
Find the condition under which charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed
Answer
548.4k+ views
Hint: The magnetic field does not work, so the kinetic energy and speed of a charged particle in a magnetic field remain constant.
In case of a positive charge Force on a charged particle due to an electric field is directed parallel to the electric field and in case of negative charge its anti -parallel
Complete Step by Step Solution:
Let us consider a case
A charged particle $q$ which is moving with velocity $v$ experiencing both magnetic $B$ and electric field $E$ perpendicular to it so it experiences the force
\[F = q(E + v \times B)\]
So now, Electric and magnetic forces are in opposite directions and the values of E and B are adjusted in such a manner that magnitudes of the two forces are equal
F = qvBsin$\theta$ where $\theta = 90^\circ $
\[
F = qvB{\text{ }} \\
{\text{F}} = qE \\
\]
We equate both considering both forces to be equal then
\[qvB = qEv = E/B\]
Then the total force on the charge is zero and the charge will move in the field undeflected.
This condition can be used to select the charged particles of a particular velocity out of a beam containing charges moving with different speed (irrespective of their charge and mass).
Only particles with speed E/B pass un deflected through the region of crossed fields.
Note So to understand it further. The electromagnetic field affects the change in behaviour of charged objects surrounding a specific area
It is a combination of electric field and a magnetic field and generally considered as the sources of the electromagnetic field. The electric field is generated by stationary charges and the magnetic field is produced by moving charges.
In case of a positive charge Force on a charged particle due to an electric field is directed parallel to the electric field and in case of negative charge its anti -parallel
Complete Step by Step Solution:
Let us consider a case
A charged particle $q$ which is moving with velocity $v$ experiencing both magnetic $B$ and electric field $E$ perpendicular to it so it experiences the force
\[F = q(E + v \times B)\]
So now, Electric and magnetic forces are in opposite directions and the values of E and B are adjusted in such a manner that magnitudes of the two forces are equal
F = qvBsin$\theta$ where $\theta = 90^\circ $
\[
F = qvB{\text{ }} \\
{\text{F}} = qE \\
\]
We equate both considering both forces to be equal then
\[qvB = qEv = E/B\]
Then the total force on the charge is zero and the charge will move in the field undeflected.
This condition can be used to select the charged particles of a particular velocity out of a beam containing charges moving with different speed (irrespective of their charge and mass).
Only particles with speed E/B pass un deflected through the region of crossed fields.
Note So to understand it further. The electromagnetic field affects the change in behaviour of charged objects surrounding a specific area
It is a combination of electric field and a magnetic field and generally considered as the sources of the electromagnetic field. The electric field is generated by stationary charges and the magnetic field is produced by moving charges.
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