a doubly ionized moleculeIe a molecule lacking two electrons

a doubly ionized molecule(I.e., a molecule lacking two electrons), moving in a magnetic force of 6.55 ×10^-16 N as it moves at 351 m/s at 69.7° to the direction of the field. Find the magnitude of the magnetic field.
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Solution

Charge of electron = 1.6×10^{_19c. Double ionized molecule = 2×1.6×10^_19c. F= 6.55×10^_16 N. V=351m/s. Theeta= 69.7°. F=qvB(sin theeta). We want to magnetic field B= F/qv(sin theeta). B= 6.55/2×1.6×10^_19×sin(69.7°). B=6.55×10^19/1.76 =3.7215×10^19T.}