# AP Physics 2: 1.1 Properties of Objects and Systems

AP Physics 2: 1.1 Properties of Objects and Systems. What is the magnitude and direction of the conventional current in this wire?

AP | AP Physics 2 |

AP Physics 2 | Properties of Objects and Systems |

AP Physics B/C | Waves and Optics |

Language | English Language |

Science Practice 1 | Using representations and models |

Test Prep | AP Physics 2 |

### Transcript

Here we go jakes learning about electric current in class

But the monotonous voice of the teacher sends jake into

a daydream Imagine shrinking down to the size of an

electron In exploring the wire jake abruptly wakes up when

his teacher asked him are you paying attention Howto electrons

behave Jake replies Well they're pretty cool They go with

the flow in his dream when a potential difference is

set up across the ends of the wire jake and

the electrons all started bumping along in the same direction

The current the wires calculated with the equation i equals

nada que where N is the number of charged particles

per unit Volume is the cross sectional area of the

wire is the drift velocity of the particles and cue

is the charge on each particle brought to you by

american express Yeah All right The cylindrical wire above has

a radius of zero point five millimeters The electrons have

a speed of be equals one millimeters per second and

a density of unequal six times ten to the twenty

eight meters to the negative third what's the magnitude and

direction of the conventional current in this wire and hear

the potential answers all right Interesting Well when we daydream

were usually in a spaceship headed to mars not wearing

clothes or we're making the winning shot at the buzzer

Yeah like that We don't dream about being an electron

and a wire but we don't judge jake here he's

going to solve the world's problems and we're grateful were

given most of what we need here were given number

of particles see were given the radius of the wire

And with that we were able to find the area

of the cross section with a little old school gym

called pi r squared remember that didi Yeah area and

because we're dealing with some crazy big numbers will make

it a little easier on ourselves and round pie down

to three Oh and will express the radius in meters

rather than millimeters keep everything in the same firm we're

given The drift velocity which is one millimeter per second

or one meter times ten to the negative third per

second We're not given the charge of each particle but

we know that we're dealing with electrons Electrons are teeny

tiny little particles with a tv tiny electrical charge Their

charges so tiny in fact that they produce the smallest

possible unit of charge called the elementary charge or just

eat elementary charge is produced by protons and electrons The

proton is all smiles with a positive elementary charged electrons

so they don't go in for all that positive attitude

stuff they're always negative Negative negative twenty four seven Yeah

like that real downer Sad mood The elementary charge of

one electron is one point six two one seven six

five seven times ten to the negative nineteen who longs

Yeah but like electricity we try to take the path

of least resistance So we'll round that up to times

ten to the negative nineteenth Cool Okay we've got our

numbers and we can plug him into our equations Quick

reminder Current equals the number of charged particles per unit

of volume times the cross sectional area of the wire

times The drift velocity the particles times the charge of

each particle Got that Now if you think and we

have our density six times ten to the twenty eighth

meters to the negative third we have our area three

times point five times ten to the negative Thirty meters

squared You have our velocity one meter times ten to

the negative Third per second And we have our charge

two times ten to the negative nineteenth cool arms putting

in the numbers just looks like this and it's pretty

ugly And we can clean that up a little by

taking care of the square in the middle making the

area equal three times point two five times ten to

the negative six meters square And when we do the

math or have our fancy pants calculated to the mat

for us we come up with an answer of a

charge of nine films per second A nine amps Look

at us making progress All right One last part of

the puzzle is knowing which direction the current is going

in this wire We're dealing in terms of conventional charge

which assumes the charge is positive Well positive charge moves

from higher potential towards lower potential Those since a has

a higher potential than be the conventional current is making

a beeline towards b b line and say we did

there get it pretty good All right now if we're

talking in terms of electron flow it'd be the opposite

But we're not talking about electron flow so don't sweat

that right now after all of this were able to

definitively state that the answer is there's A current of

nine amps headed towards you shouldn't worry if we daydream

little it's actually an important part of the scientific process

It also prepare us for a nobel prize speech Yeah 00:04:48.867 --> [endTime] now that's some great daydreaming