how do i go about answering thiss question?
particles with a mass of 7.96x10^-27 and have a (+2e)are accelerated by a voltage and kept in a magnetic field. at one stage the ion starts ar 1x10^6 m/s and exits at 2.75x10^6 m/s. immediatleyt after theey enter a magnetic feild and fallow a radius of 1.24m
the voltage in the accelerating chamber is?
What does this have to do with physics diploma?
How do you go about the question.
Take a look at your text:
Coulomb force=Eq=V/d * q
but force=mass*acceleration=mass(Vf-Vi)/time where time = distance/avgvelocity
What is avgvelocity: (Vi+Vf)/2
Solve for voltage.
There are no questions on the magnetic deflection encountered after entering the magnetic field.
a question from the old diploma
To determine the voltage in the accelerating chamber, we need to use the concepts of electric potential energy and kinetic energy.
First, let's determine the change in kinetic energy of the particle as it accelerates from 1x10^6 m/s to 2.75x10^6 m/s. The change in kinetic energy (ΔKE) is given by the equation:
ΔKE = 0.5 * m * (v_final^2 - v_initial^2)
Here, m represents the mass of the particle, and v_final and v_initial represent the final and initial velocities, respectively.
Substituting the given values into the formula, we get:
ΔKE = 0.5 * (7.96x10^-27 kg) * ((2.75x10^6 m/s)^2 - (1x10^6 m/s)^2)
Simplifying the equation:
ΔKE = 0.5 * (7.96x10^-27 kg) * (7.5625x10^12 m^2/s^2 - 1x10^12 m^2/s^2)
Next, we can calculate the change in electric potential energy (ΔPE) using the formula:
ΔPE = q * ΔV
Here, q represents the charge on the particle, and ΔV represents the change in voltage. The charge on the particle is given as +2e, where e is the elementary charge (approximately 1.6x10^-19 C).
Substituting the values, we have:
ΔPE = (+2e) * ΔV
Now, equating the change in potential energy to the change in kinetic energy:
ΔKE = ΔPE
0.5 * (7.96x10^-27 kg) * (7.5625x10^12 m^2/s^2 - 1x10^12 m^2/s^2) = (+2e) * ΔV
Simplifying the equation further, we can solve for ΔV:
ΔV = [0.5 * (7.96x10^-27 kg) * (7.5625x10^12 m^2/s^2 - 1x10^12 m^2/s^2)] / (+2e)
Calculating the value yields the voltage in the accelerating chamber.