Answer
364.2k+ views
Hint: We are going to first explain what happens to the charged particle when it enters the magnetic field and the formula for the magnetic force that the particle experiences. Then based on the force value in the perpendicular magnetic field, the unchanged parameter is chosen.
Complete step-by-step solution:
When a charged particle enters a magnetic field, there is a magnetic force that is experienced by the charged particle. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The magnetic force is given by the formula
\[\vec F = q\left( {\vec v \times \vec B} \right)\]
The magnitude of the force will be
\[F = qvB\sin \theta \]
Now, since the charged particle is moving in a direction perpendicular to the field, then the magnitude of the force becomes
\[F = qvB\sin {90^ \circ } = qvB\]
Now, also, the direction of the force is perpendicular to the direction of the velocity of the particle or the displacement, hence, the work that is done on the particle by the magnetic field is zero which means that by the Work-Energy theorem that the change in the kinetic energy is also zero which gives us the result that the kinetic energy of the particle remains the same.
The other factors like velocity, force and momentum change.
Note: It is important to note that moves in circular path direction of momentum will change but the magnitude is unchanged. Thus, the change in direction only also means a change in the velocity also. The force is applied which is the centripetal force that causes the particle to move in a circular path.
Complete step-by-step solution:
When a charged particle enters a magnetic field, there is a magnetic force that is experienced by the charged particle. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The magnetic force is given by the formula
\[\vec F = q\left( {\vec v \times \vec B} \right)\]
The magnitude of the force will be
\[F = qvB\sin \theta \]
Now, since the charged particle is moving in a direction perpendicular to the field, then the magnitude of the force becomes
\[F = qvB\sin {90^ \circ } = qvB\]
Now, also, the direction of the force is perpendicular to the direction of the velocity of the particle or the displacement, hence, the work that is done on the particle by the magnetic field is zero which means that by the Work-Energy theorem that the change in the kinetic energy is also zero which gives us the result that the kinetic energy of the particle remains the same.
The other factors like velocity, force and momentum change.
Note: It is important to note that moves in circular path direction of momentum will change but the magnitude is unchanged. Thus, the change in direction only also means a change in the velocity also. The force is applied which is the centripetal force that causes the particle to move in a circular path.
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