Given Information:
potential difference = V = 350 V
Magnetic field = B = 200 mT = 0.200 T
Mass of electron = m = 9.11x10⁻³¹ kg
charge on electron = e = 1.60x10⁻¹⁹ C
Required Information:
velocity = v = ?
radius = r = ?
Answer:
velocity = 11.1x10⁶ m/s
radius = r = 0.000316 m
Explanation:
As we know when the potential energy of the electron becomes equal to kinetic energy of electron
eV = ½mv²
re-arranging for the speed
v² = 2eV/m
v = √(2eV)/m
Where e is the charge on electron and m is the mass of electron
v = √(2*1.60x10⁻¹⁹*350)/9.11x10⁻³¹
v = 11.1x10⁶ m/s
When the electron enters in the magnetic field, then the radius of the path in which electron moves is given by
r = mv/eB
where B is the magnetic field
r = 9.11x10⁻³¹*1.11x10⁷/1.60x10⁻¹⁹*0.200
r = 0.000316 m
Answer:
a) 1.11×10^7m/s
b) radius = 3.16×10^-4m
Explanation:
Assume we have an electron accelerated using potential difference of 350v which gives the ion this speed. We set the potential energy to be equal to the kinetic energy.
eV = 1/2mv^2
Rearranging the formular gives
V = Sqrt(2eV/m)
Substituting the given values
V = Sqrt(2 (1.6×10^-19)(350)/(9.11×10^-31)
V = 1.11×10^7m/s
b) The electron enters a region of uniform magnetic feild . It moves in a circular path with raduis of:
r = mv/eB
r = (9.11×10^-31kg)(1.11×10^7m/s)/(1.6×10^'19C)(200×10^-3T)
r = 3.16×10^-4m
