Question

An electron in a cathode-ray beam passes between $2.5cm$-long parallel-plate electrodes that are $5.0mm$ apart. A $2.0mT$, $2.5-cm$-wide magnetic field is perpendicular to the electric field between the plates. The electron passes through the electrodes without being deflected if the potential difference between the plates is $600V.$

a. What is the electron’s speed?

b. If the potential difference between the plates is set to zero, what is the electron’s radius of curvature in the magnetic field?

Short answer

a. Electron's Speed; $v=$$6\times {10}^{7}m/s$

b. Radius of curvature; $r=1.708m$

Step-by-step solution

Part (a) Step 1: Given Information

We have to find out the electron's speed passing through the electrodes without being deflected

Part (a) step 2 : Detailed solution 

Distance between two long parallel plate electrodes ;

Magnetic field between the plates ;

Potential Difference between plates;

As we know,

Magnetic Force on electron ;$F_{m=}Bqv$

q is charge on electron ;

v is speed of electron ;

Also,

Electric Force ; localid="1649745706061" $F_E=qE=q(V/d)$

; E is Electric field

As we know,

$F_m=F_E$

$$\begin{gathered} \Rightarrow Bqv=qE \\ \Rightarrow Bqv=(V/d)q \\ \Rightarrow v=V/Bd=\frac{600}{2\times {10}^{-3}\times 5\times {10}^{-3}}=6\times {10}^{7}m/s \end{gathered}$$

Thus, speed of electron is$6\times {10}^{7}m/s$

part (b) step 1  Given information

We have to find out the radius of curvature of electron in magnetic field when potential difference between the plates is zero

Part(b)  step 2 : Detailed solution

Distance between two long parallel plate electrodes ;

Magnetic field between the plates ;

Potential Difference between plates;

As we know,

Centripetal force,$F_c$

localid="1649750325922" $$\begin{gathered} F_c=\frac{m{v}^2}{r}=Bqv \\ \Rightarrow r=\frac{mv}{qB}=\frac{(9.11\times {10}^{-31})(6\times {10}^7)}{(2\times {10}^{-3})(1.6\times {10}^{-19})}=1.708m \end{gathered}$$

Thus, radius of curvature in magnetic field is $1.708m$