Respuesta :
Explanation:
Given that,
Electric field, E = 7900 V/m
Magnetic field, [tex]B=9.1\times 10^{-3}\ T[/tex]
(a) Let v is the velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic fields. The magnitude of electric force and the magnetic force balance each other as :
[tex]qE=qvB[/tex]
[tex]v=\dfrac{E}{B}[/tex]
[tex]v=\dfrac{7900\ V/m}{9.1\times 10^{-3}\ T}[/tex]
[tex]v=8.68\times 10^5\ m/s[/tex]
(b) Let r is the radius of the electron orbit. The radius of the motion of electron is given by :
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r=\dfrac{9.1\times 10^{-31}\ kg\times 8.68\times 10^5\ m/s}{1.6\times 10^{-19}\ C\times 9.1\times 10^{-3}\ T}[/tex]
[tex]r=5.4\times 10^{-4}\ m[/tex]
Hence, this is the required solution.
Velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic fields is [tex]8.68\times10^{5}[/tex] m/s and radius is [tex]5.4\times10^{-4}[/tex] meters.
What is electric and magnetic field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
The magnetic field is the field in the space and around the magnet in which the magnetic field can be fill.
Given information-
The magnitude of the electric field is 7900 V/m.
The magnitude of the magnetic field is 9.1*10^-3 T.
The velocity of a beam of electrons-
As the electrons goes undeflected when moving perpendicular to an electric and magnetic fields.
It means that the electric force balance by the magnetic force, to not deflect the electron by any of the force. Therefore,
[tex]qE=qvB[/tex]
Here, [tex]qE[/tex] is the electric force and [tex]qvB[/tex] is the magnetic force.
Solve the above equation further,
[tex]qE=qvB\\E=vB\\v=\dfrac{E}{B}[/tex]
Put the values as,
[tex]v=\dfrac{7900}{9.1\times10^{-3}}\\v=8.68\times10^{5}\rm m/s[/tex]
Thus the velocity of a beam of electrons is [tex]8.68\times10^5[/tex] m/s.
The radius of the electron orbit if the electric field is turned off-
The diameter of the electron can be given as,
[tex]d=\dfrac{\dfrac{mv^2}{2}}{qvB}[/tex]
Diameter is the half of the radius. Thus,
[tex]\dfrac{r}{2}=\dfrac{\dfrac{mv^2}{2}}{qvB}\\r=\dfrac{mv}{qB}[/tex]
Here, (m) is the mass of electron and (v) is the velocity of electron and (q) is the charge of a electron.
Put the values as,
[tex]r=\dfrac{9.1\times10^{-31}\times8.68\times10^{5}}{1.6\times10^{-19}\times9.1\times10^{-3}}[/tex]
[tex]r=5.4\times10^{-4}\rm m[/tex]
Thus the radius of the electron orbit if the electric field is turned off is [tex]5.4\times10^{-4}[/tex] meters.
Hence,
- 1) The velocity of a beam of electrons is [tex]8.68\times10^{5}[/tex] m/s.
- 2) The radius of the electron orbit if the electric field is turned off is [tex]5.4\times10^{-4}[/tex] meters.
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