Chapter 31, Problem 46P

(a)

To determine

The maximum emf induced between the ends of the conductor.

Answer to Problem 46P

The maximum emf induced between the ends of the conductor is $0.802\text{V}$.

Explanation of Solution

Write the expression to obtain the magnetic flux.

    $\varphi =BA\cos \omega t$

Here, $\varphi$ is the magnetic flux, $B$ is the magnetic field, $A$ is the area enclosed in the loop, $\omega$ is the angular speed and $t$ is the time taken.

Write the expression to obtain the induced emf between the ends of the conductor.

    $\epsilon =-\frac{d\varphi }{dt}$

Here, $\epsilon$ is the induced emf in the coil and $\frac{d\varphi }{dt}$ is the rate of change of magnetic flux.

Substitute $BA\cos \omega t$ for $\varphi$ in the above equation.

    $\begin{array}{c}\epsilon =-\frac{d\left(BA\cos \omega t\right)}{dt}\\ =-BA\frac{d}{dt}\left(\cos \omega t\right)\\ =BA\omega \sin \omega t\end{array}$

Substitute $\frac{1}{2}\pi R^2$ for $A$ as flux is lined with semi circle in the above equation.

    $\epsilon =B\left(\frac{1}{2}\pi R^2\right)\omega \sin \omega t$                                                                                           (I)

Here, $R$ is the radius of the coil.

For maximum induced emf between the ends of the conductor,

Substitute $\frac{1}{2}$ for $\sin \omega t$ in equation (I).

    $\begin{array}{c}{\epsilon }_{\max }=B\left(\frac{1}{2}\pi R^2\right)\omega \left(\frac{1}{2}\right)\\ =\frac{1}{4}B\pi R^2\omega \end{array}$

Conclusion:

Substitute $1.30\text{T}$ for $B$, $0.250\text{m}$ for $R$ and $120\text{rev}/\min$ for $\omega$ in the above equation to calculate ${\epsilon }_{\max }$.

    $\begin{array}{c}{\epsilon }_{\max }=\frac{1}{4}\left(1.30\text{T}\right)\pi {\left(0.250\text{m}\right)}^2\left(120\text{rev}/\min \right)\\ =\frac{1}{4}\left(1.30\text{T}\right)\frac{22}{7}{\left(0.250\text{m}\right)}^2\left(120\text{rev}/\min \times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\text{s}}\right)\\ =\frac{1}{4}\left(1.30\text{T}\right)\frac{22}{7}{\left(0.250\text{m}\right)}^2\left(12.6\text{rad}/\text{s}\right)\\ =0.802\text{V}\end{array}$

Therefore, the maximum emf induced between the ends of the conductor is $0.802\text{V}$.

(b)

To determine

The value of the average induced emf for each complete rotation.

Answer to Problem 46P

The value of the average induced emf for each complete rotation is $0.401\text{V}$.

Explanation of Solution

Write the expression to obtain the average induced emf for each complete rotation.

    ${\epsilon }_{\text{avg}}=\frac{{\epsilon }_{\max }+{\epsilon }_{\min }}{2}$                                                                                                      (II)

Here, ${\epsilon }_{\text{avg}}$ is the average induced emf for each complete rotation, ${\epsilon }_{\max }$ is the maximum emf and ${\epsilon }_{\min }$ is the minimum emf.

Conclusion:

Substitute $0.802\text{V}$ for ${\epsilon }_{\max }$ and $0$ for ${\epsilon }_{\min }$ in equation (II) to calculate ${\epsilon }_{\text{avg}}$.

    $\begin{array}{c}{\epsilon }_{\text{avg}}=\frac{0.802\text{V}+0}{2}\\ =0.401\text{V}\end{array}$

Therefore, the value of the average induced emf for each complete rotation is $0.401\text{V}$.

(c)

To determine

The variation in answers in part (a) and (b) when change in magnetic field is allowed to extent a distance $R$ above the axis of rotation.

Answer to Problem 46P

The answer in part (a) and (b) would be doubled when change in magnetic field is allowed to extent a distance $R$ above the axis of rotation.

Explanation of Solution

Consider equation (I).

    $\epsilon =B\left(\frac{1}{2}\pi R^2\right)\omega \sin \omega t$

Substitute $1$ for $\sin \omega t$ in equation (I) in case when change in magnetic field is allowed to extent a distance $R$ above the axis of rotation.

    ${\epsilon }_{\max }=\frac{1}{2}B\pi R^2\omega$

Conclusion:

Substitute $1.30\text{T}$ for $B$, $0.250\text{m}$ for $R$ and $120\text{rev}/\min$ for $\omega$ in the above equation to calculate ${\epsilon }_{\max }$.

    $\begin{array}{c}{\epsilon }_{\max }=\frac{1}{2}\left(1.30\text{T}\right)\pi {\left(0.250\text{m}\right)}^2\left(120\text{rev}/\min \right)\\ =\frac{1}{2}\left(1.30\text{T}\right)\frac{22}{7}{\left(0.250\text{m}\right)}^2\left(120\text{rev}/\min \times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\text{s}}\right)\\ =\frac{1}{2}\left(1.30\text{T}\right)\frac{22}{7}{\left(0.250\text{m}\right)}^2\left(12.6\text{rad}/\text{s}\right)\\ =1.604\text{V}\end{array}$

Thus, the maximum emf induced between the ends of the conductor is $1.604\text{V}$.

Substitute $1.604\text{V}$ for ${\epsilon }_{\max }$ and $0$ for ${\epsilon }_{\min }$ in equation (II) to calculate ${\epsilon }_{\text{avg}}$.

    $\begin{array}{c}{\epsilon }_{\text{avg}}=\frac{1.604\text{V}+0}{2}\\ =0.802\text{V}\end{array}$

Thus, the value of the average induced emf for each complete rotation is $0.802\text{V}$.

Therefore, the answer in part (a) and (b) would be doubled when change in magnetic field is allowed to extent a distance $R$ above the axis of rotation.

(d)

To determine

The graph between emf versus time when the magnetic field is drawn as shown in the figure P31.46.

Answer to Problem 46P

The graph between emf versus time is as shown below.

Explanation of Solution

When change in magnetic field is not allowed to extent a distance $R$ above the axis of rotation, then only for the half rotation of the coil, the coil is under the influence of magnetic field and in other half rotation, there is no influence of magnetic field.

Thus, the emf is only in the coil for the half rotation and for the other half rotation, there is no induced emf in the coil.

The graph between emf versus time is as shown below.

Figure-(1)

(d)

To determine

The graph between emf versus time when the magnetic field is extended as described in part (c).

Answer to Problem 46P

The graph between emf versus time is as shown in the figure below.

Explanation of Solution

When the change in magnetic field is allowed to extent a distance $R$ above the axis of rotation, then for the full rotation of the coil, the coil is under the influence of magnetic field.

Thus, the emf is induced in the coil in the whole rotation of the coil.

The graph for this case is as shown in the figure below.

Figure-(2)