Правка разделов «Statement», «Answer»
en/10.1.29.md
+2 −2
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $10.1.29.$ [Insert the problem statement] | ||
| + | $10.1.29.$ In strong magnetic fields, an electron moves along a helical line "wound" onto the magnetic field line. Prove that in the case when the radius of the helix is so small that the field inside it can be considered homogeneous, the product of the square of the radius of the helix and the induction of the magnetic field does not change. | ||
| ### Solution | |||
| The proof here will be very brief, referring to the solution of problem 10.1.24, the value of the invariant for the spiral trajectory was derived.: | |||
| $$\frac{Be(d^2-R^2)}{2} = const$$ | |||
| Where $d$ is the distance from the side circle to the axis of the main orbit, with $d=0$ this equality inevitably follows: | |||
| $$BR^2 = const$$ | |||
| @@ -14,4 +14,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | Q.E.D. | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $10.1.29.$ [Insert the problem statement] | $10.1.29.$ In strong magnetic fields, an electron moves along a helical line "wound" onto the magnetic field line. Prove that in the case when the radius of the helix is so small that the field inside it can be considered homogeneous, the product of the square of the radius of the helix and the induction of the magnetic field does not change. | ||
| ### Solution | ### Solution | ||
| The proof here will be very brief, referring to the solution of problem 10.1.24, the value of the invariant for the spiral trajectory was derived.: | The proof here will be very brief, referring to the solution of problem 10.1.24, the value of the invariant for the spiral trajectory was derived.: | ||
| $$\frac{Be(d^2-R^2)}{2} = const$$ | $$\frac{Be(d^2-R^2)}{2} = const$$ | ||
| Where $d$ is the distance from the side circle to the axis of the main orbit, with $d=0$ this equality inevitably follows: | Where $d$ is the distance from the side circle to the axis of the main orbit, with $d=0$ this equality inevitably follows: | ||
| $$BR^2 = const$$ | $$BR^2 = const$$ | ||
| @@ -14,4 +14,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | Q.E.D. | ||