Chapter 32, Problem 70PQ

To determine

Whether the coil being slightly off-axis explains the power loss.

Answer to Problem 70PQ

If the coil is off center, the coil produces less power. The coil can produce more power by increasing the angular speed or frequency of rotation.

Explanation of Solution

Write the expression for power generated by the coil.

    $P_{\text{avg}}={\epsilon }_{\text{rms}}{I}_{\text{rms}}$                                                                                       (I)

Here, $P_{\text{avg}}$ is the power generated by the coil, ${\epsilon }_{\text{rms}}$ and $I_{\text{rms}}$ are the rms values of the emf and current respectively.

Write the expression for rms value of emf.

    ${\epsilon }_{\text{rms}}=\frac{{\epsilon }_0}{\sqrt{2}}$

Here, ${\epsilon }_0$ is the peak value of emf.

Write the expression for rms value of current.

    $I_{\text{rms}}=\frac{I_0}{\sqrt{2}}$

Here, $I_0$ is the peak value of current.

Substitute $\frac{{\epsilon }_0}{\sqrt{2}}$ for ${\epsilon }_{\text{rms}}$ and $\frac{I_0}{\sqrt{2}}$ for $I_{\text{rms}}$ in equation (I) to find $P_{\text{avg}}$.

    $P_{\text{avg}}=\frac{{\epsilon }_0{I}_0}{2}$                                                                                                  (II)

Write the expression for peak value of emf.

    ${\epsilon }_0=NBA\omega$

Here, $A$ is the area of the coil, $N$ is the number of turns of wire wound over the core, $B$ is the magnetic field strength and $\omega$ is the angular velocity of the coil.

Write the expression for peak value of current.

    $I_0=\frac{{\epsilon }_0}{R}$

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

Substitute $NBA\omega$ for ${\epsilon }_0$ in the above equation to find $I_0$.

    $\begin{array}{c}{I}_0=\frac{{\epsilon }_0}{R}\\ =\frac{NBA\omega }{R}\end{array}$

Substitute $NBA\omega$ for ${\epsilon }_0$ and $\frac{NBA\omega }{R}$ for $I_0$ in equation (II) to find $P_{\text{avg}}$.

    $P_{\text{avg}}=\frac{{\left(NBA\omega \right)}^2}{2R}$                                                                                                (III)

When the axis of rotation of the coil is at right angles to the direction of the magnetic field and to the area vector of the AC generator, the generator operates with peak efficiency.

The axis of rotation is no longer perpendicular to the area vector, if the coil is slightly off center.

The area vector is tilted at an angle $\varphi$ to the plane of the magnetic field, if the coil is off center as shown in the figure below.

Figure-(1)

Write the expression for average power if the coil is off center.

    ${P^{\prime }}_{\text{avg}}=\frac{{\left(NB{A}^{\prime }\omega \right)}^2}{2R}$

Here, $A^{\prime }$ is the area of the coil if the coil is off center and ${P^{\prime }}_{\text{avg}}$ is the average power if the coil is off-center.

Substitute $A\cos \varphi$ for $A^{\prime }$ in the above equation.

    ${P^{\prime }}_{\text{avg}}=\frac{{\left(NBA\omega \right)}^2{\cos }^2\varphi }{2R}$                                                                                     (IV)

Compare equation (III) and equation (IV).

    $P_{\text{avg}}={P^{\prime }}_{\text{avg}}{\cos }^2\varphi$

Therefore,

    ${P^{\prime }}_{\text{avg}}<P_{\text{avg}}$

Therefore, the power coil would produce, if the coil is slightly off center would be less than the power produced when the axis of the coil is in the plane of the magnetic field.

Consider equation (IV).

    ${P^{\prime }}_{\text{avg}}=\frac{{\left(NBA\omega \right)}^2{\cos }^2\varphi }{2R}$

The power can be restored to the previous value of $P_{\text{avg}}=\frac{{\left(NBA\omega \right)}^2}{2R}$, if the value of angular velocity is increased to counter the effect of the term ${\cos }^2\varphi$.

Therefore, if the coil is off center, the coil produces less power. The coil can produce more power by increasing its angular speed or frequency of rotation.