Home Page Textbooks Physics: Principles and Problems Page 659: Section Review

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

Textbook solutions

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Page 659: Section Review

Exercise 26

Solution 1

Solution 2

Step 1

1 of 2

We need to determine in which direction would be the force on the current-carrying wire that is perpendicular to the Earth’s magnetic field and runs east-west.

Step 2

2 of 2

In order to solve this problem we need to use the right-hand rule.
If the current flows **east**, right hand rule will show us that the force is acting **upwards**.
If the current flows **west**, right hand rule will shows us that the force is acting **downwards**.

Step 1

1 of 2

If we use the right-hand rule and take that electrons move east we deduce that the the force on the wire is pointing upwards.

Result

2 of 2

The force is pointing upwards.

Exercise 27

Solution 1

Solution 2

Step 1

1 of 3

We need to determine in which directions will electrons scatter if a beam of electrons in cathode ray tube approaches the deflecting magnets who have N pole at the top of the tube and the S pole at the bottom.

Step 2

2 of 3

In order to solve this problem we need to use right-hand rule.
It states that in order to determine the direction of the magnetic force on a positive moving charge, you need to point your right thumb in the direction of the velocity, your index finger in the direction of the magnetic field, and your **middle finger** will point in the direction of the the resulting **magnetic force**.

Step 3

3 of 3

If we use it in such way that the electrons are moving out of the screen and the magnetic field is pointin upwards, the positive charge would deflect left.
Since electrons are **negative** charge, they will move **right**.

Step 1

1 of 2

To solve this problem we are going to use the right-hand rule that states that if the electrons are moving out of the screen and the magnetic filed is pointing up the electrons should be deflected to the right due to their negative charge.

Result

2 of 2

Electrons’s should be deflected to the right.

Exercise 28

Solution 1

Solution 2

Step 1

1 of 3

We need to compare shown diagram of galvanometer with shown electric motor and find similarities and differences between them.

Step 2

2 of 3

Similarities can be found in a fact that both galvanometer and electric motor make use of the change of the magnetic flux as the main working principle.

Step 3

3 of 3

Main differences would be the fact that galvanometer is unable to make a full circle while electric motor is able to continously make loops.

Step 1

1 of 2

Both, galvanometer and electric motor use the change of the magnetic flux as the mechanism of work. The main difference between a galvanometer and an electric motor is that in galvanometer the loop cannot rotate more than 180$^circ$. This is achieved in electric motors by changing the direction of the current.

Result

2 of 2

They both use the change in magnetic flux as the mechanism but galvanometers cannot rotate for more than 180$^circ$.

Exercise 29

Solution 1

Solution 2

Step 1

1 of 3

We need to discuss the forces acting on a coil when the plane of the coil in a motor is perpendicular to the magnetic field and whether then the coil rotates or not.

Step 2

2 of 3

In order to see if the forces act on it we need to take a look at the equation for electromagnetic force:
$$F=IBsin (theta )$$
Where $theta =0$ in this position.
This means that the force acting on the coil in this position is **zero**.

Step 3

3 of 3

This doesn’t mean that the coil will not move. It only means that it **won’t accelerate or deccelerate** at that exact point. If ti was moving before, it will continue moving and if it wasn’t, it will not begin to move.

Step 1

1 of 2

No, since this depends on weather or not the the coil is already in motion since even if the toque is zero if the coils is already rotating, the zero torque will not bring it to rest.

Result

2 of 2

No, it doesn’t.

Exercise 30

Solution 1

Solution 2

Step 1

1 of 5

We need to calculate the total resistance of the meter and the multiplier resistor if a full-scale deflection is $V=5,,rm{V}$ and $I=180,,rm{mu A}$.

Step 2

2 of 5

Current required for full-scale deflection is achieved when maximum voltage is applied to total resistance. This means that in order to solve this problem we need to use Ohm’s Law:
$$V=IR$$

Step 3

3 of 5

From the previous equation we can extract resistance:
$$R=frac{V}{I}$$

Step 4

4 of 5

Inserting given values into the previous equation we get:
$$R=frac{5}{180cdot 10^{-6}}$$
Finally, the total resistance is
$$boxed{R=28cdot 10^3,,rm{Omega}}$$

Result

5 of 5

$$R=28cdot 10^3,,rm{Omega}$$

Step 1

1 of 2

In order to solve this problem, we are going to use Ohm’s law which says that

$$
U=IR
$$

So the resistance can be expressed as
$R=frac{U}{I}=frac{5}{180times 10^{-6}}$
Finally

$$
boxed{R=28times 10^3Omega}
$$

Result

2 of 2

$$
R=28times 10^3Omega
$$

Exercise 31

Solution 1

Solution 2

Step 1

1 of 3

We need to discuss if forces acting on parallel current-carrying wires are a result of magnetic attraction or electrostatics.

Step 2

2 of 3

If the forces were result of electrostatics, the same charges would repel.
Equivalent to a same charge would be wires with currents flowing in the same direction.
Since the wires are **not repelling**, we can conclude that the resulting force are **not a result of electrostatics**.

Step 3

3 of 3

The only option left is that the forces are a **result of acting magnetic fields**.

Step 1

1 of 2

If we assume that the currents are flowing in the same direction, it means that they are constituted of the same species of charge. By definition the same
charges repel each other but we know that these wires are attracted to each other. Therefore, this is the consequence of magnetic fields.

Result

2 of 2

If we would only have electrostatic forces the wires would repel each other but we know that they are attractive.

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