Chapter 28, Problem 39P

(a)

To determine

The flux graph through the loop.

Answer to Problem 39P

The flux graph through the loop as a function of time is shown in figure 1.

Explanation of Solution

Given:

The length of the rectangular loop, $L_{\text{side}\text{of}\text{theloop}}=10\text{cm}$

The widthof the rectangular loop, $r=5\text{cm}$

The stance of the rectangular loop, $R=2.5\Omega$

The speed of the rectangular loop, $v=2.4\text{cm}/\text{s}$

The magnetic field of the rectangular loop, $B=1.7\text{T}$

Formula used:

The formula for flux through the loop as a function of time is given by,

  ${\varphi }_m=mt+b$

The expression for the time required to enter into uniform magnetic field is given by,

  $t=\frac{L_{\text{side}\text{of}\text{theloop}}}{v}$

The expression of flux in terms of the length is given by,

  ${\varphi }_m=Br{L}_{\text{side}\text{of}\text{the loop}}$

Calculation:

The time required to enter into uniform magnetic field is calculated as,

  $\begin{array}{c}t=\frac{L_{\text{side}\text{of}\text{theloop}}}{v}\\ =\frac{\left(10\text{cm}\left(\frac{{10}^{-2}\text{m}}{\text{1}\text{cm}}\right)\right)}{2.4\frac{\text{cm}}{\text{s}}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =\frac{0.10\text{m}}{2.4\times {10}^{-2}\text{m}/\text{s}}\\ =4.17\text{s}\end{array}$

The flux $\left(4<t<17\text{s}\right)$ through the loop can be calculated as,

  $\begin{array}{c}{\varphi }_m=Br{L}_{\text{side}\text{of}\text{the loop}}\\ =\left(1.7\text{T}\left(\frac{\text{Wb}/{\text{m}}^2}{1\text{T}}\right)\right)\times \left(5\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\times \left(10\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ =\left(1.7\frac{\text{Wb}}{{\text{m}}^{\text{2}}}\right)\left(0.05\text{m}\right)\left(0.10\text{m}\right)\\ =\left(8.5\times {10}^{-3}\text{Wb}\right)\end{array}$

For, $t=8.33\text{s}$ , the flux is calculated as,

  $8.50\times {10}^{-3}\text{Wb}=m\times 8.33\text{s}+b$   ..... (1)

For, $t=12.5\text{s}$ , the flux is calculated as,

  $\begin{array}{c}{\varphi }_m=mt+b\\ \text{0}\text{Wb}=m\times 12.5\text{s}+b\end{array}$ ...... (2)

From equation (1) and (2)

  $\begin{array}{c}m=2.04\times {10}^{-3}\text{Wb}\\ b=22.5\times {10}^{-3}\end{array}$

Now,

  $\begin{array}{c}{\varphi }_m=mt+b\\ =\left(2.04\times {10}^{-3}\text{Wb}\right)t+22.5\times {10}^{-3}\end{array}$

The induced current through the loop as a function of time is shown in figure 1.

  

Figure 1

Conclusion:

The graph of induced emf through the loop as a function of time is shown in the figure 1.

(b)

To determine

The graph of induced emf and current through the loop.

Answer to Problem 39P

The graph of induced emf and current through the loop is shown in the figure 2 and 3.

Explanation of Solution

Formula used:

The formula for induced emfthrough the loop is given by,

  $\epsilon =-\frac{d{\varphi }_m}{dt}$

Calculation:

The induced emf through the loop when $t>7\text{s}$ is calculated as,

  $\begin{array}{c}\epsilon =-\frac{d{\varphi }_m}{dt}\\ =-\frac{d\left(2.04\frac{\text{mWb}}{\text{s}}\right)}{dt}\\ =-\frac{d\left(2.04\frac{\text{mWb}}{\text{s}}\left(\frac{\text{1}\text{V}\cdot \text{s}}{1\text{Wb}}\right)\right)}{dt}\\ =-2.04\text{mV}\end{array}$

Theinduced emfthrough the loop when $t>12.5\text{s}$ is calculated as,

  $\begin{array}{c}\epsilon =-t\frac{d{\varphi }_m}{dt}\\ -\frac{d}{dt}\left(\left(2.04\frac{\text{mWb}}{\text{s}}\left(\frac{\text{1}\text{V}\cdot \text{s}}{1\text{Wb}}\right)\right)+\left(25.5\times {10}^{-3}\right)\right)\\ -\frac{d}{dt}\left(\left(2.04\text{mV}\right)+\left(25.5\times {10}^{-3}\right)\right)\\ =-2.04\text{mV}\end{array}$

The current through the loop when $t>12.5\text{s}$ can be defined as,

  $\epsilon =-2.04\text{mV}$

The graph for the emf is shown in figure 2.

  

Figure 2

The expression for the value of current is given by,

  $I=\frac{\epsilon }{R}$

The graph for the current is shown in figure 3.

  

Figure 3

Conclusion:

The graph of induced emfand current through the loop is shown in the figure 2 and 3.