Chapter 23, Problem 66P

(a)

To determine

The emf induced in the loop $POQ$.

Answer to Problem 66P

The emf induced is $0.125\text{V}$ in the clockwise direction.

Explanation of Solution

Write the equation for the emf induced in the loop.

    $\epsilon =-N\frac{d{\varphi }_B}{dt}$        (I)

Here, $N$ is the number of turns in the loop, ${\varphi }_B$ is the magnetic flux through the loop $POQ$. Write the equation for the magnetic flux through the coil.

    ${\varphi }_B=B.A$        (II)

Here, $B$ is the magnetic field and $A$ is the area of the loop $POQ$. Substitute equation (I) in equation (II).

    $\epsilon =-N\frac{d}{dt}\left(BA\cos \theta \right)$        (III)

It is given that the area of the loop is $\theta a^2/2$. Substitute this in equation (III).

    $\begin{array}{c}\epsilon =-N\frac{d}{dt}\left[B\left(\theta a^2/2\right)\cos \theta \right]\\ =-N\frac{d}{dt}B\frac{\theta a^2}{2}\cos \theta \\ =-N\frac{B{a}^2}{2}\frac{d\theta }{dt}\cos \theta \\ =-\frac{1}{2}NB{a}^2\omega \cos \theta \end{array}$

Here, $\omega$ is the angular frequency of rotation of the loop.

Conclusion:

Substitute $1$ for $N$, $0.500\text{T}$ for $B$, $50.0\text{cm}$ for $a$, $2.00\text{rad/s}$ for $\omega$ and $0°$ for $\theta$.

    $\begin{array}{c}\epsilon =-\frac{1}{2}\left(1\right)\left(0.500\text{T}\right){\left(0.500\text{m}\right)}^2\left(2.00\text{rad/s}\right)\cos 0°\\ =-0.125\text{V}\end{array}$

The negative sign shows the existence of clockwise current due to the induced emf. Therefore, the emf induced is $0.125\text{V}$ in the clockwise direction.

(b)

To determine

The current in the loop $POQ$

Answer to Problem 66P

The current in the loop is $0.020\text{A}$ in the clockwise direction.

Explanation of Solution

Write the equation for the angle $\theta$ in the loop $POQ$.

    $\theta =\omega t$        (IV)

Here, $\omega$ is the angular speed of rotation of the loop and $t$ is the time.

Substitute $2.00\text{rad/s}$ for $\omega$ and $0.250\text{s}$ for $t$.

    $\begin{array}{c}\theta =\left(2.00\text{rad/s}\right)\left(0.250\text{s}\right)\\ =0.500\text{rad}\end{array}$

Write the equation for the length of the arc $PQ$.

    $PQ=a\theta$        (V)

Here, $\theta$ is the angle given in the loop $POQ$ and $a$ is the length of the rod connecting $O$ and $P$. Substitute $0.500\text{rad}$ for $\theta$ and $50.0\text{cm}$ for $a$.

    $\begin{array}{c}PQ=\left(0.500\text{m}\right)\left(0.500\text{rad}\right)\\ =0.250\text{m}\end{array}$

The length of the loop $POQ$ is the sum of the arc $PQ$, rod connecting $O$ and $P$ and the pivoted rod connecting $O$ and $Q$.

    $\begin{array}{c}POQ=0.500\text{m}\text{+}0.500\text{m}\text{+ }0.250\text{m}\\ =\text{1}\text{.25}\text{m}\end{array}$

The resistance per unit length of the loop $POQ$ is $5.00\text{Ω/m}$ and the length of the loop $POQ$ is $\text{1}\text{.25}\text{m}$. Therefore, the resistance of the loop $POQ$ is the resistance per unit length multiplied the length of the loop $POQ$.

    $\begin{array}{c}R=\left(5.00\text{Ω/m}\right)\left(\text{1}\text{.25}\text{m}\right)\\ =6.25\Omega \end{array}$

Write the equation for the induced current in the loop.

    $I=\frac{\epsilon }{R}$

Here, $\epsilon$ is the induced emf and $R$ is the resistance of the loop $POQ$.

Conclusion:

Substitute $0.125\text{V}$ for $\epsilon$ and $6.25\Omega$ for $R$.

    $\begin{array}{c}I=\frac{0.125\text{V}}{6.25\Omega }\\ =0.020\text{A}\end{array}$

Therefore, the current in the loop is $0.020\text{A}$ in the clockwise direction.