Home Page Textbooks Physics: Principles and Problems Page 721: Standardized Test Practice

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

Table of contents

Textbook solutions

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Page 721: Standardized Test Practice

Exercise 1

Solution 1

Solution 2

Step 1

1 of 2

D) the magnetic force always remains perpendicular to the velocity and is directed toward the center of the circular path

Result

2 of 2

Step 1

1 of 6

When a charged particle enters a uniform magnetic field, it is deflected in the direction determined by the right-hand rule. The centripetal force acts on the particle, and since the magnetic force is the only force acting on the body, we can say that the magnetic force has a role of centripetal force.

Step 2

2 of 6

**a)** If the force and the velocity were parallel, then the particle couldn’t move in a circular path.

Step 3

3 of 6

**b)** The keyword here is may. No, the magnetic force must be perpendicular to the velocity, and also it is impossible for it to move circularly if the force points away from the center.

Step 4

4 of 6

**c)** This is the combination of the wrong parts from the first and the second answer. Hence, this cannot be the correct answer.

Step 5

5 of 6

**d) For the particle to move in a circular path in a uniform field if it enters the field at the right angle, the force in the field is always perpendicular to the particle’s velocity, pointing towards the center of the path.**

Result

6 of 6

$$text{d)}$$

Exercise 2

Step 1

1 of 2

In order to solve this problem, we are going to use Newton’s second law which says that on a circular path (and in a magnetic field a moving charged particle will follow such a path) it has to hold that

$$
qvB=frac{mv^2}{r}
$$

From here, one can express the speed as

$$
v=frac{qrB}{m}=frac{1.6times 10^{-19}times 0.066times 0.1}{1.67times 10^{-27}}
$$

Finally, we have that

$$
boxed{v=6.3times 10^5frac{textrm{ m}}{textrm{ s}}}
$$

So the answer is A.

Result

2 of 2

$$
textrm{A) }v=6.3times 10^5frac{textrm{ m}}{textrm{ s}}
$$

Exercise 3

Step 1

1 of 2

In order to solve this problem we are going to use the definition of the speed of light in a medium

$$
v=frac{c}{sqrt{K}}=frac{3times 10^8}{sqrt{5.4}}
$$

So we have that

$$
boxed{v=1.3times 10^8frac{textrm{m}}{textrm{s}}}
$$

Which makes the correct answer D.

Result

2 of 2

$$
textrm{D) }v=1.3times 10^8frac{textrm{m}}{textrm{s}}
$$

Exercise 4

Step 1

1 of 2

To solve this problem we are going to use the definition of the frequency of EM waves which gives that

$$
f=frac{c}{lambda}=frac{3times 10^8}{2.87}
$$

Finally

$$
boxed{f=1.04times 10^8textrm{ Hz}}
$$

So the answer is C.

Result

2 of 2

$$
textrm{C) }f=1.04times 10^8textrm{ Hz}
$$

Exercise 5

Step 1

1 of 2

DC current voltage applied to a quartz crystal will not create an electromagnetic wave since DC current doesn’t change it’s intensity thus doesn’t induce magnetic field. So the correct answer is A.

Result

2 of 2

The correct answer is A.

Exercise 6

Step 1

1 of 2

$$
qvB=frac{mv^2}{r}
$$

From here, one can express the speed as

$$
v=frac{qrB}{m}=frac{1.6times 10^{-19}times 0.52times 0.45}{1.67times 10^{-27}}
$$

Finally, we have that

$$
boxed{v=2.2times 10^7frac{textrm{ m}}{textrm{ s}}}
$$

So the answer is C.

Result

2 of 2

$$
textrm{C) }v=2.2times 10^7frac{textrm{ m}}{textrm{ s}}
$$

Exercise 7

Step 1

1 of 2

$$
qvB=frac{mv^2}{r}
$$

From here, one can express the speed as

$$
v=frac{qrB}{m}=frac{1.6times 10^{-19}times 0.04times 1.5}{3.34times 10^{-27}}
$$

Finally, we have that

$$
boxed{v=2.87times 10^6frac{textrm{ m}}{textrm{ s}}}
$$

Result

2 of 2

$$
v=2.87times 10^6frac{textrm{ m}}{textrm{ s}}
$$

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Page 721: Standardized Test Practice

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