
A particle is moving in a uniform magnetic field, then
A. Its momentum changes but total energy remains the same
B. Both momentum and total energy remain the same
C. Both will change
D. Total energy changes but momentum remains the same
Answer
161.7k+ views
Hint: When a charged particle is moving in a uniform field then there is magnetic force on the moving charge. When force is applied then there will be acceleration and hence change in velocity. Newton’s 2nd law of motion defines the force as a rate of change of momentum of the body.
Formula used:
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\], here \[\vec F\]is the magnetic force vector, \[\vec v\]is the velocity of the charged particle and \[\vec B\]is the magnetic field in the region.
Complete answer:
When a magnetic field is applied in a region and charge is moving in the region, then the magnetic force acting on the charge is given as,
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\]
As we know that the vector product of two vectors is in the direction perpendicular to both the vectors simultaneously, so the magnetic force will also be perpendicular to the magnetic field and the velocity.
When a constant perpendicular force is applied on a moving particle then the path of the motion is circular.
In circular motion the magnitude of the velocity remains constant throughout the motion when a constant perpendicular force is applied.
The energy of the moving particle is due to kinetic energy which depends on the square of the magnitude of the velocity. So the energy of the particle is constant.
The momentum of the particle is a vector quantity, as the force is non-zero so the rate of change in momentum is non-zero.
Hence, the energy of the particle is unchanged but the momentum is changing.
Therefore, the correct option is (A).
Note:We should be careful while considering the momentum. The momentum is a vector quantity which is defined by the direction as well as the magnitude. Even if the magnitude is unchanged, changing the direction will change the momentum.
Formula used:
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\], here \[\vec F\]is the magnetic force vector, \[\vec v\]is the velocity of the charged particle and \[\vec B\]is the magnetic field in the region.
Complete answer:
When a magnetic field is applied in a region and charge is moving in the region, then the magnetic force acting on the charge is given as,
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\]
As we know that the vector product of two vectors is in the direction perpendicular to both the vectors simultaneously, so the magnetic force will also be perpendicular to the magnetic field and the velocity.
When a constant perpendicular force is applied on a moving particle then the path of the motion is circular.
In circular motion the magnitude of the velocity remains constant throughout the motion when a constant perpendicular force is applied.
The energy of the moving particle is due to kinetic energy which depends on the square of the magnitude of the velocity. So the energy of the particle is constant.
The momentum of the particle is a vector quantity, as the force is non-zero so the rate of change in momentum is non-zero.
Hence, the energy of the particle is unchanged but the momentum is changing.
Therefore, the correct option is (A).
Note:We should be careful while considering the momentum. The momentum is a vector quantity which is defined by the direction as well as the magnitude. Even if the magnitude is unchanged, changing the direction will change the momentum.
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