Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Table of contents
Textbook solutions
All Solutions
Page 654: Practice Problems
Exercise 16
Solution 1
Solution 2
Step 1
1 of 2
We need to discuss what is the name and the working principle of a rule that helps us determine the direction of force on a current-carrying wire at right angles to a magnetic field.
Step 2
2 of 2
The name of the rule is a **right-hand rule**.
Its working principle is 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 current, 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.
Its working principle is 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 current, 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 1
1 of 2
Once the direction of the field and direction of the current are known, one can use the third right-hand rule to determine the direction of the force.
Result
2 of 2
It is the third right-hand rule and directions of the current and the field have to be known.
Exercise 17
Solution 1
Solution 2
Step 1
1 of 4
We need to calculate the force acting on the wire that has a length of $L=0.5,,rm{m}$ and carries a current of $I=8,,rm{A}$ in a magnetic field of $B=0.4,,rm{T}$.
Step 2
2 of 4
In order to solve this problem we need to use an equation that describes a force acting on a conductor inside a magnetic field:
$$F=ILB$$
$$F=ILB$$
Step 3
3 of 4
Inserting given values into the equation we will get:
$$F=8cdot 0.5cdot 0.4$$
Finally, the force is:
$$boxed{F=1.6,,rm{N}}$$
$$F=8cdot 0.5cdot 0.4$$
Finally, the force is:
$$boxed{F=1.6,,rm{N}}$$
Result
4 of 4
$$F=1.6,,rm{N}$$
Step 1
1 of 2
In order to solve this problem we are going to use the formula that defines the force acting on the current-carrying conductor by a magnetic field
$$
F=ILB
$$
If we now plug in the values we obtain the following
$$
F=8times 0.5times 0.4
$$
Finally
$$
boxed{F=1.6textrm{ N}}
$$
Result
2 of 2
$$
F=1.6textrm{ N}
$$
F=1.6textrm{ N}
$$
Exercise 18
Solution 1
Solution 2
Step 1
1 of 5
We need to calculate the strength of the magnetic field if the wire of length $L=0.75,,rm{cm}$ carries a current of $I=6,,rm{A}$ and exerts a force of $F=0.6,,rm{N}$.
Step 2
2 of 5
In order to solve this problem we need to use an equation that describes the force acting on a current-carrying conductor positioned inside a magnetic field:
$$F=ILB$$
$$F=ILB$$
Step 3
3 of 5
From the previous equation we can extract strength of the magnetic field:
$$B=frac{F}{IL}$$
$$B=frac{F}{IL}$$
Step 4
4 of 5
We can now insert given values to get:
$$B=frac{0.6}{6cdot 0.75}$$
Finally, the magnetic field is:
$$boxed{B=0.13,,rm{T}}$$
$$B=frac{0.6}{6cdot 0.75}$$
Finally, the magnetic field is:
$$boxed{B=0.13,,rm{T}}$$
Result
5 of 5
$$B=0.13,,rm{T}$$
Step 1
1 of 2
In order to solve this problem we are going to use the formula that defines the force acting on the current-carrying conductor by a magnetic field
$$
F=ILB
$$
From here, we can express the magnetic field strength as
$$
B=frac{F}{IL}
$$
If we now plug in the values we obtain the following
$$
B=frac{0.6}{6times 0.75}
$$
Finally
$$
boxed{B=0.13textrm{ T}}
$$
Result
2 of 2
$$
B=0.13textrm{ T}
$$
B=0.13textrm{ T}
$$
Exercise 19
Solution 1
Solution 2
Step 1
1 of 7
A wire weighing $F_g=0.35,,rm{N}$ is suspended in the air using only magnetic force. We need to determine the strength of the magnetic field if the wire has $L=0.4,,rm{m}$ and carries a current of $I=6,,rm{A}$
Step 2
2 of 7
Since the wire is staying in the air we can conclude that the weight of the wire is equal to the force caused by the magnetic field:
$$F_g=F$$
$$F_g=F$$
Step 3
3 of 7
In order to solve this problem we need to use an equation that describes a force acting on a conductor positioned inside magnetic field:
$$F=ILB$$
$$F=ILB$$
Step 4
4 of 7
From the previous equation we can extract the magnetic field:
$$B=frac{F}{LB}$$
$$B=frac{F}{LB}$$
Step 5
5 of 7
Inserting the first equation into the previous one we get:
$$B=frac{F_g}{LB}$$
$$B=frac{F_g}{LB}$$
Step 6
6 of 7
Inserting values into the equation we get:
$$B=frac{0.35}{6cdot 0.4}$$
Finally, the strength of the magnetic field is:
$$boxed{B=0.15,,rm{T}}$$
$$B=frac{0.35}{6cdot 0.4}$$
Finally, the strength of the magnetic field is:
$$boxed{B=0.15,,rm{T}}$$
Result
7 of 7
$$B=0.15,,rm{T}$$
Step 1
1 of 2
In order to solve this problem we are going to use the formula that defines the force acting on the current-carrying conductor by a magnetic field
$$
F_B=ILB
$$
On the other hand, we know that in order to balance the gravitational force
$$
F_B=F_g=ILB
$$
From here, we can express the magnetic field strength as
$$
B=frac{F_g}{IL}
$$
If we now plug in the values we obtain the following
$$
B=frac{0.35}{6times 0.4}
$$
Finally
$$
boxed{B=0.15textrm{ T}}
$$
Result
2 of 2
$$
B=0.15textrm{ T}
$$
B=0.15textrm{ T}
$$
Exercise 20
Solution 1
Solution 2
Step 1
1 of 5
We need to determine the amount of current needed to produce a force of $F=0.38,,rm{N}$ on a wire $L=10,,rm{cm}$ long positioned in a magnetic field of strength of $B=0.49,,rm{T}$
Step 2
2 of 5
In order to solve this problem we need to use an equation that describes a force acting upon a current-carrying conductor positioned inside a magnetic field:
$$F=ILB$$
$$F=ILB$$
Step 3
3 of 5
From the previous equation we can extract current:
$$I=frac{F}{LB}$$
$$I=frac{F}{LB}$$
Step 4
4 of 5
Inserting given values in SI units:
$$I=frac{0.38}{0.1cdot 0.49}$$
Finally, the current is:
$$boxed{I=7.8,,rm{A}}$$
$$I=frac{0.38}{0.1cdot 0.49}$$
Finally, the current is:
$$boxed{I=7.8,,rm{A}}$$
Result
5 of 5
$$I=7.8,,rm{A}$$
Step 1
1 of 2
In order to solve this problem we are going to use the formula that defines the force acting on the current-carrying conductor by a magnetic field
$$
F_B=ILB
$$
From here, we can express the current as
$$
I=frac{F_B}{LB}
$$
If we now plug in the values we obtain the following
$$
I=frac{0.38}{0.1times 0.49}
$$
Finally
$$
boxed{I=7.8textrm{ A}}
$$
Result
2 of 2
$$
I=7.8textrm{ A}
$$
I=7.8textrm{ A}
$$
Exercise 21
Solution 1
Solution 2
Step 1
1 of 3
We need to determine in which direction does the thumb point when using the third right-hand rule for an electron moving through magnetic field.
Step 2
2 of 3
In order to solve this problem we need to be aware of the working principle of this rule:
Middle finger points at the magnetic field lines, index finger in the direction of the moving charge and thumb in the direction of the force.
Middle finger points at the magnetic field lines, index finger in the direction of the moving charge and thumb in the direction of the force.
Step 3
3 of 3
Since the electron is negatively charged, the thumb will be pointing in the **opposite direction** of the electron’s motion.
Step 1
1 of 2
The thumb is pointing in the direction of the motion for positive charges. Since electrons are negatively charged, the thumb will point to the direction opposite of the electron’s motion.
Result
2 of 2
It will point opposite of the electron’s motion.
Exercise 22
Solution 1
Solution 2
Step 1
1 of 4
We need to calculate the magnitude of the force acting on electron which passes through the magnetic field of strength $B=0.5,,rm{T}$ with velocity of $v=4cdot 10^6,,rm{m/s}$.
Step 2
2 of 4
In order to solve this problem we need to use an equation that describes magnetic force on a moving charge:
$$F=qcdot (vec{v} times vec{B} )$$
Or written the other way:
$$F=qvBsin (theta)$$
$$F=qcdot (vec{v} times vec{B} )$$
Or written the other way:
$$F=qvBsin (theta)$$
Step 3
3 of 4
We can now insert values knowin that $theta =90,,^{o}$ because charge moves at the right angle:
$$F=1.6cdot 10^{-19} cdot 4cdot 10^6cdot 0.5$$
Finally, the force is:
$$boxed{F=3.2cdot 10^{-13},,rm{N}}$$
$$F=1.6cdot 10^{-19} cdot 4cdot 10^6cdot 0.5$$
Finally, the force is:
$$boxed{F=3.2cdot 10^{-13},,rm{N}}$$
Result
4 of 4
$$F=3.2cdot 10^{-13},,rm{N}$$
Step 1
1 of 2
In order to solve this problem, we are going to use the definition of the magnetic force on a moving charge
$$
vec F=q(vec vtimes vec B)
$$
$$
F=qvBsintheta=qvB
$$
Now, we can plug in the values
$$
F=1.6times 10^{-19}times 4times 10^6times 0.5
$$
$$
boxed{F=3.2times 10^{-13}textrm{ N}}
$$
Result
2 of 2
$$
F=3.2times 10^{-13}textrm{ N}
$$
F=3.2times 10^{-13}textrm{ N}
$$
Exercise 23
Solution 1
Solution 2
Step 1
1 of 4
We need to calculate the magnitude of the force acting on each ion if a stream of doubly ionized particles (carrying net charge of two elementary charges) moves at a velocity of $v=3cdot 10^4,,rm{m/s}$ perpendicular to a magnetic field of $B=9cdot 10^{-2},,rm{T}$.
Step 2
2 of 4
In order to solve this problem we need to use the equation for magnetic force on a moving charge:
$$F=qvBsin (theta )$$
$$F=qvBsin (theta )$$
Step 3
3 of 4
In this scenario charge is $q=2e$ and $sin (theta)=1$. Which means that by inserting given values into the equation we get:
$$F=2cdot 1.6cdot 10^{-19}cdot 3cdot 10^4cdot 0.09$$
Finally, the force is:
$$boxed{F=8.6cdot 10^{-16},,rm{N}}$$
$$F=2cdot 1.6cdot 10^{-19}cdot 3cdot 10^4cdot 0.09$$
Finally, the force is:
$$boxed{F=8.6cdot 10^{-16},,rm{N}}$$
Result
4 of 4
$$F=8.6cdot 10^{-16},,rm{N}$$
Step 1
1 of 2
In order to solve this problem, we are going to use the definition of the magnetic force on a moving charge
$$
vec F=q(vec vtimes vec B)
$$
$$
F=qvBsintheta=2evB
$$
Now, we can plug in the values
$$
F=2times1.6times 10^{-19}times 3times 10^4times 0.09
$$
Finally
$$
boxed{F=8.6times 10^{-16}textrm{ N}}
$$
Result
2 of 2
$$
F=8.6times 10^{-16}textrm{ N}
$$
F=8.6times 10^{-16}textrm{ N}
$$
Exercise 24
Solution 1
Solution 2
Step 1
1 of 4
We need to calculate the magnitude of the force acting on each ion if a stream of triply ionized particles (carrying net charge of three elementary charges) moves at a velocity of $v=9cdot 10^6,,rm{m/s}$ perpendicular to a magnetic field of $B=4cdot 10^{-2},,rm{T}$.
Step 2
2 of 4
In order to solve this problem we need to use the equation for magnetic force on a moving charge:
$$F=qvBsin (theta )$$
$$F=qvBsin (theta )$$
Step 3
3 of 4
In this scenario charge is $q=3e$ and $sin (theta)=1$. Which means that by inserting given values into the equation we get:
$$F=3cdot 1.6cdot 10^{-19}cdot 9cdot 10^6cdot 0.04$$
Finally, the force is:
$$boxed{F=1.7cdot 10^{-13},,rm{N}}$$
$$F=3cdot 1.6cdot 10^{-19}cdot 9cdot 10^6cdot 0.04$$
Finally, the force is:
$$boxed{F=1.7cdot 10^{-13},,rm{N}}$$
Result
4 of 4
$$F=1.7cdot 10^{-13},,rm{N}$$
Step 1
1 of 2
In order to solve this problem, we are going to use the definition of the magnetic force on a moving charge
$$
vec F=q(vec vtimes vec B)
$$
$$
F=qvBsintheta=3evB
$$
Now, we can plug in the values
$$
F=3times1.6times 10^{-19}times 9times 10^6times 0.04
$$
Finally
$$
boxed{F=1.7times 10^{-13}textrm{ N}}
$$
Result
2 of 2
$$
F=1.7times 10^{-13}textrm{ N}
$$
F=1.7times 10^{-13}textrm{ N}
$$
Exercise 25
Solution 1
Solution 2
Step 1
1 of 4
We need to calculate the magnitude of the force acting on each particle if a stream of doubly ionized helium atoms (alpha particles) moves at a velocity of $v=4cdot 10^4,,rm{m/s}$ perpendicular to a magnetic field of $B=5cdot 10^{-2},,rm{T}$.
Step 2
2 of 4
In order to solve this problem we need to use the equation for magnetic force on a moving charge:
$$F=qvBsin (theta )$$
$$F=qvBsin (theta )$$
Step 3
3 of 4
In this scenario charge is $q=2e$ and $sin (theta)=1$. Which means that by inserting given values into the equation we get:
$$F=2cdot 1.6cdot 10^{-19}cdot 4cdot 10^4cdot 0.05$$
Finally, the force is:
$$boxed{F=6.4cdot 10^{-16},,rm{N}}$$
$$F=2cdot 1.6cdot 10^{-19}cdot 4cdot 10^4cdot 0.05$$
Finally, the force is:
$$boxed{F=6.4cdot 10^{-16},,rm{N}}$$
Result
4 of 4
$$F=6.4cdot 10^{-16},,rm{N}$$
Step 1
1 of 2
In order to solve this problem, we are going to use the definition of the magnetic force on a moving charge
$$
vec F=q(vec vtimes vec B)
$$
$$
F=qvBsintheta=2evB
$$
Now, we can plug in the values
$$
F=2times1.6times 10^{-19}times 4times 10^4times 0.05
$$
Finally
$$
boxed{F=6.4times 10^{-16}textrm{ N}}
$$
Result
2 of 2
$$
F=6.4times 10^{-16}textrm{ N}
$$
F=6.4times 10^{-16}textrm{ N}
$$
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