Chapter 22, Problem 86P

(a)

To determine

The alignment of axis of coil.

Answer to Problem 86P

The alignment of axis of coil has to be with magnetic field.

Explanation of Solution

For the best result of magnetic dipole antenna which is used to detect electromagnetic waves, the magnetic flux associated with the axis of the coil have to be the highest.

The alignment of axis of coil has to be with magnetic field so that the total magnetic flux in the coil will be the maximum.

(b)

To determine

The amplitude of the induced emf in the coil.

Answer to Problem 86P

The amplitude of the induced emf in the coil is $3.6\text{mV}$.

Explanation of Solution

Write the expression for Faraday’s law.

  $\left|\epsilon \right|=N\frac{\Delta {\Phi }_B}{\Delta t}$                                                                                        (I)

Here, $\left|\epsilon \right|$ is the induced emf in the coil, $N$ is the number of turns in coil, and $\frac{\Delta {\Phi }_B}{\Delta t}$ is the change in magnetic flux.

Write the expression to find the maximum magnetic flux.

  $\begin{array}{c}{\Phi }_B=BA\\ =B\left(\pi r^2\right)\end{array}$                                                                                         (II)

Here, $B$ is the magnetic field, $A$ is the area of the coil, and $r$ is the radius of coil.

Substitute equation (II) in (I) to find the amplitude of the induced emf in the coil.

  $\begin{array}{c}\left|\epsilon \right|=N\frac{\Delta \left(B\left(\pi r^2\right)\right)}{\Delta t}\\ =N\pi r^2\frac{\Delta B}{\Delta t}\end{array}$                                                                                       (III)

Consider that field is sinusoidal.

Write the expression to find the magnetic field.

  $B=B_m\sin \left(kz+\omega t\right)$                                                                                        (IV)

Here, $k$ is the wave number, $\omega$ is the angular velocity, and $B_m$ is the amplitude of magnetic field.

Substitute equation (IV) in (III) to find the induced emf in the coil.

  $\begin{array}{c}\left|\epsilon \right|=N\pi r^2\frac{\Delta \left(B_m\sin \left(kz+\omega t\right)\right)}{\Delta t}\\ =N\pi r^2\frac{\Delta \left(\omega B_m\cos \left(kz+\omega t\right)\right)}{\Delta t}\end{array}$

Re-write the expression considering the maximum value of $\frac{\Delta B}{\Delta t}$.

  $\left|\epsilon \right|=N\pi r^2\omega B_m$

Conclusion:

Substitute $50$ for $N$, $5.0\text{cm}$ for $r$, $870\text{kHz}$ for $\omega$, and $1.7\times {10}^{-9}\text{T}$ for $B_m$ to find the amplitude of the induced emf in the coil.

  $\begin{array}{c}\left|\epsilon \right|=\left(50\right)\pi {\left(5.0\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2\left(870\text{kHz}\frac{{10}^3\text{Hz}}{1.0\text{kHz}}\right)\left(1.7\times {10}^{-9}\text{T}\right)\\ =\left(50\right)\pi {\left(5.0\times {10}^{-2}\text{m}\right)}^2\left(870\times {10}^3\text{Hz}\right)\left(1.7\times {10}^{-9}\text{T}\right)\\ =3.6\times {10}^{-3}\text{V}\left(\frac{1\text{mV}}{{10}^{-3}\text{V}}\right)\\ =3.6\text{mV}\end{array}$

(c)

To determine

The amplitude of emf induced in an electric dipole antenna.

Answer to Problem 86P

The amplitude of emf induced in an electric dipole antenna is $25\text{mV}$.

Explanation of Solution

Write the expression to find the amplitude of emf induced in an electric dipole antenna.

  ${\epsilon }_m=E_{m}L$

Here, ${\epsilon }_m$ is the amplitude of emf induced in an electric dipole antenna, $E_m$ is the electric field amplitude, and $L$ is the length of antenna.

Conclusion:

Substitute $5.0\text{cm}$ for $L$ and $0.50\text{V/m}$ for $\lambda$ to find the amplitude of emf induced in an electric dipole antenna.

  $\begin{array}{c}{\epsilon }_m=\left(0.50\text{V/m}\right)\left(5.0\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)\\ =25\times {\text{10}}^{-3}\text{V}\left(\frac{1\text{mV}}{{\text{10}}^{-3}\text{V}}\right)\\ =25\text{mV}\end{array}$