Question

An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide magnetic field isperpendicular to the electric field between the plates. The electron passes through the electrodes without being deflected if the potential difference between the plates is 700V.
Part A:What is the electron's speed?(Express your answer to two significant figures and include the appropriate units.)

Part B: If the potential difference between the plates is set to zero, what is the electron's radius of curvature in the magnetic field?(Express your answer to two significant figures and include the appropriate units.)

Answer 1

Answer:

(a). The speed of electron is $1.56\times10^{7}\ m/s$.

(b). The radius of electron is $4.2\ cm$

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

$v=\sqrt{\dfrac{2eV}{m}}$

Put the value into the formula

$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}$

$v=15689290.81\ m/s$

$v=1.56\times10^{7}\ m/s$

(b). We need to calculate the radius of electron

Using formula of centripetal force

$\dfrac{mv^2}{r}=qvB$

$r=\dfrac{mv}{qB}$

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

$r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}$

$r=0.042\ m$

$r=4.2\ cm$

Hence, (a). The speed of electron is $1.56\times10^{7}\ m/s$.

(b). The radius of electron is 4.2 cm