Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Textbook solutions
All Solutions
Page 647: Practice Problems
Exercise 1
Solution 1
Solution 2
Step 1
1 of 3
We need to determine whether the magnets will attract or repulse of they are held:
– north pole to north pole
– north pole to south pole
– north pole to north pole
– north pole to south pole
Step 2
2 of 3
In the first scenario the magnets are brought together on the same poles.
Same poles repulse which means that the magnets will **repulse each other**.
Same poles repulse which means that the magnets will **repulse each other**.
Step 3
3 of 3
In the second scenario the magnets are brought together on the opposite poles.
Opposite poles attract which means that the magnets will **attract each other**.
Opposite poles attract which means that the magnets will **attract each other**.
Step 1
1 of 2
a)
Two north poles repulses each other.
b)
A north pole and a south pole attract each other.
Result
2 of 2
a) repulsive
b) attractive
Exercise 2
Solution 1
Solution 2
Step 1
1 of 5
In the figure we can see five magnet disks floating on each other. We need to determine the poles one top of each magnet if the magnet on top has the north pole facing up.
Step 2
2 of 5
In order for disks to float on each other they need to have **same poles facing each other**.
Step 3
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If the top-most magnet has N facing up, this means that it’s bottom side is S.
This, combined with the fact that the sides facing each other need to be of same pole, means that the second magnet has S side facing up.
This, combined with the fact that the sides facing each other need to be of same pole, means that the second magnet has S side facing up.
Step 4
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The third one then, using the same logic needs to have N side facing up.
Step 5
5 of 5
We can notice the pattern here. The magnets need to have **alternate** poles facing up. This means that the first, third and fifth magnet will have **N facing up** and second and fourth magnet will have **S pole facing up**.
Step 1
1 of 2
Since the disks are floating the forces between them are repulsive so they compensate for the gravitational force. Now, we know that the topmost disk’s topside is the north pole. Then we have that topsides of the remaining diskcs are :
2nd disk: S
3rd disk: N
4th disk: S
5th disk: N
Result
2 of 2
2nd disk: S
3rd disk: N
4th disk: S
5th disk: N
Exercise 3
Solution 1
Solution 2
Step 1
1 of 6
We need to determine which end of the nail is a south pole if a magnet with a north pole on it’s left side attracts a nail on it’s right side.
Step 2
2 of 6
The nail is made from a ferromagnetic material. This means it is not magnetized by itself but rather due to a presence of an existing magnetic field.
Step 3
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What happens inside a nail is that it’s electrons move freely from **one side to the other** and in that way make a nail become a magnet.
Step 4
4 of 6
Since we know that opposite poles attract, it can be concluded that the part of the nail which is in contact with the magnet will have **opposite pole** than the magnet at point of contact.
Step 5
5 of 6
Since that side of the magnet is S, top of the nail, which is in contact with magnet, will be N.
Step 6
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This finally means that the **bottom part of the nail will be S pole**.
Step 1
1 of 2
Since the nail is obviously made of a ferromagnetic material it will become magnetized so its head becomes the north pole, so its bottom is then the its south pole.
Result
2 of 2
Nail’s bottom is the south pole.
Exercise 4
Solution 1
Solution 2
Step 1
1 of 4
We need to think of explanations for false readings that are sometimes given by compasses.
Step 2
2 of 4
In order to solve this problem we need to be aware of the working mechanism of a compass.
Compass uses Earth’s magnetic field to orientate its needle.
Since magnetic field ‘flows’ from one pole to another, when we magnetise our needle, it will orientate itself according to the magnetic field of the Earth.
Compass uses Earth’s magnetic field to orientate its needle.
Since magnetic field ‘flows’ from one pole to another, when we magnetise our needle, it will orientate itself according to the magnetic field of the Earth.
Step 3
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Knowing this, we can conclude that any external magnetic field besides Earth’s can easily disrupt readings of a compass.
Step 4
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Example of such disruption can be any kind of electric circuit that produces strong enough magnetic field; mobile phone, tv… Also some kind of antennas as well.
Step 1
1 of 2
There are many factors that can “mess” with one’s compass. For instance, presence of natural magnets can cause compass to give a false reading. Moreover, presence of electric circuits can also cause compass to give some false readings.
Result
2 of 2
Many factors, such as natural magnets and electric circuits can cause a compass to give false readings.
Exercise 5
Solution 1
Solution 2
Step 1
1 of 3
Long, straight wire extends from north to south and conducts current.
We need to determine the direction of the current if the magnetic needle placed above the wire points its north pole towards the east.
We also need to determine which direction will the compass needle point if it’s placed underneath.
We need to determine the direction of the current if the magnetic needle placed above the wire points its north pole towards the east.
We also need to determine which direction will the compass needle point if it’s placed underneath.
Step 2
2 of 3
In order to solve this problem we need to use right-hand’s rule:
If the line of the field is pointing to the east above the wire, when we place the hand on the wire the **thumb will point north**.
Which means that the current flows to the north.
If the line of the field is pointing to the east above the wire, when we place the hand on the wire the **thumb will point north**.
Which means that the current flows to the north.
Step 3
3 of 3
Using the same rule we can see that underneath the wire our **fingers point the opposite direction** than they do above.
This means that the needle would in that scenario **point west**.
This means that the needle would in that scenario **point west**.
Step 1
1 of 2
To solve this problem, we are going to use the right-hand’s rule.
a) If we imagine our wire and grab it elongating our thumb we see that if the line of the field is pointing to the east above the wire, then the current must run $boxed{north}$.
b) We can apply the same reasoning but this time we will place the compass below the wire. Now the needle will point $boxed{west}$.
Result
2 of 2
a) North.
b) West.
Exercise 6
Solution 1
Solution 2
Step 1
1 of 7
We need to compare the strength of the magnetic field that is $r_1=1,,rm{cm}$ away from the current carrying wire to those of the fields that are $r_2=2,,rm{cm}$ and $r_3=3,,rm{cm}$ away from the same wire.
Step 2
2 of 7
In order to solve this problem we need to use equation for the magnetic field of a current carrying wire:
$$B=frac{mu_0 I}{2pi r} propto frac{1}{r}$$
Since the first part of the equation is basically constant because we are looking at the same wire with the same current, we only need to take a look at the distance $r$
$$B=frac{mu_0 I}{2pi r} propto frac{1}{r}$$
Since the first part of the equation is basically constant because we are looking at the same wire with the same current, we only need to take a look at the distance $r$
Step 3
3 of 7
Comparing the first two we get:
$$frac{B_1}{B_2}=frac{frac{1}{r_1}}{frac{1}{r_2}}$$
From this we can get:
$$frac{B_1}{B_2}=frac{r_2}{r_1}$$
$$frac{B_1}{B_2}=frac{frac{1}{r_1}}{frac{1}{r_2}}$$
From this we can get:
$$frac{B_1}{B_2}=frac{r_2}{r_1}$$
Step 4
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Inserting values we get:
$$frac{B_1}{B_2}=frac{2}{1}$$
Finally:
$$boxed{frac{B_1}{B_2}=2}$$
$$frac{B_1}{B_2}=frac{2}{1}$$
Finally:
$$boxed{frac{B_1}{B_2}=2}$$
Step 5
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Comparing the first and third we get:
$$frac{B_1}{B_3}=frac{frac{1}{r_1}}{frac{1}{r_3}}$$
From this we can get:
$$frac{B_1}{B_3}=frac{r_3}{r_1}$$
$$frac{B_1}{B_3}=frac{frac{1}{r_1}}{frac{1}{r_3}}$$
From this we can get:
$$frac{B_1}{B_3}=frac{r_3}{r_1}$$
Step 6
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Inserting values we get:
$$frac{B_1}{B_3}=frac{3}{1}$$
Finally:
$$boxed{frac{B_1}{B_3}=3}$$
$$frac{B_1}{B_3}=frac{3}{1}$$
Finally:
$$boxed{frac{B_1}{B_3}=3}$$
Result
7 of 7
$$frac{B_1}{B_2}=2$$
$$frac{B_1}{B_3}=3$$
$$frac{B_1}{B_3}=3$$
Step 1
1 of 2
To solve this problem we are going to use the formula for the magnetic field magnitude of a current-carrying wire which says
$$
B=frac{mu_0 I}{2pi r}propto frac{1}{r}
$$
a) In the first case we have that
$$
frac{B_1}{B_2}=frac{frac{1}{r_1}}{frac{1}{r_2}}=frac{r_2}{r_1}=frac{2}{1}
$$
Finally, we have that
$$
boxed{frac{B_1}{B_2}=2}
$$
So the initial field is two times stronger.
b) In the second case, we have the same procedure which gives
$$
frac{B_1}{B_2}=frac{3}{1}
$$
$$
boxed{frac{B_1}{B_2}=3}
$$
So the initial field is three times stronger.
Result
2 of 2
a) The initial field is two times stronger.
b) The initial field is three times stronger.
Exercise 7
Solution 1
Solution 2
Step 1
1 of 3
We need to determine which point of the nail with a wire wrapped around it will be the north pole. The head of the nail is facing left, and the wire is connected to the source whose left side is negative and right positive.
Step 2
2 of 3
This system can be approximated by a uniform magnetic field created by a solenoid.
This means that the magnetic field in the nail will align itself with the current flowing through the wire.
This means that the magnetic field in the nail will align itself with the current flowing through the wire.
Step 3
3 of 3
Since the current is flowing from positive to negative, the magnetic field in the nail will be facing the opposite direction.
This means that the **pointed end** of the nail will be the **north pole**.
This means that the **pointed end** of the nail will be the **north pole**.
Step 1
1 of 2
The system shown in the figure 24-13 can be approximated as a solenoid which will have a uniform magnetic field along it’s interior, in this case the nail. The magnetic field of the solenoid will align the microscopic magnetic moments of the nail which will follow the field lines and since due the direction of the current the field in this solenoid is from the head to the point, the pointed end will be the north pole of the newly obtained magnet.
Result
2 of 2
The pointed end will be the north pole of the newly obtained magnet.
Exercise 8
Solution 1
Solution 2
Step 1
1 of 3
We need to discuss which material will make a better magnet core; glass, iron or aluminum.
Step 2
2 of 3
In order to enhance strength of a magnet, it is best if the material used as a core has strong ferromagnetic properties.
Step 3
3 of 3
Looking at the three mentioned materials, the **iron** by far has the **strongest ferromagnetic properties** and would therefore make the best magnet core.
Step 1
1 of 2
When aiming to create an electromagnet one should choose the material with the strongest ferromagnetic properties. In this case that is iron and due to its microscopic structure it will make for the best electromagnet.
Result
2 of 2
The iron rod is the correct choice.
Exercise 9
Solution 1
Solution 2
Step 1
1 of 4
We need to determine whether it is possible or not to create a electromagnet with adjustable strength by putting potentiometer as a variable resistor on a magnet made of a wire wrapped around an iron rod.
Step 2
2 of 4
Strength of a magnetic field created by a wire can be calculated as:
$$B=frac{mu_0 I}{2pi r}$$
From this equation we can see that it directly depends on the current flowing through the wire.
$$B=frac{mu_0 I}{2pi r}$$
From this equation we can see that it directly depends on the current flowing through the wire.
Step 3
3 of 4
Since potentiometer works as a variable resistor, meaning a resistor with adjustable resistance, we can directly affect current in the wire. Proof for that can be found in Ohm’s law:
$$I=frac{V}{R}$$
$$I=frac{V}{R}$$
Step 4
4 of 4
Final conclusion is that adjustable electromagnet **can be made** using potentiometer since change in its resistance will change current and therefore strength of magnetic field created by it.
Step 1
1 of 2
In essence a potentiometer works as a variable resistor therefore we can use it to manipulate the current magnitude and since the induced field is proportional to the current, we should be able to to control it as well. On the other hand magnetization is proportional to the magnetic field one can indeed regulate the strength of an electromagnet with a potentiometer.
Result
2 of 2
The answer is yes.
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