Question

1. An electron having 450 eV of energy moves at right angles to a uniform magnetic field of flux density 1.50 X 10-3 T. Find the radius of its circular orbit. Assume that the specific charge (e/m) of the electron is 1.76 x 1011 C kg-1. ​

Answer 1

Explanation:

E=

2

1

mv

2

⇒v=

m

2E

r=

Be

mv

=

Be

m

m

2E

=

Be

2mE

r=

1.6×10

−19

×0.4

2×1800×1.6×10

−19

×9.1×10

−31

=3.58×10

−4

m

Answer 2

Answer:

The radius of the circular orbit is 476cm.

Explanation:

Given the energy of an electron,  E = 450 eV

The magnetic flux density, $B= 1.50\times10^{-3} T$

The specific charge of the electron is

$\dfrac{e}{m} =1.76 \times 10^{11} C kg^{-1}$

From the kinetic energy of the electron, given by

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

we get, the velocity of electron

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

Substituting the given values, the velocity of electron is

$v=\sqrt{2\times\dfrac{e}{m}\times V }$

  $=\sqrt{2\times1.76 \times 10^{11}\times 450 }=12.58 \times 10^{6}m/s$

When the magnetic field is applied, the electron moves in a circular path of radius r due to the centripetal force provided by the the magnetic force.

i.e., centripetal force = magnetic force

$\implies \dfrac{mv^{2} }{r} =Bev$

$\implies \dfrac{v }{r} =\dfrac{Be }{m} \implies r=\dfrac{v }{B \dfrac{e }{m} }$  

Substituting the given values, the radius of the circular path is

$r=\dfrac{12.58 \times 10^{6} }{1.50 \times 10^{-3}\times1.76 \times 10^{11}}$

 $=\dfrac{12.58 \times 10^{-2} }{2.64}=4.76\times 10^{-2}m=476cm$

Therefore, the radius of the circular orbit is 476cm.

For more problems:

https://brainly.in/question/18337202

https://brainly.in/question/14092455