Chapter 31, Problem 31.71AP

(a)

To determine

The maximum induced emf in the coil.

Answer to Problem 31.71AP

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

Explanation of Solution

Given Info: The number of turns in the rectangular coil is $60$, the dimensions of the coil is $0.100\text{m}$ by $0.200\text{m}$, the resistance of the coil is $10.0\Omega$, the angular speed of the coil is $30.0\text{rad}/\text{s}$ and the magnetic field along $x$ axis is $1.00\text{T}$.

The area of the rectangular coil is,

$A=l\times b$

Here,

$l$ is the length of the coil.

$b$ is the breadth of the coil.

Substitute $0.100\text{m}$ for $l$ and $0.200\text{m}$ for $b$ in the above equation.

$\begin{array}{c}A=0.100\text{m}\times 0.200\text{m}\\ =\text{0}\text{.0200}{\text{m}}^2\end{array}$

Thus, the area of the coil is $\text{0}\text{.0200}{\text{m}}^2$.

The angle between the normal area component and the magnetic field is,

$\theta =\omega t$

Here,

$\omega$ is the angular speed of the coil.

$t$ is the time.

The flux induced in the coil is,

$\begin{array}{c}\varphi =N\left(\overrightarrow{B}\cdot \overrightarrow{A}\right)\\ =NBA\cos \theta \end{array}$

Here,

$N$ is the number of turns.

$B$ is the magnetic field.

The emf induced in the coil is,

$\epsilon =-\frac{d}{dt}\left(\varphi \right)$

Substitute $NBA\cos \theta$ for $\varphi$ in the above equation.

$\begin{array}{c}\epsilon =-\frac{d}{dt}\left(NBA\cos \theta \right)\\ =-NBA\frac{d}{dt}\left(\cos \theta \right)\end{array}$

Substitute $\omega t$ for $\theta$ in the above equation.

$\begin{array}{c}\epsilon =-NBA\frac{d}{dt}\left(\cos \omega t\right)\\ =+NBA\omega \sin \omega t\end{array}$

The induced emf is maximum for the maximum value of $\sin \omega t$ and the maximum value of any sine function is $1$.

Substitute $1$ for $\sin \omega t$ in the above equation.

$\begin{array}{c}\epsilon =+NBA\omega \left(1\right)\\ =NBA\omega \end{array}$

Substitute $60$ for $N$, $\text{0}\text{.0200}{\text{m}}^2$ for $A$, $30.0\text{rad}/\text{s}$ for $\omega$ and $1.00\text{T}$ for $B$ in the above equation.

$\begin{array}{c}\epsilon =\left(60\right)\left(1.00\text{T}\right)\left(\text{0}\text{.0200}{\text{m}}^2\right)\left(30.0\text{rad}/\text{s}\right)\\ =36.0\text{V}\end{array}$

Conclusion:

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

(b)

To determine

The maximum rate of change of magnetic flux.

Answer to Problem 31.71AP

The maximum rate of change of magnetic flux is $0.600\text{T}\cdot {\text{m}}^2/\text{s}$.

Explanation of Solution

Given Info: The number of turns in the rectangular coil is $60$, the dimensions of the coil is $0.100\text{m}$ by $0.200\text{m}$, the resistance of the coil is $10.0\Omega$, the angular speed of the coil is $30.0\text{rad}/\text{s}$ and the magnetic field along $x$ axis is $1.00\text{T}$.

The induced flux is,

$\varphi =BA\cos \theta$

Substitute $\omega t$ for $\theta$ in the above equation.

$\varphi =BA\cos \omega t$

Differentiate the above equation with respect to time with respect to time for the rate of change of flux.

$\begin{array}{c}\frac{d\varphi }{dt}=\frac{d}{dt}\left(BA\cos \omega t\right)\\ =-BA\omega \sin \omega t\end{array}$

For the maximum rate of change of flux the value of $\sin \omega t$ is minimum and the minimum value of the any sine function is $\left(-1\right)$.

Substitute $\left(-1\right)$ for $\sin \omega t$ in the above equation.

$\begin{array}{c}\frac{d\varphi }{dt}=-BA\omega \left(-1\right)\\ =BA\omega \end{array}$

Substitute $\text{0}\text{.0200}{\text{m}}^2$ for $A$, $30.0\text{rad}/\text{s}$ for $\omega$ and $1.00\text{T}$ for $B$ in the above equation.

$\begin{array}{c}\frac{d\varphi }{dt}=\left(1.00\text{T}\right)\left(\text{0}\text{.0200}{\text{m}}^2\right)\left(30.0\text{rad}/\text{s}\right)\\ =0.600\text{T}\cdot {\text{m}}^2/\text{s}\end{array}$

Conclusion:

Therefore, the maximum rate of change of magnetic flux is $0.600\text{T}\cdot {\text{m}}^2/\text{s}$.

(c)

To determine

The induced emf at $t=0.050\text{s}$.

Answer to Problem 31.71AP

The induced emf at $t=0.050\text{s}$ is $35.9\text{V}$.

Explanation of Solution

Given Info: The number of turns in the rectangular coil is $60$, the dimensions of the coil is $0.100\text{m}$ by $0.200\text{m}$, the resistance of the coil is $10.0\Omega$, the angular speed of the coil is $30.0\text{rad}/\text{s}$ and the magnetic field along $x$ axis is $1.00\text{T}$.

The expression for the induced emf is,

$\epsilon =NBA\omega \sin \omega t$

Substitute $60$ for $N$, $\text{0}\text{.0200}{\text{m}}^2$ for $A$, $30.0\text{rad}/\text{s}$ for $\omega$, $0.0500\text{s}$ for $t$ and $1.00\text{T}$ for $B$ in the above equation.

$\begin{array}{c}\epsilon =\left(60\right)\left(1.00\text{T}\right)\left(0.0200{\text{m}}^2\right)\left(30.0\text{rad}/\text{s}\right)\sin \left(30.0\text{rad}/\text{s}\times 0.0500\text{s}\right)\\ =\left(36\times 0.997\right)\text{V}\\ \text{=35}\text{.9}\text{V}\end{array}$

Conclusion:

Therefore, the induced emf at $t=0.050\text{s}$ is $35.9\text{V}$.

(d)

To determine

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

Answer to Problem 31.71AP

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

Explanation of Solution

Given Info: The number of turns in the rectangular coil is $60$, the dimensions of the coil is $0.100\text{m}$ by $0.200\text{m}$, the resistance of the coil is $10.0\Omega$, the angular speed of the coil is $30.0\text{rad}/\text{s}$ and the magnetic field along $x$ axis is $1.00\text{T}$.

The formula to calculate the induced current in the coil is,

$I=\frac{{\epsilon }_{\max }}{R}$

Here,

${\epsilon }_{\max }$ is the maximum induced emf.

$R$ is the resistance of the coil.

Substitute $10.0\Omega$ for $R$ and $36.0\text{V}$ for ${\epsilon }_{\max }$ in the above equation.

$\begin{array}{c}I=\frac{36.0\text{V}}{10.0\Omega }\\ =3.60\text{A}\end{array}$

Thus, the current in the coil is $3.60\text{A}$.

The formula to calculate the magnetic moment of the coil is,

$\mu =NIA$

The formula to calculate the torque experienced by the coil is,

$\begin{array}{c}\tau =\mu \times B\\ =\mu B\sin \theta \end{array}$

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

Substitute $1$ for $\sin \theta$ in the above equation.

$\begin{array}{c}\tau =\mu B\left(1\right)\\ =\mu B\end{array}$

Substitute $NIA$ for $\mu$ in the above equation.

$\tau =NIAB$

Substitute $60$ for $N$, $0.0200{\text{m}}^2$ for $A$, $3.60\text{A}$ for $I$, $1.00\text{T}$ for $B$ in the above equation.

$\begin{array}{c}\tau =\left(60\right)\left(3.60\text{A}\right)\left(0.0200{\text{m}}^2\right)\left(1.00\text{T}\right)\\ =4.32\text{N}\cdot \text{m}\end{array}$

Conclusion:

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