A beam of electron is accelerated by a potential difference of 2.5okV. The electrons are then allowed to pass into a magnetic field of strength 10.0mT. As the electrons enter the field, the direction of the velocity is at right angles to the direction of the magnetic field. All this takes place in an evacuated chamber. What is the speed of the electron just before it enters the magnetic field?
Ans: i tried Ek= e(deltaV) and also the magnitude of force, however that is not given.
E=eV is energy, you have to convert that to KEnergy and velocity.
jnkj
To determine the speed of the electron just before it enters the magnetic field, we can use the concept of conservation of energy.
First, let's find the potential energy (PE) gained by the electron due to the potential difference (ΔV).
Potential energy gained (PE) = e * ΔV
Where e is the charge of an electron (approximately 1.6 x 10^-19 C) and ΔV is the potential difference (2.5 kV = 2.5 * 10^3 V).
PE = (1.6 x 10^-19 C) * (2.5 x 10^3 V)
Next, we can find the kinetic energy (KE) gained by the electron. Since the chamber is evacuated, we can assume that there is no resistance, and therefore the energy gained will solely be in the form of kinetic energy.
Kinetic energy gained (KE) = PE
Now, equating the PE to KE:
KE = (1/2) * m * v^2
Where m is the mass of an electron (approximately 9.1 x 10^-31 kg) and v is the speed of the electron.
(1/2) * m * v^2 = PE
Rearranging the equation to solve for v:
v = √((2 * PE) / m)
Substituting the values:
v = √((2 * (1.6 x 10^-19 C) * (2.5 x 10^3 V)) / (9.1 x 10^-31 kg))
Evaluating this expression will give you the speed of the electron just before it enters the magnetic field.