End of chapter exercises | Electromagnetism

Textbook Exercise 10.3

What did Hans Oersted discover about the relationship between electricity and magnetism?

He discovered that electricity and magnetism were related to one another. By passing an electric current through a metal wire suspended above a magnetic compass, Oersted was able to produce a definite motion of the compass needle in response to the current.

List two uses of electromagnetism.

cellular telephones, microwave ovens, radios, televisions, etc.

A uniform magnetic field of $\text{0,35}$ $\text{T}$ in the vertical direction exists. A piece of cardboard, of surface area $\text{0,35}$ $\text{m$^{2}$}$ is placed flat on a horizontal surface inside the field. What is the magnetic flux through the cardboard?

\begin{align*} \phi & = BA \cos \theta \\ & = (\text{0,35})(\text{0,35}) \\ & = \text{0,1225}\text{ Wb} \end{align*}

The one edge is then lifted so that the cardboard is now inclined at $\text{17}$$\text{°}$ to the positive $x$-direction. What is the magnetic flux through the cardboard? What will the induced emf be if the field drops to zero in the space of $\text{3}$ $\text{s}$? Why?

When the piece of cardboard is inclined the flux is:

\begin{align*} \phi & = BA \cos \theta \\ & = (\text{0,35})(\text{0,35}) \cos (\text{17}) \\ & = \text{0,117} \end{align*}

The change in flux is $\text{0,117}$ $\text{Wb}$ and the induced emf will be zero because the cardboard is not a conductor.

A uniform magnetic field of $\text{5}$ $\text{T}$ in the vertical direction exists. What is the magnetic flux through a horizontal surface of area $\text{0,68}$ $\text{m$^{2}$}$? What is the flux if the magnetic field changes to being in the positive $x$-direction?

For the first case:

\begin{align*} \phi & = BA \cos \theta \\ & = (\text{5})(\text{0,68}) \\ & = \text{3,4}\text{ Wb} \end{align*}

If the flux is in the $x$-direction the flux will be $\text{0}$.

$\text{3,4}$ $\text{Wb}$ and $\text{0}$ $\text{Wb}$

A uniform magnetic field of $\text{5}$ $\text{T}$ in the vertical direction exists. What is the magnetic flux through a horizontal circle of radius $\text{0,68}$ $\text{m}$?

\begin{align*} \phi & = BA \cos \theta \\ & = (\text{5}) \pi (\text{0,68})^{2} \\ & = \text{7,26}\text{ Wb} \end{align*}

Consider a square coil of 3 turns with a side length of $\text{1,56}$ $\text{m}$. The coil is subjected to a varying magnetic field that changes uniformly from $\text{4,38}$ $\text{T}$ to $\text{0,35}$ $\text{T}$ in an interval of $\text{3}$ $\text{minutes}$. The axis of the solenoid makes an angle of $\text{197}$$\text{°}$ to the magnetic field. Find the induced emf.

\begin{align*} \mathcal{E} &= N\frac{\Delta\phi}{\Delta t} \\ & = N\frac{\phi_{f} - \phi_{i}}{\Delta t} \\ & = N\frac{B_{f}A\cos \theta - B_{i}A \cos \theta}{\Delta t} \\ & = N\frac{A\cos \theta(B_{f} - B_{i})}{\Delta t} \\ & = 3 \left( \frac{(\text{1,56})^{2} \cos (\text{197})(\text{0,35} - \text{4,38})}{180} \right) \\ & = \text{0,16}\text{ V} \end{align*}

Consider a solenoid coil of 13 turns with radius $\text{6,8} \times \text{10}^{-\text{2}}$ $\text{m}$. The solenoid is subjected to a varying magnetic field that changes uniformly from $-\text{5}$ $\text{T}$ to $\text{1,8}$ $\text{T}$ in an interval of $\text{18}$ $\text{s}$. The axis of the solenoid makes an angle of $\text{88}$$\text{°}$ to the magnetic field.

Find the induced emf.

\begin{align*} \mathcal{E} &= N\frac{\Delta\phi}{\Delta t} \\ & = N\frac{\phi_{f} - \phi_{i}}{\Delta t} \\ & = N\frac{B_{f}A\cos \theta - B_{i}A \cos \theta}{\Delta t} \\ & = N\frac{A\cos \theta(B_{f} - B_{i})}{\Delta t} \\ & = 13 \left( \frac{\pi(\text{6,8} \times \text{10}^{-\text{2}})^{2} \cos (\text{88})(\text{1,8} - (-\text{5}))}{18} \right) \\ & = \text{0,00249}\text{ V} \end{align*}

If the angle is changed to $\text{39}$$\text{°}$, what would the radius need to be for the emf to remain the same?

\begin{align*} \mathcal{E} &= N\frac{\Delta\phi}{\Delta t} \\ & = N\frac{\phi_{f} - \phi_{i}}{\Delta t} \\ & = N\frac{B_{f}A\cos \theta - B_{i}A \cos \theta}{\Delta t} \\ & = N\frac{A\cos \theta(B_{f} - B_{i})}{\Delta t} \\ \text{0,00249}& = 13 \left( \frac{\pi(r)^{2} \cos (\text{39})(\text{1,8} - (-\text{5}))}{18} \right) \\ \text{0,04482}& = \text{215,83}(r)^{2} \\ r^{2} & = \text{0,00002077}\\ r & = \text{0,014}\text{ m} \end{align*}

Consider a solenoid with 5 turns and a radius of $\text{4,3} \times \text{10}^{-\text{1}}$ $\text{mm}$. The axis of the solenoid makes an angle of $\text{11}$$\text{°}$ to the magnetic field.

Find the change in flux if the emf is $\text{0,12}$ $\text{V}$ over a period of $\text{0,5}$ $\text{s}$.

\begin{align*} \mathcal{E} &= N\frac{\Delta\phi}{\Delta t} \\ \text{0,12} & = 5 \frac{\Delta\phi}{\text{0,5}} \\ \Delta \phi & = \text{0,012}\text{ Wb} \end{align*}

Consider a rectangular coil of area $\text{1,73}$ $\text{m$^{2}$}$. The coil is subjected to a varying magnetic field that changes uniformly from $\text{2}$ $\text{T}$ to $\text{10}$ $\text{T}$ in an interval of $\text{3}$ $\text{ms}$. The axis of the solenoid makes an angle of $\text{55}$$\text{°}$ to the magnetic field. Find the induced emf.

\begin{align*} \mathcal{E} &= N\frac{\Delta\phi}{\Delta t} \\ & = N\frac{\phi_{f} - \phi_{i}}{\Delta t} \\ & = N\frac{B_{f}A\cos \theta - B_{i}A \cos \theta}{\Delta t} \\ & = N\frac{A\cos \theta(B_{f} - B_{i})}{\Delta t} \\ & = N \left( \frac{(\text{1,73}) \cos (\text{55})(\text{10} - \text{2})}{3} \right) \\ & = \text{2,65}\text{ V} \end{align*}

10.4 Chapter summary

Table of Contents

11.1 Introduction