[RESOLVED NOW]

When i say i've been hammering my head against the problem i literally mean it.

A simple circuit is set up where a cell of potential V=1.50V is connected to itself by a wire in a circular loop of radius a=0.500m. Assume that the size of the cell is negligible compared to the length of the circuit. The wire is made out of copper and so has a resistivity ρ=1.80×10−8Ωm. Copper has a density of D=8.96×103kgm−3, a relative atomic mass of ma=63.5gmol−1 and there is one free electron per atom.

Potential of the cell: V = 1.50V

Radius of the circuit: a = 0.500m

Resistivity of copper: ρ = 1.80×10−8Ωm

Density of copper: D = 8.96×10^3kgm

Relative atomic mass of copper: ma = 63.5gmol

Avogadro's constant: nA=6.02×10^23

What is the drift velocity of electrons in this wire??

I've gotten far engouh to understand V/R = I and I should equate to n*q*A*v, I know you can get rid of A by cancelling it out with p=RA/L

by making R=pL/A, the only reason why i've been fully stuck is figuring out wth lower case n is and solving from there. The eqn i have of V/pL = D*Na*v is literally right there otherwise. (unless this entire thing is wrong too, it has happened before and franking i'm going mad)

When i say i've been hammering my head against the problem i literally mean it.

A simple circuit is set up where a cell of potential V=1.50V is connected to itself by a wire in a circular loop of radius a=0.500m. Assume that the size of the cell is negligible compared to the length of the circuit. The wire is made out of copper and so has a resistivity ρ=1.80×10−8Ωm. Copper has a density of D=8.96×103kgm−3, a relative atomic mass of ma=63.5gmol−1 and there is one free electron per atom.

Potential of the cell: V = 1.50V

Radius of the circuit: a = 0.500m

Resistivity of copper: ρ = 1.80×10−8Ωm

Density of copper: D = 8.96×10^3kgm

Relative atomic mass of copper: ma = 63.5gmol

Avogadro's constant: nA=6.02×10^23

What is the drift velocity of electrons in this wire??

I've gotten far engouh to understand V/R = I and I should equate to n*q*A*v, I know you can get rid of A by cancelling it out with p=RA/L

by making R=pL/A, the only reason why i've been fully stuck is figuring out wth lower case n is and solving from there. The eqn i have of V/pL = D*Na*v is literally right there otherwise. (unless this entire thing is wrong too, it has happened before and franking i'm going mad)

(edited 2 months ago)

the eqn i've got is V/pL = D x Na x V x n, literally what i was complaining about but i forgot to add the lower case ' n ' in

Be very frank, I have a problem understanding your notations to find out where you go wrong.

Your notation is a mess.

V/pL = D x Na x V x n

Your problem lies at the small n which is the number of electrons per unit volume and it involves molar mass (which I would write M_{r})

The small n can be found in the following 2 relationships.

$n = \dfrac{mol \times N_A}{V}$ --- Eqn (1)

$D = \dfrac{m}{V} = \dfrac{mol \times M_r}{V}$ --- Eqn (2)

where mol is the number of moles,

N_{A} is Avogadro's constant,

V is volume,

m is mass.

Your notation is a mess.

V/pL = D x Na x V x n

Your problem lies at the small n which is the number of electrons per unit volume and it involves molar mass (which I would write M

The small n can be found in the following 2 relationships.

$n = \dfrac{mol \times N_A}{V}$ --- Eqn (1)

$D = \dfrac{m}{V} = \dfrac{mol \times M_r}{V}$ --- Eqn (2)

where mol is the number of moles,

N

V is volume,

m is mass.

Original post by abuzztheuk

the eqn i've got is V/pL = D x Na x V x n, literally what i was complaining about but i forgot to add the lower case ' n ' in

You count small n twice.

Original post by Eimmanuel

Be very frank, I have a problem understanding your notations to find out where you go wrong.

Your notation is a mess.

V/pL = D x Na x V x n

Your problem lies at the small n which is the number of electrons per unit volume and it involves molar mass (which I would write M_{r})

The small n can be found in the following 2 relationships.

$n = \dfrac{mol \times N_A}{V}$ --- Eqn (1)

$D = \dfrac{m}{V} = \dfrac{mol \times M_r}{V}$ --- Eqn (2)

where mol is the number of moles,

N_{A} is Avogadro's constant,

V is volume,

m is mass.

Your notation is a mess.

V/pL = D x Na x V x n

Your problem lies at the small n which is the number of electrons per unit volume and it involves molar mass (which I would write M

The small n can be found in the following 2 relationships.

$n = \dfrac{mol \times N_A}{V}$ --- Eqn (1)

$D = \dfrac{m}{V} = \dfrac{mol \times M_r}{V}$ --- Eqn (2)

where mol is the number of moles,

N

V is volume,

m is mass.

Can't use this because I didn't tell you everything, srry about that.

Drift Velocity — Isaac Physics

here's the question if you want less ramble and more notations and what not, I don't use latex but a lot of people I know do. Basically I would use the volume in those two eqns you gave but we aren't given the cross sectional A.

What I want to find out is the velocity of the electrons in the wire and I know we can use the eqns:

p = RA/L

I = n x q x A x v

V = I x R

L is given as being pi bec the radius is 0.5 m and therefore the circumference and length of the circle is pi,

p = RA/L can be rearranged to get R = pL/A

we know V (Voltage), q (aka e, charge of an electron), R (resistance), L (length of wire), p (resistivity), Density. Na and Mr are given too.

(Potential of the cell: V = 1.50V

Radius of the circuit: a = 0.500m therefore L is pi = 3.14 ...

Resistivity of copper: ρ = 1.80×10−8Ωm

Density of copper: D = 8.96×10^3kgm

Relative atomic mass of copper: ma = 63.5gmol

Avogadro's constant: nA=6.02×10^23)

I got until

V = (n x q x A x v x p x L) / A

you can therefore remove surface area from the eqn which is correct as it isn't a given value, but then although everything else is known, n isn't and that's whats stopping me from getting to lower case v (velocity).

We don't get mass, volume or number of moles in the wire or anything like that. I know mas of the wire can be rewritten as DAL but then that would add another cross sectional area A into the eqn that can't cancel (i think).

(edited 2 months ago)

Original post by abuzztheuk

Can't use this because I didn't tell you everything, srry about that.

Drift Velocity — Isaac Physics

here's the question if you want less ramble and more notations and what not, I don't use latex but a lot of people I know do. Basically I would use the volume in those two eqns you gave but we aren't given the cross sectional A.

What I want to find out is the velocity of the electrons in the wire and I know we can use the eqns:

p = RA/L

I = n x q x A x v

V = I x R

L is given as being pi bec the radius is 0.5 m and therefore the circumference and length of the circle is pi,

p = RA/L can be rearranged to get R = pL/A

we know V (Voltage), q (aka e, charge of an electron), R (resistance), L (length of wire), p (resistivity), Density. Na and Mr are given too.

(Potential of the cell: V = 1.50V

Radius of the circuit: a = 0.500m therefore L is pi = 3.14 ...

Resistivity of copper: ρ = 1.80×10−8Ωm

Density of copper: D = 8.96×10^3kgm

Relative atomic mass of copper: ma = 63.5gmol

Avogadro's constant: nA=6.02×10^23)

I got until

V = (n x q x A x v x p x L) / A

you can therefore remove surface area from the eqn which is correct as it isn't a given value, but then although everything else is known, n isn't and that's whats stopping me from getting to lower case v (velocity).

We don't get mass, volume or number of moles in the wire or anything like that. I know mas of the wire can be rewritten as DAL but then that would add another cross sectional area A into the eqn that can't cancel (i think).

Drift Velocity — Isaac Physics

here's the question if you want less ramble and more notations and what not, I don't use latex but a lot of people I know do. Basically I would use the volume in those two eqns you gave but we aren't given the cross sectional A.

What I want to find out is the velocity of the electrons in the wire and I know we can use the eqns:

p = RA/L

I = n x q x A x v

V = I x R

L is given as being pi bec the radius is 0.5 m and therefore the circumference and length of the circle is pi,

p = RA/L can be rearranged to get R = pL/A

we know V (Voltage), q (aka e, charge of an electron), R (resistance), L (length of wire), p (resistivity), Density. Na and Mr are given too.

(Potential of the cell: V = 1.50V

Radius of the circuit: a = 0.500m therefore L is pi = 3.14 ...

Resistivity of copper: ρ = 1.80×10−8Ωm

Density of copper: D = 8.96×10^3kgm

Relative atomic mass of copper: ma = 63.5gmol

Avogadro's constant: nA=6.02×10^23)

I got until

V = (n x q x A x v x p x L) / A

you can therefore remove surface area from the eqn which is correct as it isn't a given value, but then although everything else is known, n isn't and that's whats stopping me from getting to lower case v (velocity).

We don't get mass, volume or number of moles in the wire or anything like that. I know mas of the wire can be rewritten as DAL but then that would add another cross sectional area A into the eqn that can't cancel (i think).

I agree with what you get

V = (n × q × A × <v> × ρ × L) / A = n × q × <v> × ρ × L

I mentioned in post #3 that we can use the two equations to express n in terms of other given variables, namely combined the two into a single equation where we have n, molar mass, Avogadro's constant and density.

Using Isaac Physics notation, I rewrite the two equations.

$n = \dfrac{mol \times N_A}{V_0}$ --- Eqn (1)

$D = \dfrac{m}{V_0} = \dfrac{mol \times m_a}{V_0}$ --- Eqn (2)

where

mol is number of moles,

N

V

m is mass,

m

We don't need to know the number of moles or mass.

Be careful that surface area and cross-sectional area are two different quantities.

Original post by Eimmanuel

I agree with what you get

V = (n × q × A × <v> × ρ × L) / A = n × q × <v> × ρ × L

I mentioned in post #3 that we can use the two equations to express n in terms of other given variables, namely combined the two into a single equation where we have n, molar mass, Avogadro's constant and density.

Using Isaac Physics notation, I rewrite the two equations.

$n = \dfrac{mol \times N_A}{V_0}$ --- Eqn (1)

$D = \dfrac{m}{V_0} = \dfrac{mol \times m_a}{V_0}$ --- Eqn (2)

where

mol is number of moles,

N_{A} is Avogadro's constant,

V_{0} is volume,

m is mass,

m_{a} is molar mass.

We don't need to know the number of moles or mass.

Be careful that surface area and cross-sectional area are two different quantities.

V = (n × q × A × <v> × ρ × L) / A = n × q × <v> × ρ × L

I mentioned in post #3 that we can use the two equations to express n in terms of other given variables, namely combined the two into a single equation where we have n, molar mass, Avogadro's constant and density.

Using Isaac Physics notation, I rewrite the two equations.

$n = \dfrac{mol \times N_A}{V_0}$ --- Eqn (1)

$D = \dfrac{m}{V_0} = \dfrac{mol \times m_a}{V_0}$ --- Eqn (2)

where

mol is number of moles,

N

V

m is mass,

m

We don't need to know the number of moles or mass.

Be careful that surface area and cross-sectional area are two different quantities.

I actually got it no way.

also yeah didn't notice I put surface area and cross-sectional area lol.

Thanks for the help, didn't realise what you said would work, kept on thinking volume isn't given so the eqn wouldn't work.

Original post by abuzztheuk

I actually got it no way.

also yeah didn't notice I put surface area and cross-sectional area lol.

Thanks for the help, didn't realise what you said would work, kept on thinking volume isn't given so the eqn wouldn't work.

also yeah didn't notice I put surface area and cross-sectional area lol.

Thanks for the help, didn't realise what you said would work, kept on thinking volume isn't given so the eqn wouldn't work.

I understand that you have solved the problem.

Some points to note:

•

whenever you are given Avogadro's constant, it is giving hint that you may need info/concept(s) from the ideal gas topic.

•

you should try to “derive” eqn (1) or “understand” how eqn (1) comes about.

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