Phy 122 – Assignment 5
A. 1. No . F = q v B sinθ. Since both v and B point in the same direction (+x), the angle between them is 0° making sinθ zero.
2. For the wires to go slack, the magnetic force must balance gravity:
Direction: The current has to follow the horizontal conductor, so there are only two possibilities: right or left. Consider both possibilities, and see which gives a force that points up. That turns out to be a current toward the right:
The force is perpendicular to both v and B, so it’s got to be in either the +y or –y direction. Choose between these with the right hand rule: Wrap your fingers around the y axis pointing from v toward B. Your thumb points in the direction of the force on a positive charge. That’s up the y axis for the case drawn here.
B. 1. a. Yes, yes - A charge always has an E field, and feels a force from other
E fields. b. No, No – Magnetic fields come from moving charges, and that is what feels force from other B fields.
2. Two steps: Find the wire’s field at the electron’s location. Then, find the force on the electron from being in that field.
B 0
I
2 r
( 4
2 x 10
7
)( 10 A )
(.
02 m )
1 .
00 x 10
4
T

Direction: B is in the negative z direction, by the right hand rule. (Imagine grabbing the wire with your thumb pointing in the direction the current flows.

Your fingers would wrap around the wire like the field lines. Where the electron is located, your fingers would be pointing into the page.)

Direction of F : With perpendicular to both v

 v
 in the positive x direction, and B in the negative z direction, the force points in the negative y direction. ( The force is
and

B
, so it’s got to be either +y or –y. Choose between these with the right hand rule: Wrap your fingers around the y axis pointing from v

toward

B
. Your thumb points in the direction of the force on a positive charge. That’s up the y axis. Since the charge is actually negative, it’s the other way.)
C. 1. a. No, no. Electric fields interact with charged objects and the wire has no net charge. It contains equal numbers of electrons (which happen to be moving) and positive copper ions (with fixed positions). Every time a new electron is added to one end of the wire, another electron leaves the other end, so it stays neutral. b. Yes, yes - A current (moving charge) makes a B field around itself, and feels a force from other B fields.
2. This is just like question 1, except the positive copper ions are gone.
Without them, the beam is electrically charged as well as being a current. So, a. Yes, yes. b. Yes, yes.
3. F = q vB sinθ = (1.60 x 10 -19 C)(7.00 m/s)(4.00 T)(sin 90°) = 4.48 x 10 -18
N
Direction: The force is perpendicular to both v and B, so it’s got to be either +y or –y. Choose between these with the right hand rule: Wrap your fingers around the y axis pointing from v toward B. Your thumb points in the direction of the force on a positive charge. That’s down the y axis in this case. Since the charge is actually
Ans.: 4.48 x 10 -18 negative, it’s the other way.
N in the +y direction.

D. 1. +z (out of the page toward you). The force is perpendicular to both v and B, so it’s got to be either +z or
–z. Choose between these with the right hand rule: Wrap your fingers around the z axis pointing from v toward B.
Your thumb points in the direction of the force on a positive charge. That’s in the –z direction in this case.
Since the charge is actually negative, it’s the other way.
2. c) Two steps: First, find the upper wire’s field at the lower wire’s location:
Imagine grabbing the upper wire, with your thumb pointing in the direction the current flows. Your fingers would wrap around the wire like the field lines, pointing into the page at the lower wire’s location.
Second step: Find the force on the lower wire from being in that field. With the current’s direction ( v
  of charge in the wire) in the positive x direction, and B in the negative z direction, the force points in the positive y direction. ( The force is perpendicular to both these with the right hand rule: Wrap your fingers around the y axis pointing from v


toward
B
 v

and
B
, so it’s got to be +y or –y. Choose between
. Your thumb points in the direction of the force.)
E. 1. The blow can knock its little magnetized pieces (“domains”) out of alignment with each other. Each little arrow is the field of an iron atom. Before being hit, they tend to point to the right.
After, they are all scrambled up and don’t add up to anything.
2. a. Magnetic force: F = qvB sin
= [(5)(1.60 x 10
-19
C)](8.75 x 10
3
m/s)(.55 T)sin90° = 3.85 x 10
-15
N

b. It has to be perpendicular to v

(+x direction) and

B (-z direction), so that’s either +y or –y. Choose between these with the right hand rule: Wrap your fingers around the y axis from v

toward

B . Your thumb should point in the +y direction.. c. From section 1, the electric force is given by F = qE
The question says the electric force is the same size as the magnetic force.
3.85 x 10
-15
N = [(5)(1.60 x 10
-19
E = 3.85 x 10
-15
/[(5)(1.60 x 10
-19
C)]E
)] = 4812.5 V/m = 4.81 kV/m d. The question says the electric force has the opposite direction of the magnetic force. The equation F q E says the electric field has the same direction as the electric force. So, since the magnetic force is +y, E
 is in the –y direction.