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# What is the radius(r)of the path for a singly charged ion?

The electric field between the plates of the velocity selector is 2800 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0960 T. Calculate the radius r of the path for a singly charged ion with mass m = 2.79 10-26 kg.

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- JánošíkLv 77 years agoFavorite Answer
r = (m × E) / (q × B²)

where

r = radius = ?

m = mass of the ion = 2.79E-26 kg

E = electric field = 2800 V/m

q = charge of 1 electron = 1.6E-19 C

B = magnetic field = 0.096 T

so

r = (2.79E-26 × 2800) / (1.6E-19 × 0.096²)

r = 0.05297 m = 5.3 cm

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