Question

An electron emitted by a heated cathode and accelerated through a potential difference of $2\text{kV}$ enters a region with uniform magnetic field of $0.15\text{T}$. If the magnetic field makes an angle of ${30}^{\circ }$ with the initial velocity, the radius of the hellical trajectory of electron will be: $(m_e=9.0\times {10}^{-31}\text{kg})$

• A. $1\text{mm}$
• B. $1.5\text{mm}$
• C. $2.5\text{mm}$
• D. $0.5\text{mm}$

Answer 1

The correct option is D $0.5\text{mm}$

The electron is accelerated by potential difference

$V=2\text{kV}=2000\text{volt}$

$\frac{1}{2}m{v}^2=qV$

or, $\frac{1}{2}m{v}^2=1.6\times {10}^{-19}\times 2000$
$v=\sqrt{\frac{2\times 1.6\times {10}^{-19}\times 2000}{9\times {10}^{-31}}}=\frac{8}{3}\times {10}^7\text{m/s}$




The perpendicular component of velocity will cause electron to revolve in circular path and the parallel component will make it move in a straight line.

Thus the radius of helical path will be,

$r=\frac{m{v}_{\perp }}{qB}$

$r=\frac{mv\sin \theta }{qB}$

$r=\frac{9\times {10}^{-31}\times (\frac{8}{3}\times {10}^7)\times \sin {30}^{\circ }}{1.6\times {10}^{-19}\times 0.15}$

$(\because \theta ={30}^{\circ }\&B=0.15T)$

$\Rightarrow r=0.5\times {10}^{-3}m=0.5\text{mm}$