The 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.
.01N
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A 250-milliamp current in a 1-meter wire in a 1-Tesla magnetic field.
.25N
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A 1-amp current in a 1-centimeter wire in a 1-Tesla magnetic field.
.01N
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A 250-milliamg current in a 1-centimeter wire in a .01-Tesla magnetic field.
.000025N
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`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.
East
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The current is directed toward the west and the magnetic field is downward.
South
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The current is directed toward the east and the magnetic field is directed toward the west.
Up
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The current is directed upward and the magnetic field is directed toward the south.
East
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`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?
North
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What will be the direction of the force exerted on the south side of the loop?
South
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What will be the direction of the force exerted on the east side of the loop?
East
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What will be the direction of the force exerted on the west side of the loop?
West
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What effect would these forces have on the loop?
They would cancel each other out
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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?
Not possible
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What will be the direction of the force exerted on the south side of the loop?
Not possible
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What will be the direction of the force exerted on the east side of the loop?
North
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What will be the direction of the force exerted on the west side of the loop?
North
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What effect would these forces have on the loop?
Upward effect
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`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?
****
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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?
****
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`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 charges was increased to 10 microCoulombs, with the other remaining the same?
9N
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How much force would you expect if the other force was then increased to 10 microCoulombs?
90N
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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?
(1/1) = .9N, then (1/2)*.9N =.45N
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How much force would you expect if the distance was halved to 5 cm?
2*.9N = 1.8N
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How much force would you expect if the distance was increased by a factor of 10, to 100 cm?
(1/10)*.9N = .09N
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How much force would you expect if the distance was decreased by a factor of 10, to 1 cm?
10*.9N = 9N
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What force would you expect if the charges were both 1 Coulomb and the separation was 1 meter?
(1/10)*(100*100)*.9N = 1000*.9N = 900N
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`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/s)*60s = 15 C
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How long would it take for an Avogadro's number of electrons to flow through this circuit?
1C = 6 * 10^18 fundamental charges
15 C = 15*(6 * 10^18) = 9*10^19 ele, in one min
6*10^23 ele/9*10^19 ele/min = approx. 6,666.6667min
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`q007. A Coulomb of charge flowing through 1 volt has a potential energy change of magnitude 1 Joule. What potential energy change would you expect for each of the following situations?
.25 Coulombs of charge flow through 1 volt.
.25 J
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.20 Coulombs flow through a 1.5 volt battery.
.3 J
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A current of .25 amps flows through a 1.5 volt battery for 60 seconds.
.25 amp is .25 C/s, 1.5 volts is 1.5 J/C
.25 C through 1.5 J/C = .25 C * 1.5 J = .375 J, this is amount of charge required every second
.375 J/s * 60s = 22.5 J (per min)
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`q008. How many amps must flow through a 110-volt circuit to produce 200 watts of power?
P = V*I
I = 200watts/110volts = 1.81…amps
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&#Very good work. Let me know if you have questions. &#