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

Open in App

Solution

$r = \frac{\sqrt{2 q V m}}{B q}$

$B = \frac{1}{r} \sqrt{\frac{2 V m}{q}}$
$= \frac{1}{0.18} \sqrt{\frac{2 \times 2 \times 10^{3} \times 9.1 \times 10^{- 31}}{1.6 \times 10^{- 19}}}$
$= 8.38 \times 10^{- 4} T$

Suggest Corrections

Similar questions

Q. An electron in the beam of a TV picture tube is accelerated by a potential difference 2 kV. Then it passes through a transverse magnetic field to produce a circular arc of radius 0.18m. Find the magnetic field :

Q. An electron beam in a TV picture tube is accelerated through a potential difference of

V

volts. It passes through a region of transverse magnetic induction field (

B

) and follows a circular orbit of radius

^{'} r^{'}

. The induced magnetic field (

B

) is:

Q. An electron accelerated through a potential difference

V

passes through a uniform transverse magnetic field and experiences a force

F

. If the accelerating potential is increased to

2 V

, the electron in the same magnetic field will experience a force :

An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30º with the initial velocity.

Q. When an electron is accelerated through a potential difference V,it experiences a force F through a uniform transverse magnetic field.If the potential difference is increased to 2 V,the force experienced by the electron in the same magnetic field is

Join BYJU'S Learning Program

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

r=√2qVmBqB=1r√2Vmq =10.18√2×2×103×9.1×10−311.6×10−19 =8.38×10−4T

$r = \frac{\sqrt{2 q V m}}{B q}$

$B = \frac{1}{r} \sqrt{\frac{2 V m}{q}}$
$= \frac{1}{0.18} \sqrt{\frac{2 \times 2 \times 10^{3} \times 9.1 \times 10^{- 31}}{1.6 \times 10^{- 19}}}$
$= 8.38 \times 10^{- 4} T$