Question

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 :

• A. $6\times {10}^{-4}T$
• B. $7.68\times {10}^{-4}T$
• C. $8.38\times {10}^{-4}T$
• D. $8.98\times {10}^{-4}T$

Answer 1

Answer: (C)

$8.38\times {10}^{-4}T$

Velocity $v$ acquired by the electron and accelerating potential $V$ are related by,
$\frac{1}{2}m{v}^2=eV$ where $e$ and $m$ are charge and mass of the electron
$\Rightarrow v=\sqrt{\frac{2eV}{m}}$
Radius of the circular track:
$r=\frac{mv}{eB}=\frac{m\sqrt{\frac{2eV}{m}}}{eB}$
$\Rightarrow B=\frac{\sqrt{2emV}}{er}=\sqrt{\frac{2mV}{e{r}^2}}=\sqrt{\frac{2\times 9\times {10}^{-31}\times 2\times {10}^3}{1.6\times {10}^{-19}\times {0.18}^2}}=8.34\times {10}^{-4}T$