Question

An electron in the beam of a cathode-ray tube is accelerated by a potential difference of 2.00 kV. Then it passes through a region of transverse magnetic field, where it moves in a circular arc with a radius of 0.180 m. What is the magnitude of the field?

Answer 1

Answer: The magnitude of the field is 8.384×10^-4 T

Explanation: Please see the attachments below

Answer 2

The magnitude of the magnetic field is $8.34 \times 10^{-4} T$.

Given, electron is accelerated by a potential difference of 2 KV.

Radius of circular arc is 0.18 m.

Equation of magnetic force, F = q v B.

But here electron is moving in a circular arc so there will be a centripetal force.

hence , centripetal force = magnetic force

$\frac{mv^{2} }{r}$ $= q v B$

On resolving the above equation ,

$r =\frac{mv}{qB}$.........(equation 1) here q is the charge of the electron.

We know that, Velocity v acquired  by the electron and accelerating  potential V are related by,

$\frac{1}{2} mv^{2}= e V$  where e and m are charge and mass of the electron.

$v^{2} =\frac{2eV}{m}$

$v=\sqrt[2]{\frac{2eV}{m} }$..........(equation 2)

Putting the value of v from equation 2 in equation 1, we get  $r =\frac{m\sqrt{\frac{2ev}{m} } }{qB}$

On solving we get,  $B = \frac{\sqrt{2emV} }{er}$

$B = \sqrt{ \frac{2mV}{er^{2} }$

putting the values of V, m, e and r in above equation

$B = \sqrt{\frac{2\times 9\times 10^{-31}\times 2\times 10^{3} }{1.6\times 10^{-19}\times 0.18^{2} } }$

On further solving the above equation we get $B = 8.34 \times 10^{-4} T$

therefore the magnitude of the magnetic field is $8.34 \times 10^{-4} T$.

For more details on Centripetal force follow the link below:

https://brainly.com/question/10596517