Question
Un ion con una sola carga que se mueve en un campo magnético uniforme de $7,5 \times 10^{-2} \mathrm{~T}$ completa 10 revoluciones en $3,47 \times 10^{-4} \mathrm{~s}$. Identifique el ion.
Step 1
47 \times 10^{-4} \mathrm{~s}$, we can calculate the frequency of the ion's motion using the formula: \[ f = \frac{1}{T} \] where $T$ is the time period for one revolution. \[ f = \frac{1}{3.47 \times 10^{-4}} = 2.88 \times 10^3 \mathrm{~Hz} \] Show more…
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A singly charged ion completes five revolutions in a uniform magnetic field of magnitude $5.00 \times 10^{-2} \mathrm{~T}$ in $1.50 \mathrm{~ms}$. Calculate the mass of the ion in kilograms.
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