vertical is $14\text{μWb}$ .

Explanation of Solution

Given:

The axis of coil is vertical.

Formula used:

The expression for the magnetic flux through the coil is given by,

  ${\varphi }_m=NBA\cos \theta$

Calculation:

The expression for the magnetic flux through the coil is given by,

  $\begin{array}{c}{\varphi }_m=NBA\cos \theta \\ =NB\pi r^2\cos \theta \\ =25\left(0.70\text{G}\times \frac{1\text{T}}{{10}^4\text{G}}\right)\pi {\left(5.0\times {10}^{-2}\text{m}\right)}^2\cos \theta \\ =\left(\text{13}\text{.7}\text{μWb}\right)\cos \theta \end{array}$

If the plane of the coil is horizontal, $\theta =0°$ ,

  $\begin{array}{c}{\varphi }_m=\left(13.7\text{μWb}\right)\cos 0°\\ \simeq 14\text{μWb}\end{array}$

Conclusion:

Therefore, the magnetic flux through the coil when the axis of the coil is vertical is $14\text{μWb}$ .

(c)

To determine

The magnetic flux through the coil when the axis of the coil is horizontal with the axis pointing to east.

Answer to Problem 21P

The magnetic flux through the coil when the axis of the coil is horizontal with its axis pointing east, is $0$ .

Explanation of Solution

Given:

The axis of coil is vertical.

Formula used:

The expression for the magnetic flux through the coil is given by,

  ${\varphi }_m=NBA\cos \theta$

Calculation:

The expression for the magnetic flux through the coil is given by,

  $\begin{array}{c}{\varphi }_m=NBA\cos \theta \\ =NB\pi r^2\cos \theta \\ =25\left(0.70G\times \frac{1\text{T}}{{10}^{4}G}\right)\pi {\left(5.0\times {10}^{-2}\text{m}\right)}^2\cos \theta \\ =\left(13.7\text{μWb}\right)\cos \theta \end{array}$

If the plane of the coil is horizontal with its axis pointing east, $\theta =90°$ ,

  $\begin{array}{c}{\varphi }_m=\left(13.7\text{μWb}\right)\cos 90°\\ =0\end{array}$

Conclusion:

Therefore, the magnetic flux through the coil when the axis of the coil is horizontal with its axis pointing east, is $0$ .

(d)

To determine

The magnetic flux through the coil when the axis of the coil is horizontal with the axis marking an angle of ${30}^{\circ }$ with north.

Answer to Problem 21P

The magnetic flux through the coil when the axis of the coil is horizontal with the axis marking an angle of ${30}^{\circ }$ with northis $12\text{μWb}$ .

Explanation of Solution

Given:

The axis of coil is horizontal with the axis marking an angle of ${30}^{\circ }$ with north.

Formula used:

The expression for the magnetic flux through the coil is given by,

  ${\varphi }_m=NBA\cos \theta$

Calculation:

The expression for the magnetic flux through the coil is given by,

  $\begin{array}{c}{\varphi }_m=NBA\cos \theta \\ =NB\pi r^2\cos \theta \\ =25\left(0.70\text{G}\times \frac{1\text{T}}{{10}^4\text{G}}\right)\pi {\left(5.0\times {10}^{-2}\text{m}\right)}^2\cos \theta \\ =\left(\text{13}\text{.7}\text{μWb}\right)\cos \theta \end{array}$

If the plane of the coil is horizontal with its axis making an angle of $30°$ with north, $\theta =90°$ ,

  $\begin{array}{c}{\varphi }_m=\left(1.37\text{μWb}\right)\cos 30°\\ =\left(\frac{\sqrt{3}}{2}\right)\left(1.37\text{μWb}\right)\\ =12\text{μWb}\end{array}$

Conclusion:

Therefore, the magnetic flux through the coil when the axis of the coil is is horizontal with the axis marking an angle of $30°$ with north is $12\text{μWb}$ .