Question

Two particles of masses $m_1$ and $m_2$ having charges $q_1$ and $q_2$ respectively are projected with the same velocity in a region of uniform magnetic field $\vec{B}$ pointing vertically upward. If they describe circular paths as shown in the figure, one may conclude that :

• A. $\frac{m_1}{m_2} > \frac{q_1}{q_2}$
• B. $\frac{m_1}{m_2} > \frac{q_2}{q_1}$
• C. $\frac{m_1}{m_2} < \frac{q_1}{q_2}$
• D. $\frac{m_1}{m_2} < \frac{q_2}{q_1}$

Hint:

For particles moving with the same velocity in the same magnetic field, trajectory radius is directly proportional to specific mass-to-charge ratio ($r \propto \frac{m}{q}$). Larger radius means larger $\frac{m}{q}$.

Answer 1

The Correct Option is A

Concept:
• When a charged particle moves perpendicularly through a uniform magnetic field $\vec{B}$, magnetic Lorentz force provides the necessary centripetal force: $q v B = \frac{m v^2}{r}$.

• The radius of the resulting circular trajectory is $r = \frac{m v}{q B}$.

Step 1: Express radius in terms of mass-to-charge ratio
The trajectory radius is:
\[ r = \frac{m v}{q B} \]
Since velocity $v$ and magnetic field $B$ are identical for both particles:
\[ r \propto \frac{m}{q} \implies \frac{m}{q} = r \left(\frac{B}{v}\right) \]

Step 2: Compare radii from the figure
Observing the given circular trajectories from the figure:
The path of particle 1 has a larger radius of curvature than particle 2:
\[ r_1 > r_2 \]

Step 3: Derive inequality
Substitute $r_1 > r_2$ into the radius relation:
\[ \frac{m_1 v}{q_1 B} > \frac{m_2 v}{q_2 B} \]
Canceling common non-zero terms $v$ and $B$:
\[ \frac{m_1}{q_1} > \frac{m_2}{q_2} \]
Rearranging terms by cross-multiplying $m_2$ and $q_1$:
\[ \frac{m_1}{m_2} > \frac{q_1}{q_2} \]

Step 4: Conclusion
The ratio of masses satisfies $\frac{m_1}{m_2} > \frac{q_1}{q_2}$, corresponding to option (A).