Home Page Textbooks Physics: Principles and Problems Page 675: Practice Problems

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

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Page 675: Practice Problems

Exercise 1

Step 1

1 of 2

In order to solve this problem we are going to use the definition of the electro-motive force
$EMF=BLv$
a) Now, we can insert the values given in the problem

$$
EMF=0.4times 0.5times 20
$$

which gives that

$$
boxed{EMF=4textrm{V}}
$$

b) Now, can proceed to find the induced current in the circuit by employing Ohm’s law

$$
I=frac{EMF}{R}=frac{4}{6}
$$

so we get that

$$
boxed{I=0.67textrm{A}}
$$

Result

2 of 2

$$
textrm{a) }EMF=4textrm{V}
$$

$$
textrm{b) }I=0.67textrm{A}
$$

Exercise 2

Step 1

1 of 2

In order to solve this problem we are going to use the definition of the electro-motive force
$EMF=BLv$
Now, we can insert the values given in the problem

$$
EMF=5times 10^{-5}times 25times 125
$$

which gives that

$$
boxed{EMF=0.16textrm{V}}
$$

Result

2 of 2

$$
EMF=0.16textrm{V}
$$

Exercise 3

Step 1

1 of 6

wire + magnet + motion = induced EMF

general formula for induced EMF

Step 2

2 of 6

$$
EMF_{ind}= -BLvsintheta
$$

law of emf induced in straight wire

Step 3

3 of 6

the negative is from heinrich lenz rule

Heinrich Lenz rule : whenever an electromotive force (EMF) is induced the induced current must be in a direction such as to resist the change in magnetic flux producing it

Step 4

4 of 6

$EMF_{ind} = 1 times 30 times2 times sin(90)$ = 60 v

angle = 90 since the wire is perpendicular to magnetic flux lines

Step 5

5 of 6

$I = dfrac{V}{R} = dfrac{60}{15}$ = 4 A

from ohm’s law we get the current intensity
Ohm’s law (V=I*R)

Result

6 of 6

a) induced EMF = 60 v
b) current intensity = 4 A

Exercise 4

Step 1

1 of 2

By using the fourth-right hand rule we get that the north of the magnet has to be at the bottom. Exercise scan

Result

2 of 2

The north pole is at the bottom.

Exercise 5

Step 1

1 of 2

In order to solve this problem we are going to use the formulas for the effective voltage, effective current and Ohm’s law. So, let’s do it.

a) We start from the formula for the effective voltage which says

$$
V_{eff}=frac{sqrt{2}}{2}V_{max}
$$

So after we plug the values

$$
V_{eff}=frac{sqrt{2}}{2}170
$$

$boxed{V_{eff}=120textrm{V}}$
b) The same procedure we repeat to find the effective current

$$
I_{eff}=frac{sqrt{2}}{2}I_{max}
$$

So after we plug the values

$$
I_{eff}=frac{sqrt{2}}{2}0.7
$$

Finally, we have that
$boxed{I_{eff}=49textrm{A}}$
c) At the end, we use Ohm’s law to find the circuit resistance

$$
R=frac{V_{eff}}{I_{eff}}=frac{120}{49}
$$

$$
boxed{R=240Omega}
$$

Result

2 of 2

$$
textrm{a) }V_{eff}=120textrm{V}
$$

$$
textrm{b) }I_{eff}=49textrm{A}
$$

$$
textrm{c) }R=240Omega
$$

Exercise 6

Step 1

1 of 2

In order to solve this problem we are going to use the formulas for the effective voltage and effective current. So, let’s do it. We start from the formula for the effective (RMS) voltage which says

$$
V_{RMS}=frac{sqrt{2}}{2}V_{max}
$$

We can express the maximum voltage as follows

$$
V_{max}=sqrt{2}V_{RMS}
$$

So after we plug the values

$$
V_{max}=sqrt{2}times 117
$$

$$
boxed{V_{max}=165textrm{V}}
$$

The same procedure we repeat to find the maximum current

$$
I_{RMS}=frac{sqrt{2}}{2}I_{max}
$$

$$
I_{max}=sqrt{2}I_{RMS}
$$

So after we plug the values

$$
I_{max}=sqrt{2}times 5.5
$$

$$
boxed{I_{max}=7.76textrm{A}}
$$

Result

2 of 2

$$
V_{max}=165textrm{V}
$$

$$
I_{max}=7.76textrm{A}
$$

Exercise 7

Step 1

1 of 2

In order to solve this problem we are going to use the formulas for the effective voltage, effective current and Ohm’s law. So, let’s do it.

a) We start from the formula for the effective voltage which says

$$
V_{eff}=frac{sqrt{2}}{2}V_{max}
$$

So after we plug the values

$$
V_{eff}=frac{sqrt{2}}{2}425
$$

$boxed{V_{eff}=300textrm{V}}$
b) Now, we can use Ohm’s law to find the effective current

$$
I_{eff}=frac{V_{eff}}{R}=frac{300}{500}
$$

$$
boxed{I_{eff}=0.6textrm{A}}
$$

Result

2 of 2

$$
textrm{a) }V_{eff}=300textrm{V}
$$

$$
textrm{b) }I_{eff}=0.6textrm{A}
$$

Exercise 8

Step 1

1 of 2

In order to solve this problem we are going to use the formulas for the average/effective voltages and currents. Since the power is given as

$$
P=Itimes V
$$

then the effective power is given as

$$
P_{eff}=I_{eff}V_{eff}=frac{sqrt{2}}{2}I_{max}times frac{sqrt{2}}{2}V_{max}
$$

This gives that

$$
P_{eff}=frac{P_{max}}{2}
$$

Now, we can express the maximum power as

$$
P_{max}=2P_{eff}=2times 75
$$

Finally,

$$
boxed{P_{max}=150textrm{W}}
$$

Result

2 of 2

$$
P_{max}=150textrm{W}
$$

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Page 675: Practice Problems

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