Return to 6a. The singly charged negative ion is replaced with a triply charged

Question

Return to 6a. The singly charged negative ion is replaced with a triply charged negative (-3e) ion that has the same mass m(kg) and velocity v(m/s). If the radius of the path of the ion in 6a is R(m), the radius of the path of this new ion in the magnetic field will be _______________. Give your answer in terms of R in the form "R(m)/n" or "nxR" where n is a factor dividing R(m) or multiplying R(m). For example, if your answer is 9 times smaller than R(m) then you would input R(m)/9. If your amnswer is 9 times larger then you would input 9R(m).

Explanation / Answer

A charge placed in the magnetic field moves in a circular orbit and in this situation, essentially, the centripetal force is equal to the magnetic force experienced by the charge.

In case of of the singly charged negative ion, magnitude of the charge is, |q| = e, mass is m (in kg) and velocity is v (in m/s). Magnetic field be B (in T). Then,

mv2/R = |q|vB

where R (in m) is the radius of the circular path of the singly charged ion.

Thus,

R = mv/|q|B = mv/eB

Now, we replace this singly charged ion with a triply charged ion, with mass m, velocity v remaining unaltered. Thus, magnitude of the charge is |q| = 3e

Hence, radius of the circular path of this triply charged ion is,

R-3e = mv/|q|B

R-3e = mv/[(3e)B]

R-3e = (1/3)R = R/3

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