When a charged particle moves perpendicular to the uniform magnetic field, then
A. Its momentum changes, total energy is same
B. Both momentum and total energy remains the same
C. Both momentum and its total energy will change
D. Total energy changes, momentum remains same
Answer
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Hint: To answer this question, formula for Lorentz force and the formula for kinetic energy of a charged particle can be used. Using the Lorentz force, we can find the variation in velocity of the particle when the particle moves perpendicular to the magnetic field. With the help of variation in the velocity, we can find what happens to momentum. Then, we use this same variation in formula for the kinetic energy. Thus, we can know whether the energy remains constant or changes.
Formula used:
$F= q (v \times B)$
$K.E. =\dfrac {1}{2}m{v}^{2}$
Complete answer:
Force on a charged particle is given by,
$F= q (v \times B)$ …(1)
Where, v is the velocity of the particle
q is the charge on the particle
B is the magnetic field
Using equation. (1), when a charged particle moves perpendicular to the uniform magnetic field, the magnetic force will act perpendicular to the velocity at every instant which makes the particle move in a circular path with constant velocity v. Thus, the magnitude of velocity remains same, but the direction changes. As the direction of the particle changes, its momentum also changes.
Kinetic energy of a charged particle is given by.
$K.E. =\dfrac {1}{2}m{v}^{2}$ …(2)
Where, m is the mass of the particle
v is the velocity of particle
From the equation. (2), it can be inferred that the energy is proportional to the velocity of the particle. So, when the velocity of the particle is constant its energy will also remain constant. Thus, we can say, when a charged particle moves perpendicular to the uniform magnetic field, then the momentum changes while the total energy remains the same.
So, the correct answer is “Option A”.
Note:
If the velocity is not perpendicular to the magnetic field, then we compare each component of the velocity individually with the magnetic field. The velocity component perpendicular to the magnetic field, makes the particle move in a circular path. But, the velocity component parallel to the field makes the particle move along a straight line. The resulting motion is helical. While moving from this helical path, there can be regions of non-uniform magnetic fields.
Formula used:
$F= q (v \times B)$
$K.E. =\dfrac {1}{2}m{v}^{2}$
Complete answer:
Force on a charged particle is given by,
$F= q (v \times B)$ …(1)
Where, v is the velocity of the particle
q is the charge on the particle
B is the magnetic field
Using equation. (1), when a charged particle moves perpendicular to the uniform magnetic field, the magnetic force will act perpendicular to the velocity at every instant which makes the particle move in a circular path with constant velocity v. Thus, the magnitude of velocity remains same, but the direction changes. As the direction of the particle changes, its momentum also changes.
Kinetic energy of a charged particle is given by.
$K.E. =\dfrac {1}{2}m{v}^{2}$ …(2)
Where, m is the mass of the particle
v is the velocity of particle
From the equation. (2), it can be inferred that the energy is proportional to the velocity of the particle. So, when the velocity of the particle is constant its energy will also remain constant. Thus, we can say, when a charged particle moves perpendicular to the uniform magnetic field, then the momentum changes while the total energy remains the same.
So, the correct answer is “Option A”.
Note:
If the velocity is not perpendicular to the magnetic field, then we compare each component of the velocity individually with the magnetic field. The velocity component perpendicular to the magnetic field, makes the particle move in a circular path. But, the velocity component parallel to the field makes the particle move along a straight line. The resulting motion is helical. While moving from this helical path, there can be regions of non-uniform magnetic fields.
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