Chapter 31, Problem 67AP

(a)

To determine

The induced emf in the closed loop $POQ$.

Answer to Problem 67AP

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

Explanation of Solution

Write the expression for emf induced in the coil due to change in magnetic field.

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

Here, $\epsilon$ is the induced emf, ${\Phi }_{\text{B}}$ is the magnetic flux through loop and $t$ is the time taken.

Write the expression for magnetic flux through loop.

    ${\Phi }_{\text{B}}=BAN\cos \theta$                                                                                                      (II)

Here, $B$ is the magnetic field, $N$ is the number of turns in the coil, $\theta$ is the angle between the magnetic field and the normal vectors of the loop and $A$ is the area of the loop.

Substitute $1$ for $N$, $\frac{\theta a^2}{2}$ for $A$ and $0°$ for $\theta$ in equation (II) to calculate ${\Phi }_{\text{B}}$.

    $\begin{array}{c}{\Phi }_{\text{B}}=B\left(\frac{\theta a^2}{2}\right)\left(1\right)\cos 0°\\ =B\left(\frac{\theta a^2}{2}\right)\end{array}$

Substitute $B\left(\frac{\theta a^2}{2}\right)$ for ${\Phi }_{\text{B}}$ in equation (I) to calculate $\epsilon$.

    $\begin{array}{c}\epsilon =-\frac{d\left(B\left(\frac{\theta a^2}{2}\right)\right)}{dt}\\ =-B\frac{d\left(\frac{\theta a^2}{2}\right)}{dt}\\ =-\frac{B{a}^2}{2}\left(\frac{d\theta }{dt}\right)\end{array}$                                                                                                 (III)

Write the expression for angular velocity.

    $\omega =\frac{d\theta }{dt}$                                                                                                                  (IV)

Here, $\omega$ is the angular velocity.

Substitute $\omega$ for $\frac{d\theta }{dt}$ in equation (III) to calculate $\epsilon$.

    $\epsilon =-\frac{B{a}^2}{2}\left(\omega \right)$                                                                                                           (V)

Conclusion:

Substitute $50.0\text{cm}$ for $a$, $0.500\text{T}$ for $B$ and $2.00\text{rad}/\text{s}$ for $\omega$ in equation (V) to calculate $\epsilon$.

    $\begin{array}{c}\epsilon =-\frac{\left(0.500\text{T}\right){\left(50.0\text{cm}\left(\frac{1\text{m}}{1\times {10}^2\text{cm}}\right)\right)}^2}{2}\left(2.00\text{rad}/\text{s}\right)\\ =-\frac{\left(0.500\text{T}\right){\left(0.5\text{m}\right)}^2}{2}\left(2.00\text{rad}/\text{s}\right)\\ =-\frac{0.25}{2}\text{V}\\ \text{=}-0.125\text{V}\end{array}$

Therefore, the emf induced in the loop is $0.125\text{V}$ in clockwise direction.

(b)

To determine

The induced current in the loop $POQ$.

Answer to Problem 67AP

The induced current in the loop is $0.02\text{A}$.

Explanation of Solution

The following figure shows the loop $POQ$ and the length of the sides.

Figure-(1)

From figure (1) it is shown that the length of the arc $PQ$ is $l\theta$ and length of the sides $PO$ and $OQ$ is $l$.

Rewrite equation (IV) for $\theta$.

    $\begin{array}{c}\omega =\frac{\theta }{t}\\ \theta =\omega t\end{array}$                                                                                                                    (VI)

Total length of the loop is,

    $L=OP+OQ+PQ$

Substitute $l\theta$ for $PQ$ and l for $PO$ and $OQ$ in the above equation.

    $L=l+l+l\theta$                                                                                                           (VII)

Resistance per length for the conductor is given as.

    $\frac{R}{L}=5.00\Omega /\text{m}$                                                                                                      (VIII)

Write the expression for induced current in the loop.

    $I=\frac{\epsilon }{R}$                                                                                                                      (IX)

Here, $I$ is the induced current.

Conclusion:

Substitute $2.00\text{rad}/\text{s}$ for $\omega$ and $0.250\text{s}$ for $t$ in Equation (VI) to calculate $\theta$.

    $\begin{array}{c}\theta =\left(2.00\text{rad}/\text{s}\right)\left(0.250\text{s}\right)\\ =0.5\text{rad}\end{array}$

Substitute $50.0\text{cm}$ for $l$ and $0.5\text{rad}$ for $\theta$ in Equation (VII) to calculate $L$.

    $\begin{array}{c}L=\left[\left(50.0\text{cm}\right)+\left(50.0\text{cm}\right)+\left(50.0\text{cm}\right)\left(0.5\text{rad}\right)\right]\left(\frac{1\text{m}}{1\times {10}^2\text{cm}}\right)\\ =\left(0.5\text{m}\right)+\left(0.5\text{m}\right)+\left(0.5\text{m}\right)\left(0.5\text{rad}\right)\\ =1\text{m}+\text{0}\text{.25}\text{m}\\ =\text{1}\text{.25}\text{m}\end{array}$

Substitute $\text{1}\text{.25}\text{m}$ for $L$ in Equation (VIII) to calculate $R$.

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

Substitute $0.125\text{V}$ for $\epsilon$ and $6.25\Omega$ for $R$ in Equation (IX) to calculate $I$.

    $\begin{array}{c}I=\frac{0.125\text{V}}{6.25\Omega }\\ =0.02\text{A}\end{array}$

Therefore, the induced current in the loop is $0.02\text{A}$.