
A charge moves in a circle perpendicular to a magnetic field. The time period of revolution is independent of
A. Magnetic field
B. Charge
C. Mass of the particle
D. Velocity of the particle
Answer
161.1k+ views
Hint: If a charge moves in a circle perpendicular to the magnetic field, then the direction between the magnetic field and the velocity vector is 90°. And so the magnetic force will be perpendicular to the velocity, the speed of the charge will be constant.
Formula used:
\[{F_m} = qVB\], here \[{F_m}\]is the magnetic field acting on the charge particle q moving with speed v in a magnetic field strength B where the motion is perpendicular to the magnetic field.
\[{F_c} = \dfrac{{m{v^2}}}{r}\], here \[{F_c}\]is the centrifugal force acting on a body of mass m moving with speed v in a circular path of radius r.
Complete answer:
Let the mass of the charge is m, the charge is q and the velocity of the charge in the magnetic field is v.
The velocity is perpendicular to the magnetic field, so the magnitude of the magnetic force is,
\[F = qvB\sin 90^\circ \]
The magnetic force acting on the moving charge is,
\[{F_m} = qvB \ldots \ldots \left( i \right)\]
The direction of the magnetic force is inward to the circular path.
Then the centrifugal force acting on the charge is,
\[{F_C} = \dfrac{{m{v^2}}}{r} \ldots \ldots \left( {ii} \right)\]
To balance the outward centrifugal force, the magnetic force on the charge should be inward and have magnitude equal to the outward force.
For the charge to be in circular motion, the two forces must balance each other.
On balancing the force, we get
\[\dfrac{{m{v^2}}}{r} = evB\]
\[r = \dfrac{{mv}}{{Bq}}\]
So, the radius of the circular path of the charge is \[r = \dfrac{{mv}}{{Bq}}\]
The time period of the charge in circular motion will be the time taken to complete one revolution,
\[T = \dfrac{{2\pi r}}{v}\]
\[T = \dfrac{{2\pi \left( {\dfrac{{mv}}{{Bq}}} \right)}}{v}\]
\[T = \dfrac{{2\pi m}}{{Bq}}\]
So, from the obtained expression of the time period of charge, the time period is independent of the velocity of the charge.
Therefore, the correct option is (D).
Note:We have assumed that the mass of the particle is constant and there is no conversion of mass to the energy while the particle is in circular motion otherwise the time period will change.
Formula used:
\[{F_m} = qVB\], here \[{F_m}\]is the magnetic field acting on the charge particle q moving with speed v in a magnetic field strength B where the motion is perpendicular to the magnetic field.
\[{F_c} = \dfrac{{m{v^2}}}{r}\], here \[{F_c}\]is the centrifugal force acting on a body of mass m moving with speed v in a circular path of radius r.
Complete answer:
Let the mass of the charge is m, the charge is q and the velocity of the charge in the magnetic field is v.
The velocity is perpendicular to the magnetic field, so the magnitude of the magnetic force is,
\[F = qvB\sin 90^\circ \]
The magnetic force acting on the moving charge is,
\[{F_m} = qvB \ldots \ldots \left( i \right)\]
The direction of the magnetic force is inward to the circular path.
Then the centrifugal force acting on the charge is,
\[{F_C} = \dfrac{{m{v^2}}}{r} \ldots \ldots \left( {ii} \right)\]
To balance the outward centrifugal force, the magnetic force on the charge should be inward and have magnitude equal to the outward force.
For the charge to be in circular motion, the two forces must balance each other.
On balancing the force, we get
\[\dfrac{{m{v^2}}}{r} = evB\]
\[r = \dfrac{{mv}}{{Bq}}\]
So, the radius of the circular path of the charge is \[r = \dfrac{{mv}}{{Bq}}\]
The time period of the charge in circular motion will be the time taken to complete one revolution,
\[T = \dfrac{{2\pi r}}{v}\]
\[T = \dfrac{{2\pi \left( {\dfrac{{mv}}{{Bq}}} \right)}}{v}\]
\[T = \dfrac{{2\pi m}}{{Bq}}\]
So, from the obtained expression of the time period of charge, the time period is independent of the velocity of the charge.
Therefore, the correct option is (D).
Note:We have assumed that the mass of the particle is constant and there is no conversion of mass to the energy while the particle is in circular motion otherwise the time period will change.
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