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# 6 Physics problems: Calculating ion field paths

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1. A singly charged positive ion has a mass of 2.5*10^-26kg. After being accelerated through a potential difference of 250V, the ion enter a magnetic field of 0.50T, in a direction perpendicular to the field. Find the radius of the path of the ion in the field.

2. Two ions with masses of 5.1*10^-26kg move out of the slit of a mass spectrometer and into a region in which the magnetic field is 0.04T. Each has a speed of 1.44*10^6m/s, but one ion is singly charged while the other is doubly charged. I. Find the radius of the circular path followed by each in the field, and II. the distance of separation when they have move through one half their circular path and strike a piece of photographic paper.

3. A swimmer is in a deep pool with her eyes a horizontal distance R=1.5m from the edge. How far below the surface are her eyes if she is just able to see the full height of a lifeguard who is standing on the pool's edge?? The index of refraction of water is 1.33.

4. A submarine is 314m horizontally out from the shore and 111m beneath the surface of the water. A laser beam is sent from the sub so that it strikes the surface of the water at a point 214m from the shore. If the beam just strkes the top of a building standing directly at the water's edge, find the height of the building.

5. The radius of the galaxy is 3*10^18m I. How fast would a spaceship have to travel to cross the entire galaxy in 441 years, as measured from within the spaceship?? II. How much time would elapse on Earth?

6. If it takes 3750MeV of work to accelerate a proton from rest to a speed of v, determine v.

https://brainmass.com/physics/galaxies/6-physics-problems-calculating-ion-field-paths-7982

#### Solution Preview

1.) mass of ion (m) = 2.5*10^(-26) kg
charge (q) = 1.6 * 10^(-19) Coulomb (Because, singly ionised)
Potential difference (V) = 250 Volt
Magnetic field intensity (B) = 0.5 Tesla
Because ion is accelerated by a potential difference of V, therefore,
q*V = energy gained = kinetic energy = (1/2)m*v^2
=>v = sqrt(2*q*V/m) ............(1)
when ion eneters a magnetic field, perpendicular to the direction of field, because of charge and motion of the ion it will trace a circular path. The centripetal force is obtained from magnetic (Lorentz)force:
q*v*B = m*v^2/r
where r is the radius of circular path traced by ion.
=> q*B = m*v/r => r = m*v/(q*B)
substitute the value of v from equation (1), we get,
r = m*sqrt(2*q*V/m)/(q*B)
=> r = sqrt(2*m*V/q)/B
=> r = sqrt[2*2.5*10^(-26) * 250/{1.6*10^(-19)}]/0.5
=> r = sqrt[2*25*250*10^(-26+19)/16]/0.5
=> r = 2*sqrt(2*25*25*10^(-7+1)/16]
=> r = 2*5*5*10^(-3)*sqrt(2)/4
=> r = 12.5*sqrt(2)*10^(-3) meter
=> r = 17.67 *10^(-3) meter = 1.76 Cm Answer

2.) mass m1 = ...

#### Solution Summary

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