Chapter 32, Problem 62PQ

To determine

The difference between the two rotations and check whether they are indistinguishable.

Answer to Problem 62PQ

When the coil is parallel to the field and is rotated to a position where it is perpendicular to the field, the emf induced is found to be negative. The emf is found to be positive, when the coil is rotated back to its initial position. Both the rotations are distinguishable.

Explanation of Solution

Faraday’s law states that, when the magnetic flux changes an emf is induced in the coil.

The direction of the induced emf is given by Lenz law. Lenz law states that the change in flux is opposed by the current induced in the circuit due to a change in magnetic field. The current induced in the circuit, exerts a mechanical force as well.

Write the expression for induced emf from Faraday’s and Lenz law.

    $\epsilon =-\frac{d{\varphi }_{\text{B}}}{dt}$                                                                                                            (I)

Here, $\epsilon$ is the induced emf, ${\varphi }_{\text{B}}$ is the magnetic flux linked with the coil and $t$ is the time during which the flux linked with the coil changes.

Write the expression for magnetic flux.

    ${\varphi }_{\text{B}}=\overrightarrow{B}\cdot \overrightarrow{A}$

Here, $\overrightarrow{B}$ is the strength of magnetic field and $\overrightarrow{A}$ is the area vector of the coil.

Write the expression for magnitude of the vector $\overrightarrow{A}$.

    $A=\pi r^2$

Here, $A$ is the area of the loop and $r$ is the radius of the coil.

Write the expression for magnitude of the magnetic flux.

    ${\varphi }_{\text{B}}=BA\cos \theta$

Here, $B$ is the magnetic field and $\theta$ is the angle between $\overrightarrow{B}$ and $\overrightarrow{A}$.

Write the expression for initial flux linked with the coil.

    ${\left({\varphi }_{\text{B}}\right)}_{\text{i}}=BA\cos {\theta }_{\text{i}}$

Here, ${\theta }_{\text{i}}$ is the initial angle between $\overrightarrow{B}$ and $\overrightarrow{A}$.

Substitute $90.0°$ for ${\theta }_{\text{i}}$ in the above equation to find ${\left({\varphi }_{\text{B}}\right)}_{\text{i}}$.

    $\begin{array}{c}{\left({\varphi }_{\text{B}}\right)}_{\text{i}}=BA\cos 90.0°\\ =0\end{array}$

Write the expression for final flux linked with the coil.

    ${\left({\varphi }_{\text{B}}\right)}_{\text{f}}=BA\cos {\theta }_{\text{f}}$

Here, ${\left({\varphi }_{\text{B}}\right)}_{\text{f}}$ is the flux linked with the coil after it is rotated by an angle of $90°$ and ${\theta }_{\text{f}}$ is the final angle between the vectors $\overrightarrow{B}$ and $\overrightarrow{A}$.

Substitute $0°$ for ${\theta }_{\text{f}}$ in the above equation to find ${\left({\varphi }_{\text{B}}\right)}_{\text{f}}$.

    $\begin{array}{c}{\left({\varphi }_{\text{B}}\right)}_{\text{f}}=BA\cos 0°\\ =BA\end{array}$

Substitute $\pi r^2$ for $A$ in the above equation to find ${\left({\varphi }_{\text{B}}\right)}_{\text{f}}$.

    ${\left({\varphi }_{\text{B}}\right)}_{\text{f}}=B\left(\pi r^2\right)$                                                                                                      (II)

Write the equation for change in magnetic flux.

    $d{\varphi }_{\text{B}}={\left({\varphi }_{\text{B}}\right)}_{\text{f}}-{\left({\varphi }_{\text{B}}\right)}_{\text{i}}$                                                                                              (III)

The coil is now rotated back to its initial position. The coil now rotates from the position where the magnetic flux linked with it is $0.294\text{T}\cdot {\text{m}}^2$, to the final position when the flux linked with it becomes $0$.

Therefore, $d{{\varphi }^{\prime }}_{\text{B}}=-d{\varphi }_{\text{B}}$.

Write the expression for induced emf due to second rotation.

    ${\epsilon }^{\prime }=-\frac{d{{\varphi }^{\prime }}_{\text{B}}}{dt}$                                                                                                           (IV)

Conclusion:

Substitute $1.50\text{T}$ for $B$ and $0.25\text{m}$ for $r$ in equation (II) to find ${\left({\varphi }_{\text{B}}\right)}_{\text{f}}$.

    $\begin{array}{c}{\left({\varphi }_{\text{B}}\right)}_{\text{f}}=1.50\text{T}\times \text{3}\text{.14}\times \text{0}\text{.25}{\text{m}}^2\\ =0.294\text{T}\cdot {\text{m}}^{\text{2}}\end{array}$

Substitute $0.294\text{T}\cdot {\text{m}}^2$ for ${\left({\varphi }_{\text{B}}\right)}_{\text{f}}$ and $0$ for ${\left({\varphi }_{\text{B}}\right)}_{\text{f}}$ in equation (III) to find $d{\varphi }_{\text{B}}$.

    $\begin{array}{c}d{\varphi }_{\text{B}}=0.294\text{T}\cdot {\text{m}}^{\text{2}}-0\\ =0.294\text{T}\cdot {\text{m}}^2\end{array}$

Substitute $0.294\text{T}\cdot {\text{m}}^2$ for $d{\varphi }_{\text{B}}$ and $0.200\text{s}$ for $dt$ in equation (I) to find $\epsilon$.

    $\begin{array}{c}\epsilon =-\frac{0.294\text{T}\cdot {\text{m}}^2}{0.200\text{s}}\\ =-1.470\text{V}\end{array}$

Therefore, the current in the coil flows in the counter clockwise direction.

Substitute $-0.294\text{T}\cdot {\text{m}}^2$ for $d{{\varphi }^{\prime }}_{\text{B}}$ and $0.200\text{s}$ for $dt$ in equation (IV) to find ${\epsilon }^{\prime }$.

    $\begin{array}{c}{\epsilon }^{\prime }=-\frac{\left(-0.294\text{T}\cdot {\text{m}}^2\right)}{0.200\text{s}}\\ =1.470\text{V}\end{array}$

The induced current flows in the clockwise direction.

Therefore, when the coil is parallel to the field and is rotated to a position where it is perpendicular to the field, the emf induced is found to be negative. The emf is found to be positive, when the coil is rotated back to its initial position. Both the rotations are distinguishable.