Chapter 31, Problem 71AP

(a)

To determine

The maximum induced emf in the coil.

Answer to Problem 71AP

The maximum induced emf in the coil is $36.0\text{V}$.

Explanation of Solution

Let $\theta$ be the angle between the normal to the coil and the magnetic field.

At $t=0$, $\theta =0$ and $\theta =\omega t$ at later times.

Write the expression for the emf induced in the coil.

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

Here, $N$ is the number of the loops, $A$ is the area of the loop, $B$ is the magnetic field and $\theta$ is angle between normal to the area vector and magnetic field.

Substitute $\theta =\omega t$ in the above expression to find $\epsilon$.

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

Write the expression for the area of the loop.

    $A=l.b$                                                                                                                   (II)

Here, $l$ is the length of the loop and $b$ is the breadth of the loop.

The maximum value of $\sin \theta$ is $1$

Conclusion:

Substitute $lb$ for $A$ in equation (I) to find $\epsilon$.

    $\epsilon =NlbB\omega \sin \omega t$                                                                                              (III)

Substitute $60$ for $N$, $0.100\text{m}$ for $l$ , $0.200\text{m}$ for $b$, $1.00\text{T}$ for $B$, $30.0\text{rad}/\text{s}$ for $\omega$ and $1$ for $\sin \omega t$ in equation (III) to find $\epsilon$.

    $\begin{array}{c}\epsilon =60\left(0.100\text{m}\right)\left(0.200\text{m}\right)\left(1.00\text{T}\right)\left(30.0\text{rad}/\text{s}\right)\\ =36.0\text{V}\end{array}$

Therefore, the maximum induced emf in the coil is $36.0\text{V}$.

(b)

To determine

The maximum rate of change of magnetic flux through coil.

Answer to Problem 71AP

The maximum rate of change of magnetic flux through coil is $0.60\text{Wb}/\text{s}$

Explanation of Solution

Write the expression for the rate of change of magnetic flux.

    $\begin{array}{c}\frac{d{\varphi }_{\text{B}}}{dt}=\frac{d\left(BA\cos \theta \right)}{dt}\\ =-Blb\omega \sin \omega t\end{array}$                                                                                            (IV)

The minimum value of $\sin \theta$ is $-1$.

Conclusion:

Substitute , $0.100\text{m}$ for $l$ , $0.200\text{m}$ for $b$, $1.00\text{T}$ for $B$, $30.0\text{rad}/\text{s}$ for $\omega$ and $-1$ for $\sin \omega t$ in equation (IV) to find $\frac{d{\varphi }_{\text{B}}}{dt}$.

    $\begin{array}{c}\frac{d{\varphi }_{\text{B}}}{dt}=\left(0.100\text{m}\right)\left(0.200\text{m}\right)\left(1.00\text{T}\right)\left(30.0\text{rad}/\text{s}\right)\left(-1\right)\\ =0.60\text{T}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\frac{1\text{Wb}}{1\text{T}\cdot {\text{m}}^{\text{2}}}\right)\\ =0.60\text{Wb}/\text{s}\end{array}$

Therefore, the maximum rate of change of magnetic flux through coil is $0.60\text{Wb}/\text{s}$

(c)

To determine

The emf induced at $t=0.05\text{s}$.

Answer to Problem 71AP

The emf induced at $t=0.05\text{s}$ is $35.9\text{V}$

Explanation of Solution

At $t=0.05\text{s}$,

    $\begin{array}{c}\theta =\left(30.0\text{rad}/\text{s}\right)\left(0.05\text{s}\right)\\ =1.5\text{rad}\\ \text{=1}\text{.5}\times \text{57}\text{.2958}\\ =85.9°\end{array}$

Conclusion:

Substitute $60$ for $N$, $0.100\text{m}$ for $l$ , $0.200\text{m}$ for $b$, $1.00\text{T}$ for $B$, $30.0\text{rad}/\text{s}$ for $\omega$ and $85.9°$ for $\omega t$ in equation (III) to find $\epsilon$.

    $\begin{array}{c}\epsilon =60\left(0.100\text{m}\right)\left(0.200\text{m}\right)\left(1.00\text{T}\right)\left(30.0\text{rad}/\text{s}\right)\sin \left(85.9°\right)\\ =35.9\text{V}\end{array}$

Therefore, the emf induced at $t=0.05\text{s}$ is $35.9\text{V}$

(d)

To determine

The torque exerted on the coil by the magnetic field when the emf is maximum.

Answer to Problem 71AP

The torque exerted on the coil by the magnetic field when the emf is maximum is $4.32\text{N}\cdot \text{m}$.

Explanation of Solution

The emf induced is maximum when $\theta =90°$.

Write the expression for the torque.

    $\begin{array}{c}\tau =BINA\sin \theta \\ ={\rm N}{\epsilon }_{\max }\frac{lbB}{R}\end{array}$                                                                                                    (V)

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

Conclusion:

Substitute $60$ for $N$, $36.0\text{V}$ for ${\epsilon }_{\max }$, $0.100\text{m}$ for $l$, $0.200\text{m}$ for $b$, $1.00\text{T}$ for $B$ and $10.0\Omega$ for $R$ in the equation (V) to find $\tau$.

    $\begin{array}{c}\tau =60\left(36.0\text{V}\right)\frac{\left(0.100\text{m}\right)\left(\text{0}\text{.200}\text{m}\right)\left(1.00\text{T}\right)}{10.0\Omega }\\ =4.32\text{N}\cdot \text{m}\end{array}$

Therefore, the torque exerted on the coil by the magnetic field when the emf is maximum is $4.32\text{N}\cdot \text{m}$.