Homework Helpers: Physics

7 Magnetism


Answer Key

The actual answers will be shown in brackets, followed by an explanation. If you don’t understand an explanation that is given in this section, you may want to go back and review the lesson that the question came from.

Lesson 7–1 Review

1. [magnetic domain]—In a magnetized substance, the magnetic domains are lined up in the same direction.

2. [ferromagnetic]—Recall, the magnetic properties of these elements are believed to result from the configurations of individual atoms.

3. [north]—If the magnetic south pole is located near the geographic North, it stands to reason that the magnetic north pole is located near the geographic South.

Lesson 7–2 Review

1. [south to north]—Outside of the solenoid, the lines come out of the south and into the north. Within the solenoid, they travel from the south pole to the north.

2. [counterclockwise]—Using the right-hand rule, if you hold the thumb of your right hand pointing out of the page, the other fingers of your right hand curl counterclockwise.

3. [into the page]—The thumb of your right hand would be pointing to the right, so the other fingers of your hand would go into the page below the wire, as you motioned to grab it.

Lesson 7–3 Review

1. [into the page]—Using the right-hand rule, we point the index finger of our right hand in the direction of the magnetic field, to the left of the page. We point our thumb in the direction of the velocity of the positive particle, toward the bottom of the page. Our middle finger is at a right angle to both of the other fingers when it is directed into the page.



Lesson 7–4 Review

1. [Magnetic flux]—The formula for calculating magnetic flux is ϕ = BA cosθ.



Chapter 7 Examination

1. [e. volts]—Emf is often simply represented with the symbol ε.

2. [d. weber]—Remember: 1 weber (Wb) = 1 T · m2.

3. [f. tesla]—1 tesla (T) = 1 N/A × m

4. [c. Joseph Henry].

5. [a. Hans Christian Oersted]—The link between electricity and magnetism was established.

6. [g. magnetic flux]—Magnetic flux is represented with the symbol, ϕ.

7. [c. aluminum]—Aluminum foil is not strongly attracted to magnets.

8. [b. moving protons]—Magnetic fields are produced by charges in motion. A neutron has no charge, so it won’t produce a magnetic field.

9. [c. twice as great]—The formula F = qvB shows us that the magnetic force experienced by a moving charged particle in a magnetic field is proportional to the charge (q) on the object. The oxide ion has a charge that is twice as great as the charge on the electron.

10. [c. 90°]—the extended right-hand rule shows the direction of the force, in relation to the field.

11. [b. double]—The formula, F = qvB, indicates that the magnetic force experienced by a charged particle in motion through it is proportional to the velocity of the particle. If you double the velocity, you double the force.

12. [a. Lenz’s law]—Lenz’s law states that the law of conservation of energy is not violated by magnetic induction.

13. [b. Faraday’s law]—More formally, this is called “Faraday’s law of magnetic induction.”

14. [d. number of loops]—The number of loops of wire in a coil

15. [a. 0°]—Cosine 0° = 1, giving the maximum value for magnetic flux from the equation, ϕ = BA cosθ.

16. [2.8 × 109 T]—First, we can see that the alpha particle, which is a helium ion, has two excess protons. That gives the particle a charge (q) of (q = n × e = 2(+1.60 × 10-19 C)) = +3.20 × 10-19 C.
We then solve as shown.

17. [0 N]—Be prepared for this type of question, which is designed to check if you are paying attention to the angle between the field lines and the wire. Sine 0° = 0, so the answer to our calculation will also be zero.

F = ILBsinθ = (2.0 A)(3.50 m)(2.30 T)(sin 0°) = 0 N

18. [99 V]—That question is quite a mouthful, so you may have had trouble deciding where to start. The wire loop is being pulled out of the magnetic field, so the final magnetic flux will be zero. We need to find the initial magnetic flux to find the change in flux. Then we can use Faraday’s law to determine the induced emf.