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 C

Step 1: Radius of the circular path:
The magnetic force supplies the centripetal force, so \( qvB = \dfrac{mv^2}{r} \).
This gives \( r = \dfrac{mv}{qB} \).
Both particles have the same \( v \) and move in the same \( B \), so \( r \propto \dfrac{m}{q} \).

Step 2: Compare the two radii from the figure:
Both particles start from the same point and enter in the same direction.
Path 1 bends away more sharply than path 2, so path 1 has the smaller radius.
Hence \( r_1 < r_2 \).

Step 3: Form the inequality:
\[ \frac{m_1 v}{q_1 B} < \frac{m_2 v}{q_2 B} \]
Cancel \( v \) and \( B \):
\[ \frac{m_1}{q_1} < \frac{m_2}{q_2} \]
Multiply both sides by \( \dfrac{q_1}{m_2} \) (all quantities are taken as magnitudes, so they are positive).
\[ \frac{m_1}{m_2} < \frac{q_1}{q_2} \]

Step 4: Check the options:
Option (A) \( \dfrac{m_1}{m_2} > \dfrac{q_1}{q_2} \) is the reverse of our result, so it is wrong.
Options (B) and (D) contain \( \dfrac{q_2}{q_1} \). Our result has \( \dfrac{q_1}{q_2} \), so both are wrong.
Option (C) \( \dfrac{m_1}{m_2} < \dfrac{q_1}{q_2} \) matches.

Final Answer:
\( \dfrac{m_1}{m_2} < \dfrac{q_1}{q_2} \), which is option (C).