Chapter 31, Problem 58PQ

To determine

The net current through the loop that must be present.

Answer to Problem 58PQ

The net current through the loop that must be present is $\underset{\_}{124.34\text{A}}$.

Explanation of Solution

Write the expression for the Ampere circuital law as.

    ${\displaystyle \oint B\cdot dl={\mu }_{0}I}$

Rearrange the above equation for $I$.

  $I=\frac{{\displaystyle \oint B.dl}}{{\mu }_0}$                                                                                                     (I)

Here, $B$ is the magnetic field , $l$ is length of the wire, ${\mu }_0$ is the permeability of free space and $I$ is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

  $\text{Circulation integral=}{\displaystyle \oint B\cdot dl}$                                                                        (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

  ${\displaystyle \oint B\cdot dl}={\left({\displaystyle \oint B\cdot dl}\right)}_x+{\left({\displaystyle \oint B\cdot dl}\right)}_y$                                                              (III)

Here, ${\left({\displaystyle \oint B\cdot dl}\right)}_x$ is the circulation integral in the x-direction, ${\left({\displaystyle \oint B\cdot dl}\right)}_y$ is the circulation integral in the y-direction and ${\displaystyle \oint B\cdot dl}$ is the total circulation integral.

Conclusion:

The length of the segment in the X-direction is $2.50\text{cm}$.

Substitute ${\left({\displaystyle \oint B\cdot dl}\right)}_x$ for ${\displaystyle \oint B\cdot dl}$ , $\left(0.125y\widehat{i}-0.125x\widehat{j}\right)\text{T}$ for $B$ and $dx\widehat{i}$ for $dl$ in equation (II).

  $\begin{array}{c}{\left({\displaystyle \oint B\cdot dl}\right)}_x={\displaystyle \oint \left\{\left(0.125y\widehat{i}-0.125x\widehat{j}\right)\text{T}\cdot dx\widehat{i}\right\}}\\ =0.125y{\displaystyle \underset{0}{\overset{2.50}{\int }}dx}\\ =0.125y{\left[x\right]}_0^{2.50}\\ =0.125y(2.50\text{cm})\end{array}$

Substitute $2.50\text{cm}$ for $y$ in above equation.

  $\begin{array}{c}{\left({\displaystyle \oint B\cdot dl}\right)}_x=0.125\left(2.50\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)\left(2.50\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)\\ =7.8125\times {10}^{-5}\text{T}\cdot \text{m}\end{array}$

The length of the segment in the Y-direction is $2.50\text{cm}$.

Substitute ${\left({\displaystyle \oint B\cdot dl}\right)}_y$ for ${\displaystyle \oint B\cdot dl}$ , $\left(0.125y\widehat{i}-0.125x\widehat{j}\right)\text{T}$ for $B$ and $dy\widehat{j}$ for $dl$ in equation (II).

  $\begin{array}{c}{\left({\displaystyle \oint B\cdot dl}\right)}_y={\displaystyle \oint \left\{\left(0.125y\widehat{i}-0.125x\widehat{j}\right)\text{T}\cdot dy\widehat{j}\right\}}\\ =0.125x{\displaystyle \underset{0}{\overset{2.50}{\int }}dy}\\ =0.125x{\left[y\right]}_0^{2.50}\\ =0.125x\left(2.50\text{ cm}\right)\end{array}$

Substitute $2.50\text{cm}$ for $x$ in above equation.

    $\begin{array}{c}{\left({\displaystyle \oint B\cdot dl}\right)}_y=0.125\left(\text{2}\text{.50}\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)\left(\text{2}\text{.50}\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)\\ =7.8125\times {10}^{-5}\text{T}\cdot \text{m}\end{array}$

Substitute $7.8125\times {10}^{-5}\text{T}\cdot \text{m}$ for ${\left({\displaystyle \oint B\cdot dl}\right)}_x$ and $7.8125\times {10}^{-5}\text{T}\cdot \text{m}$${\left({\displaystyle \oint B\cdot dl}\right)}_y$ in equation (III).

    $\begin{array}{c}{\displaystyle \oint B\cdot dl}=\left(7.8125\times {10}^{-5}\text{T}\cdot \text{m}\right)+\left(7.8125\times {10}^{-5}\text{T}\cdot \text{m}\right)\\ =1.5625\times {10}^{-4}\text{T}\cdot \text{m}\end{array}$

Substitute $1.5625\times {10}^{-4}\text{T}\cdot \text{m}$ for ${\displaystyle \oint B\cdot dl}$ and $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$ in equation (I).

  $\begin{array}{c}I=\frac{\left(1.5625\times {10}^{-4}\text{T}\cdot \text{m}\right)}{\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)}\\ =124.34\text{A}\end{array}$

Thus, the net current through the loop that must be present is $\underset{\_}{124.34\text{A}}$.