Edits to “Statement”, “Answer”
en/10.1.11.md
+0 −11
| @@ -1,11 +1,5 @@ | |||
| ### Statement | |||
| − | $10.1.11.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| 10.1.11. A particle of mass $m$ and charge $q$ enters a uniform magnetic field of induction $B$ at an angle $\alpha$ to the field with velocity $v$. Find the radius and pitch of the helical path along which the particle moves. | |||
| ### Solution | |||
| Let us decompose the particle's velocity into two components — parallel to the magnetic field: $v_{\parallel} = v\cos\alpha$ and perpendicular to the magnetic field: $v_{\perp} = v\sin\alpha$. | |||
| The perpendicular component causes circular motion due to the action of the Lorentz force, while the parallel one causes uniform translational motion along the magnetic field lines. The combined action of these two motions forms a helical path. | |||
| Let us write Newton's second law: | |||
| $$qv_{\perp}B = m\frac{v_{\perp}^2}{R}$$ | |||
| $$R = \frac{mv_{\perp}}{qB} = \frac{mv\sin\alpha}{qB}$$ | |||
| The period of revolution (time of one turn): | |||
| $$T = \frac{2\pi R}{v_{\perp}} = \frac{2\pi \frac{mv\sin\alpha}{qB}}{v\sin\alpha} = \frac{2\pi m}{qB}$$ | |||
| The pitch of the helical path (distance along the field per one turn): | |||
| $$h = v_{\parallel} \cdot T = v\cos\alpha \cdot \frac{2\pi m}{qB} = \frac{2\pi mv\cos\alpha}{qB}$$ | |||
| @@ -27,8 +21,3 @@Solution | |||
| #### Answer | |||
| $$R = \frac{mv\sin\alpha}{qB}$$ | |||
| $$h = \frac{2\pi mv\cos\alpha}{qB}$$ | |||
| − | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $10.1.11.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| 10.1.11. A particle of mass $m$ and charge $q$ enters a uniform magnetic field of induction $B$ at an angle $\alpha$ to the field with velocity $v$. Find the radius and pitch of the helical path along which the particle moves. | 10.1.11. A particle of mass $m$ and charge $q$ enters a uniform magnetic field of induction $B$ at an angle $\alpha$ to the field with velocity $v$. Find the radius and pitch of the helical path along which the particle moves. | ||
| ### Solution | ### Solution | ||
| Let us decompose the particle's velocity into two components — parallel to the magnetic field: $v_{\parallel} = v\cos\alpha$ and perpendicular to the magnetic field: $v_{\perp} = v\sin\alpha$. | Let us decompose the particle's velocity into two components — parallel to the magnetic field: $v_{\parallel} = v\cos\alpha$ and perpendicular to the magnetic field: $v_{\perp} = v\sin\alpha$. | ||
| The perpendicular component causes circular motion due to the action of the Lorentz force, while the parallel one causes uniform translational motion along the magnetic field lines. The combined action of these two motions forms a helical path. | The perpendicular component causes circular motion due to the action of the Lorentz force, while the parallel one causes uniform translational motion along the magnetic field lines. The combined action of these two motions forms a helical path. | ||
| Let us write Newton's second law: | Let us write Newton's second law: | ||
| $$qv_{\perp}B = m\frac{v_{\perp}^2}{R}$$ | $$qv_{\perp}B = m\frac{v_{\perp}^2}{R}$$ | ||
| $$R = \frac{mv_{\perp}}{qB} = \frac{mv\sin\alpha}{qB}$$ | $$R = \frac{mv_{\perp}}{qB} = \frac{mv\sin\alpha}{qB}$$ | ||
| The period of revolution (time of one turn): | The period of revolution (time of one turn): | ||
| $$T = \frac{2\pi R}{v_{\perp}} = \frac{2\pi \frac{mv\sin\alpha}{qB}}{v\sin\alpha} = \frac{2\pi m}{qB}$$ | $$T = \frac{2\pi R}{v_{\perp}} = \frac{2\pi \frac{mv\sin\alpha}{qB}}{v\sin\alpha} = \frac{2\pi m}{qB}$$ | ||
| The pitch of the helical path (distance along the field per one turn): | The pitch of the helical path (distance along the field per one turn): | ||
| $$h = v_{\parallel} \cdot T = v\cos\alpha \cdot \frac{2\pi m}{qB} = \frac{2\pi mv\cos\alpha}{qB}$$ | $$h = v_{\parallel} \cdot T = v\cos\alpha \cdot \frac{2\pi m}{qB} = \frac{2\pi mv\cos\alpha}{qB}$$ | ||
| @@ -27,8 +21,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$R = \frac{mv\sin\alpha}{qB}$$ | $$R = \frac{mv\sin\alpha}{qB}$$ | ||
| $$h = \frac{2\pi mv\cos\alpha}{qB}$$ | $$h = \frac{2\pi mv\cos\alpha}{qB}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||