Edits to “Statement”, “Solution”, “Conclusion”
en/10.2.7.md
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| @@ -1,10 +1,10 @@ | |||
| ### Statement | |||
| − | $10.2.7.$ [Insert the problem statement] | ||
| + | $10.2.7.$ Prove that a charged particle in crossed electric and magnetic fields orbits with a frequency $\omega = \frac{qB}{m}$ around a center that moves with a drift velocity (and therefore the velocity of the particle at any given time is equal to the vector sum of the linear velocity of rotation around the instantaneous center and the drift velocity). | ||
| − | ### Solution | ||
| + |  | ||
| − | ### Solution | ||
| + | ### Solution | ||
| 1. Equation of motion in the laboratory frame | |||
| Let us write Newton's second law for a charged particle in crossed fields in our stationary (laboratory) reference frame: | |||
| $$m \frac{d\vec{v}}{dt} = q\vec{E} + q[\vec{v} \times \vec{B}]$$ | |||
| 2. Transition to a moving reference frame | |||
| Let us try to find an inertial reference frame $K'$ moving with some constant velocity $\vec{u}$ relative to the laboratory, in which the particle's motion will look simple. | |||
| According to the classical Galilean velocity addition law, the particle's velocity in the laboratory frame $\vec{v}$ is the vector sum of its velocity in the new frame $\vec{v}'$ and the velocity of the frame itself $\vec{u}$: | |||
| $$\vec{v} = \vec{v}' + \vec{u}$$ | |||
| Let's substitute this velocity into Newton's second law. Since frame $K'$ moves uniformly ($\vec{u} = \text{const}$), the acceleration is identical in both frames: $\frac{d\vec{v}}{dt} = \frac{d\vec{v}'}{dt}$. | |||
| $$m \frac{d\vec{v}'}{dt} = q\vec{E} + q[(\vec{v}' + \vec{u}) \times \vec{B}]$$ | |||
| Expanding the cross product: | |||
| $$m \frac{d\vec{v}'}{dt} = q\vec{E} + q[\vec{v}' \times \vec{B}] + q[\vec{u} \times \vec{B}]$$ | |||
| Let us group the constant terms that do not depend on the particle's current velocity $\vec{v}'$: | |||
| $$m \frac{d\vec{v}'}{dt} = q(\vec{E} + [\vec{u} \times \vec{B}]) + q[\vec{v}' \times \vec{B}]$$ | |||
| 3. Condition for the drift velocity | |||
| Now for the main physical trick: we can choose the velocity vector $\vec{u}$ to be whatever we want. Let us select a frame velocity $\vec{u}$ such that the constant bracket in our equation becomes strictly zero: | |||
| $$\vec{E} + [\vec{u} \times \vec{B}] = 0$$ | |||
| Since the fields are crossed by the problem's condition (i.e., $\vec{E} \perp \vec{B}$), the velocity vector $\vec{u}$ must be perpendicular to both of them so that the cross product $[\vec{u} \times \vec{B}]$ is directed perfectly opposite to the vector $\vec{E}$. In terms of magnitude, this means: | |||
| $$E = uB \implies u = \frac{E}{B}$$ | |||
| This is exactly the drift velocity $v_{\text{dr}} = \frac{E}{B}$. | |||
| 4. Rotation in the new reference frame | |||
| If we transition to the reference frame moving with this drift velocity, the constant force disappears (the effective electric field is nullified purely mechanically, without any Lorentz transformations!). The equation of motion for the particle in this new frame takes the simplest form: | |||
| $$m \frac{d\vec{v}'}{dt} = q[\vec{v}' \times \vec{B}]$$ | |||
| This is the standard equation of motion for a charge solely in a uniform magnetic field. As is well known, under the action of the Lorentz force, the particle undergoes uniform circular motion. | |||
| Let's write Newton's second law for this rotation (where the centripetal acceleration is $a_n = \omega^2 R$, and the linear rotational velocity is $v' = \omega R$): | |||
| $$m \omega^2 R = q (\omega R) B$$ | |||
| @@ -45,11 +45,7 @@Solution | |||
| $$m \omega = qB$$ | |||
| $$\omega = \frac{qB}{m}$$ | |||
| − | |||
| + | #### Conclusion: | ||
| We have rigorously proven that the particle's motion is a superposition of two simple motions: | |||
| 1. Translational rectilinear motion (drift) with velocity $u = E/B$. | |||
| 2. Rotation around this drifting center with the cyclotron frequency $\omega = qB/m$. | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,10 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $10.2.7.$ [Insert the problem statement] | $10.2.7.$ Prove that a charged particle in crossed electric and magnetic fields orbits with a frequency $\omega = \frac{qB}{m}$ around a center that moves with a drift velocity (and therefore the velocity of the particle at any given time is equal to the vector sum of the linear velocity of rotation around the instantaneous center and the drift velocity). | ||
| ### Solution |  | ||
| ### Solution |
### Solution | ||
| 1. Equation of motion in the laboratory frame | 1. Equation of motion in the laboratory frame | ||
| Let us write Newton's second law for a charged particle in crossed fields in our stationary (laboratory) reference frame: | Let us write Newton's second law for a charged particle in crossed fields in our stationary (laboratory) reference frame: | ||
| $$m \frac{d\vec{v}}{dt} = q\vec{E} + q[\vec{v} \times \vec{B}]$$ | $$m \frac{d\vec{v}}{dt} = q\vec{E} + q[\vec{v} \times \vec{B}]$$ | ||
| 2. Transition to a moving reference frame | 2. Transition to a moving reference frame | ||
| Let us try to find an inertial reference frame $K'$ moving with some constant velocity $\vec{u}$ relative to the laboratory, in which the particle's motion will look simple. | Let us try to find an inertial reference frame $K'$ moving with some constant velocity $\vec{u}$ relative to the laboratory, in which the particle's motion will look simple. | ||
| According to the classical Galilean velocity addition law, the particle's velocity in the laboratory frame $\vec{v}$ is the vector sum of its velocity in the new frame $\vec{v}'$ and the velocity of the frame itself $\vec{u}$: | According to the classical Galilean velocity addition law, the particle's velocity in the laboratory frame $\vec{v}$ is the vector sum of its velocity in the new frame $\vec{v}'$ and the velocity of the frame itself $\vec{u}$: | ||
| $$\vec{v} = \vec{v}' + \vec{u}$$ | $$\vec{v} = \vec{v}' + \vec{u}$$ | ||
| Let's substitute this velocity into Newton's second law. Since frame $K'$ moves uniformly ($\vec{u} = \text{const}$), the acceleration is identical in both frames: $\frac{d\vec{v}}{dt} = \frac{d\vec{v}'}{dt}$. | Let's substitute this velocity into Newton's second law. Since frame $K'$ moves uniformly ($\vec{u} = \text{const}$), the acceleration is identical in both frames: $\frac{d\vec{v}}{dt} = \frac{d\vec{v}'}{dt}$. | ||
| $$m \frac{d\vec{v}'}{dt} = q\vec{E} + q[(\vec{v}' + \vec{u}) \times \vec{B}]$$ | $$m \frac{d\vec{v}'}{dt} = q\vec{E} + q[(\vec{v}' + \vec{u}) \times \vec{B}]$$ | ||
| Expanding the cross product: | Expanding the cross product: | ||
| $$m \frac{d\vec{v}'}{dt} = q\vec{E} + q[\vec{v}' \times \vec{B}] + q[\vec{u} \times \vec{B}]$$ | $$m \frac{d\vec{v}'}{dt} = q\vec{E} + q[\vec{v}' \times \vec{B}] + q[\vec{u} \times \vec{B}]$$ | ||
| Let us group the constant terms that do not depend on the particle's current velocity $\vec{v}'$: | Let us group the constant terms that do not depend on the particle's current velocity $\vec{v}'$: | ||
| $$m \frac{d\vec{v}'}{dt} = q(\vec{E} + [\vec{u} \times \vec{B}]) + q[\vec{v}' \times \vec{B}]$$ | $$m \frac{d\vec{v}'}{dt} = q(\vec{E} + [\vec{u} \times \vec{B}]) + q[\vec{v}' \times \vec{B}]$$ | ||
| 3. Condition for the drift velocity | 3. Condition for the drift velocity | ||
| Now for the main physical trick: we can choose the velocity vector $\vec{u}$ to be whatever we want. Let us select a frame velocity $\vec{u}$ such that the constant bracket in our equation becomes strictly zero: | Now for the main physical trick: we can choose the velocity vector $\vec{u}$ to be whatever we want. Let us select a frame velocity $\vec{u}$ such that the constant bracket in our equation becomes strictly zero: | ||
| $$\vec{E} + [\vec{u} \times \vec{B}] = 0$$ | $$\vec{E} + [\vec{u} \times \vec{B}] = 0$$ | ||
| Since the fields are crossed by the problem's condition (i.e., $\vec{E} \perp \vec{B}$), the velocity vector $\vec{u}$ must be perpendicular to both of them so that the cross product $[\vec{u} \times \vec{B}]$ is directed perfectly opposite to the vector $\vec{E}$. In terms of magnitude, this means: | Since the fields are crossed by the problem's condition (i.e., $\vec{E} \perp \vec{B}$), the velocity vector $\vec{u}$ must be perpendicular to both of them so that the cross product $[\vec{u} \times \vec{B}]$ is directed perfectly opposite to the vector $\vec{E}$. In terms of magnitude, this means: | ||
| $$E = uB \implies u = \frac{E}{B}$$ | $$E = uB \implies u = \frac{E}{B}$$ | ||
| This is exactly the drift velocity $v_{\text{dr}} = \frac{E}{B}$. | This is exactly the drift velocity $v_{\text{dr}} = \frac{E}{B}$. | ||
| 4. Rotation in the new reference frame | 4. Rotation in the new reference frame | ||
| If we transition to the reference frame moving with this drift velocity, the constant force disappears (the effective electric field is nullified purely mechanically, without any Lorentz transformations!). The equation of motion for the particle in this new frame takes the simplest form: | If we transition to the reference frame moving with this drift velocity, the constant force disappears (the effective electric field is nullified purely mechanically, without any Lorentz transformations!). The equation of motion for the particle in this new frame takes the simplest form: | ||
| $$m \frac{d\vec{v}'}{dt} = q[\vec{v}' \times \vec{B}]$$ | $$m \frac{d\vec{v}'}{dt} = q[\vec{v}' \times \vec{B}]$$ | ||
| This is the standard equation of motion for a charge solely in a uniform magnetic field. As is well known, under the action of the Lorentz force, the particle undergoes uniform circular motion. | This is the standard equation of motion for a charge solely in a uniform magnetic field. As is well known, under the action of the Lorentz force, the particle undergoes uniform circular motion. | ||
| Let's write Newton's second law for this rotation (where the centripetal acceleration is $a_n = \omega^2 R$, and the linear rotational velocity is $v' = \omega R$): | Let's write Newton's second law for this rotation (where the centripetal acceleration is $a_n = \omega^2 R$, and the linear rotational velocity is $v' = \omega R$): | ||
| $$m \omega^2 R = q (\omega R) B$$ | $$m \omega^2 R = q (\omega R) B$$ | ||
| @@ -45,11 +45,7 @@Solution | |||
| $$m \omega = qB$$ | $$m \omega = qB$$ | ||
| $$\omega = \frac{qB}{m}$$ | $$\omega = \frac{qB}{m}$$ | ||
| #### Conclusion: | |||
| We have rigorously proven that the particle's motion is a superposition of two simple motions: | We have rigorously proven that the particle's motion is a superposition of two simple motions: | ||
| 1. Translational rectilinear motion (drift) with velocity $u = E/B$. | 1. Translational rectilinear motion (drift) with velocity $u = E/B$. | ||
| 2. Rotation around this drifting center with the cyclotron frequency $\omega = qB/m$. | 2. Rotation around this drifting center with the cyclotron frequency $\omega = qB/m$. | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||