Question

An electron travels in a circular path of radius 20 cm in a magnetic field 2×10^-3 T. Calculate the speed and V

Answer 1

Answer:

Explanation:

r= 0.2m

B= 2x10-3T

m= 9.1x10-31 kg

q= 1.6x10-19 C

thus, v= qrB/m= 7.032x107 ms-1

to aceelerate an electron to this speed-

eV=(1/2)mv2

V=(1/2)mv2/e=3.516x103 volt

Answer 2

Explanation:

It is given that,

Radius of the circular path, r = 20 cm = 0.2 m

Magnetic field, $B=2\times 10^{-3}\ T$

The centripetal force is balanced by the magnetic force acting on the electron such that,

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

$v=\dfrac{qBr}{m}$

$v=\dfrac{1.6\times 10^{-19}\times 2\times 10^{-3}\times 0.2}{9.1\times 10^{-31}}$

$v=7.03\times 10^7\ m/s$

Using the relation as :

$eV=\dfrac{1}{2}mv^2$

$V=\dfrac{1}{2e}mv^2$

$V=\dfrac{1}{2\times 1.6\times 10^{-19}}\times 9.1\times 10^{-31}\times (7.03\times 10^7)^2$

$V=14054.06\ volts$

Hence, this is the required solution.