Question

A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.33×10−16 N as it moves at 369 m/s at 68.5∘ to the direction of the field. Find the magnitude of the magnetic field.

Answer #1

We have given that molecule is doubly ionized

So charge
q=2e=2×1.6×10^{-19}=3.2×10^{-19}c

Magnetic force is given
by F = 5.33×10^{-16}N

velocity is given as v = 369 m/sec

Angle =68.5

We know that magnetic force is given by

F=qvBsin

Now magnetic field B=F/qvsin

Substituting all the required values in above equation.

Therefore
B=5.33×10^{-16}/(3.2×10^{-19}×369×sin68.5)

B=4.851T

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