First slide

Motion of a charged particle

Question

An electron gun G emits electrons of energy 2 keV travelling in the positive x-direction. The electrons are required to hit the spot S where GS=0.1 m, and the line GS makes an angle of $60^{0}$ with the x-axis as shown in figure. A uniform magnetic field $\vec{B}$ parallel to GS exists in the region outside the electron gun. Find the minimum value of ‘B’ needed to make the electron hit S.

Moderate

A

$4.73 \times 10^{- 3} T$
B

$47.3 \times 10^{- 3} T$
C

$473 \times 10^{- 3} T$
D

$0.473 \times 10^{- 3} T$

Solution

Kinetic energy of electron, $K = \frac{1}{2} m v^{2} = 2 k e V$
Speed of electron, $v = \sqrt{\frac{2 K}{m}} = \sqrt{\frac{2 \times 2 \times 1.6 \times 10^{- 16}}{9.1 \times 10^{- 31}}} m s^{- 1}$ $= 2.65 \times 10^{7} m s^{- 1}$
Since the velocity $(\vec{v})$ of the electron makes an angle of $θ = 60^{0}$ with the magnetic field $\vec{B}$ ,
the path will be a helix.
So, the particle will hit ‘S’ if $G S = n p$ , Here, $n = 1, 2, 3, ......$ and $p =$ pitch of helix $= \frac{2 π m}{q B} v \cos θ$
$\Rightarrow G S = \frac{n 2 π m v \cos θ}{q B}$
$\Rightarrow B = \frac{n 2 π m v \cos θ}{q (G S)}$
But for ‘B’ to be minimum, $n = 1$
$B_{\min} = \frac{2 π m v \cos θ}{q (G S)}$
Substituting the values, we have
$B_{\min} = \frac{2 π (9.1 \times 10^{- 31}) (2.65 \times 10^{7}) (\frac{1}{2})}{(1.6 \times 10^{- 19}) (0.1)} = 4.73 \times 10^{- 3} T$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App