Question #113855

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?

Expert's answer

Given: U=100VU=100 V ; d=5mm=5.0⋅10−3Vd=5mm=5.0\cdot 10^{-3}V; B=0.01T=10−2TB=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 F⃗c=e⋅E⃗\vec F_c=e\cdot \vec E and Lorentz F⃗l=e⋅[v⃗хB⃗]\vec F_l=e\cdot[\vec vх\vec B] forces. Since an electron beam passes un deflected with uniform velocity v⃗\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 k^\hat k (E⃗=E⋅k^\vec E=E\cdot \hat k ), the magnetic field is directed along j^\hat j (B⃗=B⋅j^\vec B=B\cdot \hat j ) parallel plates and the velocity of electrons is directed along parallel plates too, say v⃗=vxi^+vyj^\vec v=v_x\hat i+v_y\hat j . If we calculate the Lorentz force,

(1) F⃗l=e⋅(i^j^k^vxvy00B0)=e⋅(vxB⋅k^)\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 vy≠0v_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 vy=0v_y=0 and the movement of electrons occurs only along X direction. Subject to compensation of forces we have

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

(2) E=vxBE=v_xB or vx=EBv_x=\frac{E}{B}

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

(3)E=Ud=100V5⋅10−3m=2⋅104V/mE=\frac {U}{d}=\frac{100V}{5\cdot 10^{-3}m}=2\cdot10^4 V/m

Substituting (3) into (2) we get finally

vx=2⋅104V/m10−2T=2⋅106m/sv_x=\frac{2\cdot 10^4V/m}{10^{-2}T}=2\cdot 10^6 m/s

Answer: the velocity of beam is 2⋅106m/s2\cdot 10^6 m/s


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