Question

Four particles A, B, C and D of masses $m_A$, $m_B$, $m_C$ and $m_D$ respectively, follow the paths shown in the figure, in a uniform magnetic field. Each particle moving with same speed. $Q_A$, $Q_B$, $Q_C$ and $Q_D$ are the specific charge of particles A, B, C and D respectively (assume that the motion of each particle is in the same plane perpendicular to the magnetic field).

• A. $Q_A < Q_B < Q_C < Q_D$

Answer 1

The statement $Q_A < Q_B < Q_C < Q_D$ is FALSE.

Answer 2

To determine the order of specific charges, we first establish the relationship between the radius of the circular path and the specific charge of a particle moving in a uniform magnetic field.

When a charged particle of mass $m$, charge $q$, and speed $v$ moves perpendicular to a uniform magnetic field $B$, the magnetic force provides the necessary centripetal force for circular motion.

The magnetic force is given by:

$F_B = |q|vB$

The centripetal force is given by:

$F_c = \frac{mv^2}{r}$

Equating these two forces:

$|q|vB = \frac{mv^2}{r}$

From this, the radius of the circular path is:

$r = \frac{mv}{|q|B}$

The specific charge $Q$ is defined as the ratio of the magnitude of charge to mass, i.e., $Q = \frac{|q|}{m}$.

Substituting this into the radius formula:

$r = \frac{v}{QB}$

Rearranging this equation to find the specific charge:

$Q = \frac{v}{rB}$

We are given that all particles move with the same speed ($v$) and are in a uniform magnetic field ($B$). Therefore, $v$ and $B$ are constants.

This implies that the specific charge $Q$ is inversely proportional to the radius $r$ of the path:

$Q \propto \frac{1}{r}$

Now, let's analyze the paths of the four particles from the figure:

1. Particle A: Follows a circular path with radius $R_A = R$. Specific charge $Q_A = \frac{v}{RB}$.

2. Particle B: Follows a circular path with radius $R_B = 2R$. Specific charge $Q_B = \frac{v}{2RB}$.

3. Particle C: Follows a straight line path. A charged particle moves in a straight line in a magnetic field if its velocity is parallel to the magnetic field (not applicable here as motion is perpendicular) or if it is uncharged. Since the problem states the motion is perpendicular to the field, particle C must be uncharged. If $q_C = 0$, then its specific charge $Q_C = \frac{|q_C|}{m_C} = 0$. Alternatively, a straight line path can be considered a circle with infinite radius ($R_C = \infty$). Specific charge $Q_C = \frac{v}{\infty \cdot B} = 0$.

4. Particle D: Follows a circular path with radius $R_D = R$. Specific charge $Q_D = \frac{v}{RB}$.

Let's compare the specific charges:

Let $k = \frac{v}{RB}$.

$Q_A = k$

$Q_B = k/2$

$Q_C = 0$

$Q_D = k$

Now, let's order them from smallest to largest:

$Q_C = 0$

$Q_B = k/2$

$Q_A = k$

$Q_D = k$

Therefore, the correct order of specific charges is:

$Q_C < Q_B < Q_A = Q_D$

The statement given in the question is $Q_A < Q_B < Q_C < Q_D$.

Comparing our derived order with the given statement, we find that the statement is false.