Chapter 20, Problem 44E

(a)

To determine

The magnitude of flux due to the current in the strip.

Answer to Problem 44E

The elemental flux through the strip is $\frac{2c{k}^{\prime }i}{r}dr$.

Explanation of Solution

Write the expression for flux in the through strip.

    $d\varphi =B\cdot dA$        (1)

Here, $d\varphi$ is the small amount of flux, $B$ is the magnetic field and $dA$ is the elemental area.

Write the expression for the magnetic field in the strip.

    $B=\frac{2k^{\prime }i}{r}$

Here, $k^{\prime }$ is the magnetic constant, $i$ is the current in the wire and $r$ is the distance of the strip from the wire.

Write the expression for the area of the strip.

    $dA=cdr$

Here, $c$ is the length of the loop and $dr$ is small elemental width.

Conclusion:

Substitute $\frac{2k^{\prime }i}{r}$ for $B$ and $cdr$ for $dA$ in equation (1).

    $\begin{array}{c}d\varphi =\left(\frac{2k^{\prime }i}{r}\right)\cdot \left(cdr\right)\\ =\frac{2c{k}^{\prime }i}{r}dr\end{array}$

Thus, the elemental flux through the strip is $\frac{2c{k}^{\prime }i}{r}dr$.

(b)

To determine

The magnitude of the flux through the loop.

Answer to Problem 44E

The flux through the loop is $2c{k}^{\prime }i\left[ln\left(\frac{b-a}{a}\right)\right]$.

Explanation of Solution

Write the expression for the elemental flux through the strip in the loop.

    $\begin{array}{c}d\varphi =\left(\frac{2k^{\prime }i}{r}\right)\cdot \left(cdr\right)\\ =\frac{2c{k}^{\prime }i}{r}dr\end{array}$        (2)

Conclusion:

Integrate equation (2).

    ${\displaystyle \int d\varphi }=2c{k}^{\prime }i{\displaystyle \underset{a}{\overset{b-a}{\int }}\frac{dr}{r}}$

Here, $a$ is the distance of the first side of the loop from the wire and $b$ is the distance of the last side of the loop from the wire.

Solve the above integration.

    $\begin{array}{c}\varphi =2c{k}^{\prime }i\left[ln\left(b-a\right)-ln\left(a\right)\right]\\ =2c{k}^{\prime }i\left[ln\left(\frac{b-a}{a}\right)\right]\end{array}$

Thus, the flux through the loop is $2c{k}^{\prime }i\left[ln\left(\frac{b-a}{a}\right)\right]$.

(c)

To determine

The magnitude of the induced EMF through the loop.

Answer to Problem 44E

The induced EMF through the loop is $\left(-2c{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}\right)$.

Explanation of Solution

Write the expression for induced EMF.

    $\epsilon =-\frac{d\varphi }{dt}$        (3)

Here, $\epsilon$ is the induced EMF.

Write the expression for the flux through the loop.

    $\varphi =2c{k}^{\prime }i\left[ln\left(\frac{b-a}{a}\right)\right]$        (4)

Conclusion:

Differentiate equation (4) with respect to $t$.

    $\frac{d\varphi }{dt}=2c{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}$

Here, $\frac{di}{dt}$ is the change in current and $\frac{d\varphi }{dt}$ is the change in flux.

Substitute $\left(2c{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}\right)$ for $\frac{d\varphi }{dt}$ in equation (3).

    $\epsilon =-2c{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}$

Thus, the induced EMF through the loop is $\left(-2c{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}\right)$.

(d)

To determine

The induced current in the loop.

Answer to Problem 44E

The induced current in the loop is $\left(-2\frac{c}{R}{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}\right)$.

Explanation of Solution

Write the expression for induced EMF in the loop.

    $\epsilon =-2c{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}$        (5)

Write the expression for the induced current in the loop.

    $i_1=\frac{\epsilon }{R}$        (6)

Here, $i_1$ is the induced current and $R$ is the resistance of the loop.

Conclusion:

Substitute $\left(-2c{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}\right)$ for $\epsilon$ in equation (6).

    $i_1=-2\frac{c}{R}{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}$

Thus, the induced current in the loop is $\left(-2\frac{c}{R}{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}\right)$.

(e)

To determine

The direction of $i_1$ for increasing in $i$..

Answer to Problem 44E

The direction of the induced current is clockwise.

Explanation of Solution

Write the expression for $i_1$.

    $i_1=-2\frac{c}{R}{k}^{\prime }\left[ln\left(\frac{b-a}{a}\right)\right]\frac{di}{dt}$

Conclusion:

The induced current is proportional to the change in $i$.

Thus, the direction of the induced current is clockwise.