#$&*
course PHY 242
3/31/11; 15:45
110321 Physics IIThe following can all be answered using general knowledge of basic quantities like Newtons, Joules, watts, and general knowledge of first-semester physics. The main focus here is on reasoning. Formulas are unnecessary and should not be used. You shouldn't have to, but if you forget what a watt is, or a Joule, you can look up those definitions.
A 1-amp current in a straight 1-meter wire in a uniform 1-Tesla magnetic field perpendicular to the wire experiences a force of 1 Newton, in a direction perpendicular to both the wire and the magnetic field. The direction of the force is found by the right-hand rule, taking the cross product of the current flow vector with the magnetic field vector.
To begin to make sense of all this statement work through the following:
`q001. How much force would you expect to result from each of the following, assuming the wire to be straight, and the current and magnetic field vectors to be perpendicular?
A 1-amp current in a 1-meter wire in a .01-Tesla magnetic field.
****
1T = 1N/(A*m)
1A * 1m x .01T = .01Am*T = .01 N
#$&*
A 250-milli-amp current in a 1-meter wire in a 1-Tesla magnetic field.
****.25N
#$&*
A 1-amp current in a 1-centimeter wire in a 1-Tesla magnetic field.
****.01N
#$&*
A 250-milliamg current in a 1-centimeter wire in a .01-Tesla magnetic field.
****.25*.01*.01 = 25*10^-6 N
#$&*
`q002. What would be the direction of the force on the current in each of the following scenarios?
The current is directed toward the north and the magnetic field is upward.
****if the current is in the positive y, j, direction and the magnetic field is in the positive z, k, direction => the force is in the pos x direction, i.
#$&*
The current is directed toward the west and the magnetic field is downward.
****south
#$&*
The current is directed toward the east and the magnetic field is directed toward the west.
****no direction the magnetic field is parallel with the current flow. [they are going in opp directions but that has no effect if the cross product =0
#$&*
The current is directed upward and the magnetic field is directed toward the south.
****
east
#$&*
`q003. A square loop of wire is balanced on a thin wood beam, with the beam in the north-south direction and two of the sides of the loop parallel to the beam. A uniform magnetic field is directed downward. A switch is thrown, causing a current to run through the loop in the clockwise direction. The force exerted on the loop is much less than its weight, so the loop will not lift off of the beam.
The four sides of the loop will be referred to as the north, south, east and west sides.
What will be the direction of the force exerted on the north side of the loop?
****west
#$&*
@& A clockwise current on the north side of the loop would be to the east, resulting in a force to the north.*@
@& You might be using the direction of the current to label the sides, but the current flows to the north on the west side of the loop.
No problem, though, since I think I know what you're thinking.*@
What will be the direction of the force exerted on the south side of the loop?
****east
#$&*
What will be the direction of the force exerted on the east side of the loop?
****north
#$&*
What will be the direction of the force exerted on the west side of the loop?
****south
#$&*
What effect would these forces have on the loop?
****the west and east sides are getting forced outward while the north and south sides are contracting. This forms a rectangle.
#$&*
Answer the same questions, assuming that the magnetic field is now directed toward the west.
What will be the direction of the force exerted on the north side of the loop?
****up[ initially]
#$&*
What will be the direction of the force exerted on the south side of the loop?
****down[ initially]
#$&*
What will be the direction of the force exerted on the east side of the loop?
****none [ initially]
#$&*
What will be the direction of the force exerted on the west side of the loop?
****none [ initially]
#$&*
What effect would these forces have on the loop?
****a turning effect, raising on the north side and lowering on the south side.
Then the west and east side would be affected.
#$&*
`q004. A simple pendulum of mass .1 gram and length 10 cm is pulled back 1 cm from its equilibrium position.
How much force is required?
****
The work involved would be the distance times the weight = .95 cm * .1 g * 9.81 m/s^2 = 0.93195 g * cm *m/ s^2 = 9.32 micro J.
The force is simply the weight * the constant of gravity = .98 milli-Newtons
#$&*
@& It's not clear where you're getting .95 cm.
The force constant for a pendulum is m g / L. Multiply this by the pullback distance to get the work.
Or use similar triangles as in class notes (and statics).
*@
If this force is the result of a .15 amp current flowing through a 2-cm length of straight wire, in a magnetic field perpendicular to the wire, what is the magnetic field?
****F = .98 mlli-N
F = I*L x T_magnetic field
.98 mlli-N = .15A * .02m x T, because the magnetic field is perpendicular => the whole magnetic field is being applied => T = .98 mlli-N/ (.15A * .02m) = .33 T
#$&*
`q005. The Coulomb is a unit of electric charge. Two charges, each of 1 microCoulomb and each confined to a very small sphere which can be considered to be a point, with the spheres separated by 10 cm, will exert a force of magnitude .9 Newtons. The force decreases as the distance between the spheres increases, and increases as that distance decreases. The force is inversely proportional to the distance between the spheres. The interaction is completely analogous to the gravitational interaction between two masses, except that the force can be one of attraction or repulsion, depending on the signs of the two charges (i.e., whether each is positive or negative).
How much force would you expect if one of the forces was increased to 10 microCoulombs, with the other remaining the same?
****
r = .1m
F_0 = .9N
F is inversely proportional to r [I believe it is inversely proportional to the square of the distance.]
10x the force would be the result of ten times the charge of on charge if the other remained constant.
=> 9N
#$&*
How much force would you expect if the other force was then increased to 10 microCoulombs?
****90N
#$&*
Returning to the original 1 microCoulomb charges and 10 cm distance:
How much force would you expect if the distance was doubled to 20 cm?
****
it would be halfed according to the original definition given above, but I believe it is actually fourth = .9N/4 = .225 N
#$&*
How much force would you expect if the distance was halved to 5 cm?
****
four times as much = .9*4 N = 3.6N
#$&*
How much force would you expect if the distance was increased by a factor of 10, to 100 cm?
****
1 / 100 times the force = .009N
#$&*
How much force would you expect if the distance was decreased by a factor of 10, to 1 cm?
****
100x the force = 90N
#$&*
What force would you expect if the charges were both 1 Coulomb and the separation was 1 meter?
****the distance change would make it 1/100 the force but the charge change would make the force (1 * 10^6)^2
= .9N / 100 * 10^12 = 9 * 10^9N
#$&*
@& ... and this is where k = 9 * 10^9 N m^2 / C^2 comes from ...
*@
`q006. An amp (short for 'Ampere') is a flow of 1 Coulomb of charge per second. 1 Coulomb of charge is about 6 * 10^18 fundamental charges, where the fundamental charge is the magnitude of the charge on an electron.
How many Coulombs of charge flow in 1 minute through a circuit in which the current is .25 amps?
****
.25 C/sec * 60 sec/min = 15 C
#$&*
How long would it take for an Avogadro's number of electrons to flow through this circuit?
**** if A = C/s then s = C/A and if avagodros number is about 6.02*10^23 and the number of fundamental charges per C is (6*10^18) = > 6.02*10^23/(6*10^18) = about 10^5 C of charge in one mole. Know that the rate of .25 C/s => (10^5 charges) / .25 C = 4 * 10^5 = 4* 10 ^ 5 sec = about 111 hours
#$&*
`q007. A Coulomb of charge flowing through 1 volt has a potential energy change of 1 Joule. What potential energy change would you expect for each of the following situations?
.25 Coulombs of charge flow through 1 volt.
****
.25 Joules, determined via simple ratio.
#$&*
.20 Coulombs flow through a 1.5 volt battery.
****.3 Joules
#$&*
A current of .25 amps flows through a 1.5 volt battery for 60 seconds.
****A = C/s
=>A*t = C = .25 C/s * 60 s = 15C * 1.5 volts = 22.5 Joules
#$&*
`q008. How many amps must flow through a 110-volt circuit to produce 200 watts of power?
****
V*C = J
W = V*A
=>A=W/V = (200 J/s) / (110 J/C) = 1.8 C/s
#$&*
"
@& See my notes on a couple of potential glitches, but well done nonetheless.
*@