Answer:
4.4345× 10^-7V
Explanation:
The computation of the half voltage for a 1.2T magnetic field applied is shown below
The volume of one mole of copper is
v = m ÷p
= 63.5 ÷ 8.92
= 7.12cm
Now the density of free electrons in copper is
n = Na ÷ V
= 6.02 × 10^23 ÷ 7.12
= 8.456× 10^28/m^3
Now the half voltage is
= IB ÷ nqt
= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)
= 4.4345× 10^-7V
Answer:
I would say all of the above.
Explanation:
Look below for more examples
Answer:
The correct answer is
Explanation:
The formula for the electron drift speed is given as follows,
where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.
Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is . Converting this number to m³ using very elementary unit conversion we get
. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,
if we convert the area from mm³ to m³ we get .So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,
which is our final answer.