To determine the specific charge (the ratio of charge to mass) of the particle, we will analyze the motion of the particle in both the electric and magnetic fields. Let's break down the problem step by step:

1. Electric Field Region (Acceleration of the Particle)

The particle moves through a potential difference $V$, gaining kinetic energy due to the electric field. The work done by the electric field on the particle is converted into its kinetic energy.

The kinetic energy gained by the particle is:

$\text{Kinetic Energy} = eV$

where:

• $e$ is the charge of the particle,
• $V$ is the potential difference.

The kinetic energy of the particle is also given by:

$\text{Kinetic Energy} = \frac{1}{2}mv^2$

where:

• $m$ is the mass of the particle,
• $v$ is the velocity of the particle after it is accelerated by the electric field.

Equating the two expressions for kinetic energy:

$eV = \frac{1}{2}mv^2$

Solving for the velocity $v$ of the particle after passing through the potential difference $V$:

$v = \sqrt{\frac{2eV}{m}}$

2. Magnetic Field Region (Circular Motion)

Once the particle enters the magnetic field, it moves in circular trajectories due to the Lorentz force acting as the centripetal force.

The Lorentz force in a magnetic field is:

$F_{\text{mag}} = evB$

This force provides the centripetal force required to keep the particle in circular motion:

$F_{\text{centripetal}} = \frac{mv^2}{R}$

where:

• $R$ is the radius of the circular trajectory,
• $B$ is the magnetic field induction.

Equating the magnetic force to the centripetal force:

$evB = \frac{mv^2}{R}$

Solving for the velocity $v$:

$v = \frac{eBR}{m}$

3. Combining the Two Regions

Now we have two expressions for the velocity $v$:

From the electric field region:

$v = \sqrt{\frac{2eV}{m}}$

From the magnetic field region:

$v = \frac{eBR}{m}$

Equating these two expressions:

$\sqrt{\frac{2eV}{m}} = \frac{eBR}{m}$

Square both sides to eliminate the square root:

$\frac{2eV}{m} = \frac{e^2B^2R^2}{m^2}$

Multiply both sides by $m^2$ to simplify:

$2eVm = e^2B^2R^2$

Solve for the specific charge $\frac{e}{m}$:

$\frac{e}{m} = \frac{2V}{B^2R^2}$

4. Conclusion

The specific charge (the ratio of charge to mass) of the particle is given by:

$\boxed{\frac{e}{m} = \frac{2V}{B^2R^2}}$

This formula shows that the specific charge depends on the potential difference $V$, the magnetic field $B$, and the radius of the circular trajectory $R$.

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Do you want more details or have any questions on the steps involved?

Here are 5 relative questions to consider:

1. How does the specific charge change if the magnetic field strength is doubled?
2. What happens to the radius of the circular trajectory if the particle's charge increases?
3. How does the velocity of the particle change with an increasing potential difference?
4. What assumptions are made about the homogeneity of the electric and magnetic fields?
5. How would the trajectory change if the magnetic field were not perpendicular to the particle's motion?

Tip: Always check the units of your final answer to ensure consistency in physical problems.