Question

An electron accelerated by the potential difference U = 103V enters into a magnetic field

An electron accelerated by the potential difference U = 10^3V enters into a magnetic field

is perpendicular to its direction of motion. Inductive

B = 1,19.10^-3T

. Find:

a. The curvature radius of the ellectic orbit

B T 1,19.10

. Find:

a. The curvature radius of the ellectic orbit

Answer 1

Using the Newton’s Second Law of Motion we can write:

$$qvB=\dfrac{mv^2}{r},$$ $$r=\dfrac{mv}{qB}.$$

We can find the velocity of the electron from the law of conservation of energy:

$$qV=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{2qV}{m}}.$$

Substituting $v$ into the formula for the radius of the electron orbit, we get:

$$r=\dfrac{\sqrt{2Vm}}{\sqrt{q}B}$$ $$r=\dfrac{\sqrt{2\cdot10^3\ V\cdot 9.1\cdot10^{-31}\ kg}}{\sqrt{1.6\cdot10^{-19}\ C}\cdot 1.19\cdot10^{-3}\ T}=0.089\ m.$$