- Home
- Standard 12
- Physics
4.Moving Charges and Magnetism
medium
A proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha - $particle (mass = $4m$ and charge = $+ 2e),$ so that it can revolve in the path of same radius.......$MeV$
A
$1$
B
$4$
C
$2$
D
$0.5$
Solution
(a) $r = \frac{{\sqrt {2mK} }}{{qB}}$$⇒$ $K \propto \frac{{{q^2}}}{m}$ $⇒$ $\frac{{{K_p}}}{{{K_\alpha }}} = {\left( {\frac{{{q_p}}}{{{q_\alpha }}}} \right)^2} \times \frac{{{m_\alpha }}}{{{m_p}}}$
$⇒$$\frac{1}{{{K_\alpha }}} = {\left( {\frac{{{q_p}}}{{2{q_p}}}} \right)^2} \times \frac{{4{m_p}}}{{{m_p}}} = 1$ $==>$ ${K_\alpha } = 1\,MeV$.
Standard 12
Physics
