Chapter 28, Problem 26P

(a)

To determine

The magnetic flux through the coil.

Answer to Problem 26P

The magnetic flux through the coilis $30.2\text{mWb}$ .

Explanation of Solution

Given:

The number of turn of coil is $N=15.0$ .

The uniform magnetic fieldis $\overrightarrow{B}=\left(4.00\text{kG}\right)\widehat{i}$ .

The radius of coilis $r=4.00\text{cm}$ .

The unit normal to the plane of the coilis $\widehat{n}=\widehat{i}$ .

Formula used:

The expression for magnetic flux through coil is given by,

  ${\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2$

Calculation:

The magnetic flux through coil is calculated as,

  $\begin{array}{c}{\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2\\ =\left(15.0\right)\left[\left(4.00\text{kG}\widehat{i}\right)\cdot \widehat{i}\right]\pi {\left(4.00\text{cm}\right)}^2\\ =\left(15.0\right)\left[\left(\left(\left(4.00\text{kG}\right)\left(\frac{{10}^{-1}\text{T}}{1\text{kG}}\right)\right)\widehat{i}\right)\cdot \widehat{i}\right]\pi {\left(\left(4.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\\ =30.2\times {10}^{-3}\text{Wb}\end{array}$

Solve further as,

  $\begin{array}{c}{\varphi }_m=30.2\times {10}^{-3}\text{Wb}\left(\frac{{10}^3\text{mWb}}{1\text{Wb}}\right)\\ =30.2\text{mWb}\end{array}$

Conclusion:

Therefore, the magnetic flux through the coilis $30.2\text{mWb}$ .

(b)

To determine

The magnetic flux through the coil.

Answer to Problem 26P

The magnetic flux through the coilis $0$ .

Explanation of Solution

Given:

The unit normal to the plane of the coilis $\widehat{n}=\widehat{j}$ .

Formula used:

The expression for magnetic flux through coil is given by,

  ${\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2$

Calculation:

The magnetic flux through coil is calculated as,

  $\begin{array}{c}{\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2\\ =\left(15.0\right)\left[\left(4.00\text{kG}\widehat{i}\right)\cdot \widehat{j}\right]\pi {\left(4.00\text{cm}\right)}^2\\ =\left(15.0\right)\left[\left(\left(\left(4.00\text{kG}\right)\left(\frac{{10}^{-1}\text{T}}{1\text{kG}}\right)\right)\widehat{i}\right)\cdot \widehat{j}\right]\pi {\left(\left(4.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\\ =0\end{array}$

Conclusion:

Therefore, the magnetic flux through the coilis $0$ .

(c)

To determine

The magnetic flux through the coil.

Answer to Problem 26P

The magnetic flux through the coilis $21.35\text{mWb}$ .

Explanation of Solution

Given:

The unit normal to the plane of the coilis $\widehat{n}=\left(\widehat{i}+\widehat{j}\right)/\sqrt{2}$ .

Formula used:

The expression for magnetic flux through coil is given by,

  ${\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2$

Calculation:

The magnetic flux through coil is calculated as,

  $\begin{array}{c}{\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2\\ =\left(15.0\right)\left[\left(4.00\text{kG}\widehat{i}\right)\cdot \left(\left(\widehat{i}+\widehat{j}\right)/\sqrt{2}\right)\right]\pi {\left(4.00\text{cm}\right)}^2\\ =\left(15.0\right)\left[\left(\left(\left(4.00\text{kG}\right)\left(\frac{{10}^{-1}\text{T}}{1\text{kG}}\right)\right)\widehat{i}\right)\cdot \left(\left(\widehat{i}+\widehat{j}\right)/\sqrt{2}\right)\right]\pi {\left(\left(4.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\\ =\left(15.0\right)\left[\frac{\left(\left(\left(4.00\text{kG}\right)\left(\frac{{10}^{-1}\text{T}}{1\text{kG}}\right)\right)\widehat{i}\right)\cdot \widehat{i}}{\sqrt{2}}+0\right]\pi {\left(\left(4.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\end{array}$

Solve further as,

  $\begin{array}{c}{\varphi }_m=21.35\times {10}^{-3}\text{Wb}\\ =21.35\times {10}^{-3}\text{Wb}\left(\frac{{10}^3\text{mWb}}{1\text{Wb}}\right)\\ =21.35\text{mWb}\end{array}$

Conclusion:

Therefore, the magnetic flux through the coilis $21.35\text{mWb}$ .

(d)

To determine

The magnetic flux through the coil.

Answer to Problem 26P

The magnetic flux through the coilis $0$ .

Explanation of Solution

Given:

The unit normal to the plane of the coilis $\widehat{n}=\widehat{k}$ .

Formula used:

The expression for magnetic flux through coil is given by,

  ${\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2$

Calculation:

The magnetic flux through coil is calculated as,

  $\begin{array}{c}{\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2\\ =\left(15.0\right)\left[\left(4.00\text{kG}\widehat{i}\right)\cdot \widehat{k}\right]\pi {\left(4.00\text{cm}\right)}^2\\ =\left(15.0\right)\left[\left(\left(\left(4.00\text{kG}\right)\left(\frac{{10}^{-1}\text{T}}{1\text{kG}}\right)\right)\widehat{i}\right)\cdot \widehat{k}\right]\pi {\left(\left(4.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\\ =0\end{array}$

Conclusion:

Therefore, the magnetic flux through the coilis $0$ .

(e)

To determine

The magnetic flux through the coil.

Answer to Problem 26P

The magnetic flux through the coilis $18.12\text{mWb}$ .

Explanation of Solution

Given:

The unit normal to the plane of the coilis $\widehat{n}=0.60\widehat{i}+0.80\widehat{j}$ .

Formula used:

The expression for magnetic flux through coil is given by,

  ${\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2$

Calculation:

The magnetic flux through coil is calculated as,

  $\begin{array}{c}{\varphi }_m=N\left(\overrightarrow{B}\cdot \widehat{n}\right)\pi r^2\\ =\left(15.0\right)\left[\left(4.00\text{kG}\widehat{i}\right)\cdot \left(0.60\widehat{i}+0.80\widehat{j}\right)\right]\pi {\left(4.00\text{cm}\right)}^2\\ =\left(15.0\right)\left[\left(\left(\left(4.00\text{kG}\right)\left(\frac{{10}^{-1}\text{T}}{1\text{kG}}\right)\right)\widehat{i}\right)\cdot \left(0.60\widehat{i}+0.80\widehat{j}\right)\right]\pi {\left(\left(4.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\\ =\left(15.0\right)\left[\left(\left(\left(4.00\text{kG}\right)\left(\frac{{10}^{-1}\text{T}}{1\text{kG}}\right)\right)\widehat{i}\right)\cdot \left(0.60\widehat{i}\right)+0\right]\pi {\left(\left(4.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\end{array}$

Solve further as,

  $\begin{array}{c}{\varphi }_m=18.12\times {10}^{-3}\text{Wb}\\ =18.12\times {10}^{-3}\text{Wb}\left(\frac{{10}^3\text{mWb}}{1\text{Wb}}\right)\\ =18.12\text{mWb}\end{array}$

Conclusion:

Therefore, the magnetic flux through the coilis $18.12\text{mWb}$ .