Physics
Physics
1st Edition
Walker
ISBN: 9780133256925
Textbook solutions

All Solutions

Page 831: Lesson Check

Exercise 20
Step 1
1 of 2
The electrical generator is a device that is designed to convert mechanical work into electrical power. The mechanical work rotates the generator turbine and this change in the magnetic field of the generator induct the emf on the other side.$$
Result
2 of 2
Mechanical work. $$
Exercise 21
Step 1
1 of 2
The most important difference between the electric motor and generator is that the generator doesn’t need electrical power for work. By converting the mechanical work, the generator produces electricity which can be used for the supply of other electrical devices.

On the other hand, the motor can’t be used without electricity. The motor converts the electrical power to mechanical work which can be used for moving some objects, lifting an object, grinding…
In general, the motor and the generator have a very similar construction, so we can say that the motor represents the generator that work’s in a reverse. $$

Result
2 of 2
Construction of the motor and generator is very similar.

Principe of work is completely reverse, this is the main difference.

Exercise 22
Step 1
1 of 2
The maximum emf of the generator is given by the equation:

$$
begin{align}
varepsilon_{max}=Ncdot{B}cdot{A}cdotomega
end{align}
$$

Where the $N$ stands for the number of loops, $A$ stands for the area of each individual loop, $B$ represents the magnitude of the magnetic field and $omega$ is the angular speed.

The emf is directly proportional to an angular speed.

$$
boxed{text{If we increase the angular speed, the emf will also increase.}}
$$

Result
2 of 2
If we increase the angular speed, the emf will also increase.$$
Exercise 23
Step 1
1 of 2
The maximum emf of the generator is given by the equation:

$$
begin{align}
varepsilon_{max}=Ncdot{B}cdot{A}cdotomega
end{align}
$$

Where the $N$ stands for the number of loops, $A$ stands for the area of each individual loop, $B$ represents the magnitude of the magnetic field and $omega$ is the angular speed.

The emf is directly proportional to a number of loops.

$$
boxed{text{If we increase the coil loops number, the emf will also increase.}}
$$

Result
2 of 2
If we increase the coil loops number, the emf will also increase.$$
Exercise 24
Step 1
1 of 2
The maximum emf of the generator is given by the equation:

$$
begin{align}
varepsilon_{max}=Ncdot{B}cdot{A}cdotomega
end{align}
$$

Where the $N$ stands for the number of loops, $A$ stands for the area of each individual loop, $B$ represents the magnitude of the magnetic field and $omega$ is the angular speed.

Let’s express the magnitude of the magnetic field from the first equation:

$$
begin{align*}
B&=frac{varepsilon_{max}}{Ncdot{A}cdotomega}
end{align*}
$$

If we substitute all given:

$$
begin{align*}
B&=frac{75text{ V}}{95cdot{0.0044text{ m}^2}cdot{220 frac{text{rad}}{text{s}}}}
end{align*}
$$

Now we just have to compute the value:

$$
boxed{B=0.816text{ T}}
$$

Result
2 of 2
$$
B=0.816text{ T}
$$
Exercise 25
Step 1
1 of 2
The maximum emf of the generator is given by the equation:

$$
begin{align}
varepsilon_{max}=Ncdot{B}cdot{A}cdotomega
end{align}
$$

Where the $N$ stands for the number of loops, $A$ stands for the area of each individual loop, $B$ represents the magnitude of the magnetic field and $omega$ is the angular speed.

Let’s substitute all given:

$$
begin{align*}
varepsilon_{max}=55cdot{0.95text{ T}}cdot{0.0085text{ m}^2}cdot{310 frac{text{rad}}{text{s}}}
end{align*}
$$

Now we just have to compute the value:

$$
boxed{varepsilon_{max}=137.7text{ V}}
$$

Result
2 of 2
$$
varepsilon_{max}=137.7text{ V}
$$
Exercise 26
Step 1
1 of 2
The maximum emf of the generator is given by the equation:

$$
begin{align}
varepsilon_{max}=Ncdot{B}cdot{A}cdotomega
end{align}
$$

Where the $N$ stands for the number of loops, $A$ stands for the area of each individual loop, $B$ represents the magnitude of the magnetic field and $omega$ is the angular speed.

Let’s express the area of the loop from the first equation:

$$
begin{align*}
A&=frac{varepsilon_{max}}{Ncdot{B}cdotomega}
end{align*}
$$

If we substitute all given:

$$
begin{align*}
A&=frac{22text{ V}}{27cdot{0.82text{ T}}cdot{290 frac{text{rad}}{text{s}}}}
end{align*}
$$

Now we just have to compute the value:

$$
boxed{A=0.00343text{ m}^2}
$$

Result
2 of 2
$$
A=0.00343text{ m}^2
$$
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