Ch.29 1 5 16 19 25 32 36 53

Quiz 11

Answers
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Chapter summary: 
(A) Use the right hand rule or RHR to find the direction of magnetic force vector-F on a moving charge.

In the 2 examples below, assume q>0.


See problem 1 and figure P29.1

(B) Force on a wire: Use RHR.


(C) Use the RHR to find the direction of the dipole moment:


(D) The torque on a magnetic moment tends to rotate it so it points in the same direction as the magnetic field-B

The torque that causes the rotation is :

(E) Circular motion of a charge in a B-field. Assume charge >0. Use RHR everywhere on circle to find the direction of the force, which points to the center of the circle !

Radius  of the orbit is:
r = mv/qB

 
1. In each case for fig. P29.1, analyze the direction of the force using the right-hand-rule as we did in class. For the vector force,


You wrap the v-vector into the B-vector with your right fingers. Your thumb will point in the direction of the vector-F. Here are two hints on the direction of the force. For part (a): up. For part (b): out. As indicated in part (b), if the charge is negative, the force direction will be opposite your thumb direction. Be careful !
5. The magnitude of the force is F = ma, where m is the mass of the proton and a is the acceleration. Solve for F. Now, from the right hand rule, the magnetic field vector is in the y-z plane. For simplicity, the problem assumes that the magnetic field vector is perpendicular to the velocity vector, which is directed along the positive z-axis. Thus,


Solve for B. Clearly, the magnetic field vector is directed along the y-axis. Use the right hand rule, and determine whether it points in the positive y direction or the negative y direction. Review Ch. 11 on cross products.
16. Study fig. P29.16. The force of gravity points down. Thus the magnetic force points up. The magnetic field vector points in. Use the formula,

.

Clearly, the current i is in the plane of the paper and is perpendicular to the vector magnetic field, which, again, points into the paper (in). The vector length points in the same direction as the current i. Thus,

Use the right hand rule for the cross product:

to determine whether the current (which, again, is in the same direction as the vector length) points to the left or to the right.

19. Let the x-direction be along the wire in the direction of the current, the z direction be upward, and the y direction be rightward in fig. P29.19. Thus, if I represents the current,


The Cartesian unit vectors are given by:

Review Chapter 3 and Ch. 11.

Now, from Ch. 7, the work done by the magnetic force equals the change in kinetic energy. And the kinetic energy has two components, transnational and rotational. Review Ch. 11.
Thus,

.

Here, I represents the moment of inertia and v = the linear speed of the center of mass. Go to Ch. 10  and look up the moment of inertia of a solid cylinder of mass m and radius R. Also, the magnitude of the force is:
,
where here I = the current. Solve for v.
25. Use the definition for the torque vector on a loop in a magnetic field:

This is the cross product between the magnetic moment vector and the B-vector. Study fig. 29.14. Wrap your right fingers around the loop in the direction of the current, and your thumb will point perpendicular to the plane of the loop. In the case of fig P29.25, your thumb will point at an angle of 60 degrees with the B-vector. Thus, the magnitude of the torque is:

.

Here,
32. Please read sec. 29.4. You also will need to review the Ch. 25 quiz on the relationship between the change in kinetic energy and the potential:

Solve for the speed.  Also,

[Note that from lecture in  discussion section, a negative particle will travel in the opposite direction (i.e. clockwise) of the  positive particle (traveling counterclockwise in fig. 29.17). These opposite directions form the basis of the technology of high energy physics accelerators, in which oppositely charged particles under the influence of a magnetic field travel oppositely in a circle, and  undergo a "head-on" collision. The reaction by-products are studied to investigate the fundamental forces of nature and the origins of the universe! For more information, click here, take a look at the page, and then explore this very interesting site .]
Solve the above equation for r.
36. (a):
.
Thus,
qrB = mv (Eqn. 1)
Note that here q is the magnitude of the electron charge. Review sec. 11.3, ex 11.6, page 314 (Vol 1 of 4th Edition) for the definition of the magnitude of the angular momentum:
L = mvr (Eqn. 2). Substitute Eqn. 1 into Eqn. 2 to eliminate v. Finally, solve for r in terms of L, q, and B.
(b) Use Eqn. 2 above to solve for v.
 53. Please study fig. 29.23 for a picture of a charged particle entering a region of a uniform magnetic field. The particle enters with a velocity perpendicular to the boundary, and moves along a half circle of radius r until it reemerges from the region with a velocity pointed in the opposite direction. Thus, the maximum depth of penetration is the radius r.

(a):
.
Here e = the magnitude of the electron charge.
Now,


(Eqn. 1)

Solve for t (Hint: What is the angle for a half-circle ?)
(b) :


(Eqn. 2)

Solve for the speed using Eqn. 1 in part (a). Then substitute into Eqn. 2 to get the kinetic energy.