Question

An electron after being accelerated through a potential difference of 100v enters a uniform magnetic field of 0.004t perpendicular to its direction of motion.Calculate radius of path described by electron

Answer 1

Answer:

Explanation:

Given An electron after being accelerated through a potential difference of 100 v enters a uniform magnetic field of 0.004 t perpendicular to its direction of motion.Calculate radius of path described by electron

We are given v = 100 v, B = 0.004 T r = ?

Let v be the velocity acquired by electron when accelerated under a potential difference v

So eV = 1/2 mv^2

Since the electron describes a circular path of radius r in a magnetic field,

Beν = mv^2 / r

r = mν / Be

 = m / Be √2eV / m

 = √ 2 meV / Be

 = √ 2 x 9.1 x 10^-31 x 1.6 x 10^-19 x 100 / 0.004 x 1.6 x 10^-19

 = 8.4 x 10^-3 m

= 8.4 mm

Answer 2

The radius of path described by electron is 8.4 mm.

Explanation:

Given potential difference, V = 100 V.

The magnetic field, B = 0.004 T.

The charge on electron,$e=1.6\times10^-^1^9C$.

We have to calculate the radius, r (let).

As electron when accelerated through a pot. differnce, so  

$eV=\frac{1}{2}mv^2$                            (1)

Here, v is the velocity of the electron.

The path described by the electron, is circular in the perpendicular magnetic field, so

magnetic force =  centripetal force

$evB=\frac{mv^2}{r}$                                  (2)

From equations, (1) and (2)

$r=\frac{\sqrt{2meV} }{eB}$

substitute the given values, we get

$r=\frac{\sqrt{2\times9.1\times10^-^3^1\times100V} }{1.6\times10^-^1^9\times0.004 T}$

r = 8.4 mm.

Thus, the radius of path described by electron is 8.4 mm.

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