If one were to apply the Bohr model to a particle of mass $$m$$ and charge $$q$$ moving in a plane under the influence of a magnetic field 'B', the energy of the charged particle in the $$n^{th}$$ level will be:

We are applying the Bohr model to a charged particle of mass $$m$$ and charge $$q$$ moving in a plane under a constant magnetic field $$B$$. In the Bohr model, the angular momentum is quantized. Specifically, for the $$n$$th orbit, the angular momentum $$L$$ is given by:

$$ L = n \frac{h}{2\pi} $$

where $$h$$ is Planck's constant and $$n$$ is the quantum number (1, 2, 3, ...).

Since the particle is moving in a magnetic field perpendicular to the plane, the force is perpendicular to the velocity, causing circular motion. The magnetic force provides the centripetal force:

$$ q v B = \frac{m v^2}{r} $$

where $$v$$ is the speed and $$r$$ is the radius of the circular path. Simplifying this equation:

$$ q v B = \frac{m v^2}{r} $$

Divide both sides by $$v$$ (assuming $$v \neq 0$$):

$$ q B = \frac{m v}{r} $$

Rearranging, we get:

$$ m v = q B r $$

The angular momentum $$L$$ for a particle in circular motion is $$L = m v r$$. Substituting $$m v$$ from above:

$$ L = (q B r) \cdot r = q B r^2 $$

But from the quantization condition, $$L = n \frac{h}{2\pi}$$. Setting them equal:

$$ q B r^2 = n \frac{h}{2\pi} $$

Solving for $$r^2$$:

$$ r^2 = \frac{n h}{2\pi q B} $$

Now, the energy of the charged particle is purely kinetic because the magnetic field does no work. Thus:

$$ E = \frac{1}{2} m v^2 $$

From the force equation, we have $$m v = q B r$$, so solving for $$v$$:

$$ v = \frac{q B r}{m} $$

Substitute this into the energy equation:

$$ E = \frac{1}{2} m \left( \frac{q B r}{m} \right)^2 = \frac{1}{2} m \cdot \frac{q^2 B^2 r^2}{m^2} = \frac{q^2 B^2 r^2}{2m} $$

Now substitute $$r^2 = \frac{n h}{2\pi q B}$$:

$$ E = \frac{q^2 B^2}{2m} \cdot \frac{n h}{2\pi q B} $$

Simplify by canceling one $$q$$ and one $$B$$:

$$ E = \frac{q B}{2m} \cdot \frac{n h}{2\pi} = \frac{q B n h}{4\pi m} $$

Therefore, the energy in the $$n$$th level is:

$$ E_n = n \left( \frac{h q B}{4\pi m} \right) $$

Comparing with the options, this matches option B.

Hence, the correct answer is Option B.