Question

An electron after being accelerated through a potential difference of 100 V enters a uniform magnetic field of 0*004T, perpendicular to its direction of motion. Calculate the radius of the path described by the electron.

Answer 1

Answer:

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

Answer : 8.43 mm

Explanation:

Given that

Potential Difference, V = 100 V

Magnetic Field , B = 0.004 T

As we know that

Force, F = qV = 1/2 (mv²)

⇒ V = $\sqrt{\frac{2qv}{m} }$

and for radius of path,

mv² / R = qvB

⇒ R = mv / qB = m / qB x $\sqrt{\frac{2qv}{m} }$

after putting the value

⇒ R = 843.17 x 10⁻⁵m = 8.43 mm

Therefore the radius of the path described by the electron will be 8.43 mm.

Answer 2

Given :

The potential difference of electron = 100 volt

The magnetic field = 0.004 Tesla

To Find :

The radius of the path described by the electron

Solution :

A electron enters a uniform magnetic field, perpendicular to its direction of motion .

mass of electron = m = 9.1 × $10^{-31}$ kg

charge of electron = q = 1.6 × $10^{-19}$

∵ Magnetic force = F = charge × potential

i.e       F = q × v         ..........1

And  For moving electron , force acting on it = F = $\dfrac{1}{2}$ m V²          ..........2

        where m = mass of electron

        And     V = velocity of electron

So,   from eq 1 and eq 2

       $\dfrac{1}{2}$ m V²  =  q × v  

Or,    m V²  = 2 q v

Or,    V²  = $\dfrac{2qv}{m}$

i.e  Velocity  = $\sqrt{\dfrac{2qv}{m} }$    m/s

Again

For Velocity of particle in circular path

   $\dfrac{mV^{2} }{r}$  = q v B

i.e    m V  =  q v B × r

Or,    m ($\sqrt{\dfrac{2qv}{m} }$)  =  q B × r

9.1 × $10^{-31}$ × $\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}}$  =  1.6 × $10^{-19}$ × 0.004  × r

Or,  9.1 × $10^{-31}$ × 5929994.5 = 6.4  × $10^{-22}$  × r

∴    radius = r = 8.43  × $10^{-3}$ meters

i.e     radius of path = 8.43 mm

Hence, The radius of the circular  path described by the electron is 8.43 millimeters  Answer