Question

Two charged particles of the same charge but different masses \(m_1\) and \(m_2\) are projected with the same velocity \(V\) normally into the uniform magnetic field \(B\) as shown in the figure. The maximum separation of the particles is: 

• A. \( \frac{|(m_2 - m_1)|V}{|qB|} \)
• B. \( \frac{(m_2 - m_1)V}{|2qB|} \)
• C. \( \frac{2(m_2 - m_1)V}{|qB|} \)
• D. \( \frac{|(m_2 - m_1)|V}{|4qB|} \)

Hint:

The maximum separation between two particles in a magnetic field depends on their masses and the magnetic field strength.

Answer 1

The Correct Option is C

1. Radius of Circular Path:
The magnetic force ($F_m$) on a charged particle is given by: $F_m = qVB$
This force provides the centripetal force ($F_c$): $F_c = \frac{mv^2}{r}$
Equating the two forces: $qVB = \frac{mv^2}{r}$
Solving for the radius (r): $r = \frac{mV}{qB}$
2. Radii for the Two Particles:
Radius for particle 1 ($m_1$): $r_1 = \frac{m_1V}{qB}$
Radius for particle 2 ($m_2$): $r_2 = \frac{m_2V}{qB}$
3. Maximum Separation:
The maximum separation ($d$) between the particles will occur when they complete a half-circle (a semicircle).
The separation will be the difference in the diameters of their circular paths.
Diameter of particle 1: $2r_1 = \frac{2m_1V}{qB}$
Diameter of particle 2: $2r_2 = \frac{2m_2V}{qB}$
Maximum separation ($d$) = $2r_2 - 2r_1 = \frac{2m_2V}{qB} - \frac{2m_1V}{qB}$
$d = \frac{2(m_2 - m_1)V}{qB}$
Therefore, the maximum separation of the particles is $\frac{2(m_2 - m_1)V}{qB}$, which matches option (3).