Magnetism
Intermediate
Electric Particles in Magnetic Field
Source: High school physics (Chinese)
Problem Sets:
magnetic field
Problem
A proton ($P$), a deuteron ($D$), and an alpha particle ($\alpha$), all with the same kinetic energy, enter a uniform magnetic field $\vec{B}$ perpendicular to their velocities. Their masses satisfy $m_\alpha = 2 m_D = 4 m_P$, and their charges are $q_P = q_D = e$, $q_\alpha = 2e$. Let $r_P$, $r_D$, $r_\alpha$ be their respective orbital radii.
- Find $r_D / r_P$.
- Find $r_\alpha / r_P$.
$r_D / r_P = \sqrt{2} \approx 1.41$; $r_\alpha / r_P = 1$.
With kinetic energy $E_k$, the speed is $v = \sqrt{2E_k/m}$ and the orbital radius is
$$r = \dfrac{mv}{qB} = \dfrac{\sqrt{2mE_k}}{qB} \propto \dfrac{\sqrt{m}}{q}.$$Taking ratios at fixed $E_k$ and $B$:
$$\dfrac{r_D}{r_P} = \dfrac{\sqrt{m_D}\,/\,q_D}{\sqrt{m_P}\,/\,q_P} = \sqrt{\dfrac{m_D}{m_P}} \cdot \dfrac{q_P}{q_D} = \sqrt{2} \cdot 1 = \sqrt{2}.$$ $$\dfrac{r_\alpha}{r_P} = \sqrt{\dfrac{m_\alpha}{m_P}} \cdot \dfrac{q_P}{q_\alpha} = \sqrt{4} \cdot \dfrac{1}{2} = 1.$$