Hence the expression of ω in terms of m and k is
Given the expressions;
Equating both expressions we will have;
Divide both equations by 2π
Square both sides
Take the square root of both sides
Hence the expression of ω in terms of m and k is
an electron accelerated from rest through a voltage of 770 v enters a region of constant magnetic field. part a if the electron
1 answer:
The magnitude of the magnetic field on which an electron accelerated from rest through a voltage of 770v enters a region of constant magnetic field is 3.744 * T
From the conservation of energy, we have
=m/s
At Equilibrium, Centripetal force = magnetic force
B=mv/rq
by putting, m=9.11* 10^-31 kg
v=1.664*10^7m/s
r=25*10^-2m
q=1.6*10^-19 C
We get, B=3.774* 10^-4 T
Hence the magnitude of the magnetic field, B=3.774* 10^-4 T
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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.