Question

An electron beam passes un deflected with
uniform velocity v through two parallel
plates, when a magnetic field of 0 01 T is
applied perpendicular to an electric field
between the plates produced by a p d of
100 V The separation of the plates is 5 mm
Calculate the velocity?

Answer 1

Given: $U=100 V$ ; $d=5mm=5.0\cdot 10^{-3}V$; $B=0.01T=10^{-2}T$

Solution: In the problem, the motion of electrons occurs in electric and magnetic fields. In this case, two forces exert on a charged particle, these are Coulomb $\vec F_c=e\cdot \vec E$ and Lorentz $\vec F_l=e\cdot[\vec vх\vec B]$ forces. Since an electron beam passes un deflected with uniform velocity $\vec v$ these two forces compensate each other. We may choose a coordinate system so that the electric field is directed perpendicular to plates along $\hat k$ ($\vec E=E\cdot \hat k$ ), the magnetic field is directed along $\hat j$ ($\vec B=B\cdot \hat j$ ) parallel plates and the velocity of electrons is directed along parallel plates too, say $\vec v=v_x\hat i+v_y\hat j$ . If we calculate the Lorentz force,

(1) $\vec F_l=e\cdot \begin{pmatrix} \hat i & \hat j & \hat k\\ v_x & v_y &0\\0& B &0 \end{pmatrix}=e\cdot(v_x B\cdot \hat k)$ we will see that the Lorentz force disappears for movement along the magnetic field. Therefore, if the electrons had a velocity component $v_y\not =0$ they would move in the Y direction without Lorentz force, and a Lorentz force could not compensate the electric force. Therefore, we must put $v_y=0$ and the movement of electrons occurs only along X direction. Subject to compensation of forces we have

$\vec F_c=\vec F_l$ The charge of partical is reduced, the directions of forces are coincide and we get

(2) $E=v_xB$ or $v_x=\frac{E}{B}$

The field $E$ of parallel plates condenser can be calculated by the formula

(3)$E=\frac {U}{d}=\frac{100V}{5\cdot 10^{-3}m}=2\cdot10^4 V/m$

Substituting (3) into (2) we get finally

$v_x=\frac{2\cdot 10^4V/m}{10^{-2}T}=2\cdot 10^6 m/s$

Answer: the velocity of beam is $2\cdot 10^6 m/s$