Question

A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.31×10−16 N as it moves at 265 m/s at 69.1∘ to the direction of the field. Find the magnitude of the magnetic field. T

Answer 1

Answer:

So magnetic field will be 0.0067 T

Explanation:

We have given that molecule is doubly ionized

So charge $q=2e=2\times 1.6\times 10^{-19}=3.2\times 10^{-19}C$

Magnetic force is given by F = $5.31\times 10^{-16}N$

velocity is given as v = 265 m/sec

Angle $\Theta =69.1^{\circ}$

We know that magnetic force is given by

$F=qvBsin\Theta$

$B=\frac{F}{qvsin\Theta }=\frac{5.31\times 10^{-16}}{3.2\times 10^{-16}\times 265\times sin69.1^{\circ}}=0.0067T$

Answer 2

The magnitude of the magnetic field through which the doubly ionized molecule moves is obtained to be 6.7 Tesla.

Lorentz Force

We know that the Lorentz force experienced by a charge moving in a magnetic field is given by;

$F = q\,(\vec{v}\times \vec{B})=qvB\,sin\,\theta$

Therefore, the magnitude of the magnetic field can be given by;

$B=\frac{F}{qv\,sin\,\theta}$

Given that;

$q=2e=2\times 1.6\times 10^{-19}\,C\\v=265\,m/s\\\theta = 69.1^\circ\\F=5.31\times 10^{-16}\,N$

Substituting all the known values in the equation for 'B' we get;

$B=\frac{5.31\times 10^{-16}\,N}{(2\times 1.6\times 10^{-19}C)\times (265\,m/s)\times sin\,69.1^\circ}$

$\implies B= 6.7\,T$

Learn more about Lorentz force here:

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