Chapter 27, Problem 38P

(a)

To determine

The magnetic field at the unoccupied corner.

Answer to Problem 38P

The magnetic field at the unoccupied corner is $B=\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x+\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y$ .

Explanation of Solution

Given:

Thecurrent carried by each of the wires is $I$ .

The length of each wire is same that is $L$ .

Formula used:

The expression for the magnetic field of wirein terms of vector form is given by,

  $B=\frac{{\mu }_{o}I}{2\pi L}\widehat{a}$

Calculation:

All the currents are in to the page. Say the direction of each current is $-{\widehat{a}}_x$ .

The magnitude of magnetic field by current $I_1$ in vector form is calculated as,

  $\begin{array}{c}{\overrightarrow{B}}_{I_1}=\frac{{\mu }_o{I}_1}{2\pi L}\left(-{\widehat{a}}_x\times {\widehat{a}}_y\right)\\ =\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z\end{array}$

The magnitude of magnetic field by current $I_2$ in vector form is calculated as,

  $\begin{array}{c}{\overrightarrow{B}}_{I_2}=\frac{{\mu }_o{I}_2}{2\pi \left(\sqrt{2}L\right)}-{\widehat{a}}_x\times \left({\widehat{a}}_y\times {\widehat{a}}_z\right)\\ =\frac{-{\mu }_o{I}_2}{2\sqrt{2}\pi L}{\widehat{a}}_z\times {\widehat{a}}_y\\ =\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x\end{array}$

The magnitude of magnetic field by current $I_3$ in vector form is calculated as,

  $\begin{array}{c}{\overrightarrow{B}}_{I_3}=\frac{{\mu }_o{I}_3}{2\pi L}\left(-{\widehat{a}}_x\times {\widehat{a}}_z\right)\\ =\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y\end{array}$

Thus, the net magnetic field is calculated as,

  $B=\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x+\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y$

Conclusion:

Therefore, themagnetic field at the unoccupied corner is $\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x+\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y$ .

(b)

To determine

The magnetic field at the unoccupied corner

Answer to Problem 38P

The magnetic field at the unoccupied corner is $\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x+\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y$

Explanation of Solution

Formula used:

The expression for the magnetic field of wire in terms of vector form is given by,

  $B=\frac{{\mu }_{o}I}{2\pi L}\widehat{a}$

Calculation:

The direction of $I_1\text{and}{I}_3$ is $-{\widehat{a}}_x$ and $I_2$ is $+{\widehat{a}}_x$ .

Thus, the net magnetic field is calculated as,

  $\begin{array}{c}B=B_1+B_2+B_3\\ =\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x\times \left({\widehat{a}}_y\times {\widehat{a}}_z\right)+\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y\\ =\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x+\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y\end{array}$

Conclusion:

Thereforethe magnetic field at the unoccupied corner is $\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x+\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y$ .

(c)

To determine

The magnetic field at the unoccupied corner

Answer to Problem 38P

The magnetic field at the unoccupied corner is $\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x-\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y$

Explanation of Solution

Formula used:

The expression for the magnetic field of wire in terms of vector form is given by,

  $B=\frac{{\mu }_{o}I}{2\pi L}\widehat{a}$

Calculation:

The direction of $I_1\text{and}{I}_2$ is in to the paper say their direction is $-{\widehat{a}}_x$ and $I_3$ is out of the paper and the direction is $+{\widehat{a}}_x$ .

Thus, the netmagnetic field is calculated as,

  $\begin{array}{c}B=B_1+B_2+B_3\\ =\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x+\frac{{\mu }_o{I}_3}{2\pi L}\left({\widehat{a}}_x\times {\widehat{a}}_z\right)\\ =\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x-\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y\end{array}$

Conclusion:

Therefore, the magnetic field at the unoccupied corner is $\frac{-{\mu }_o{I}_1}{2\pi L}{\widehat{a}}_z+\frac{{\mu }_o{I}_2}{2\pi \sqrt{2}L}{\widehat{a}}_x-\frac{{\mu }_o{I}_3}{2\pi L}{\widehat{a}}_y$.