Bohr model is applied to a particle of mass 'm' and charge 'q' moving in a plane under the influence of a transverse magnetic field B. The energy of the charged particle in the $n^{th}$ level will be (h=Planck's constant)

For a charged particle moving in a transverse magnetic field, the magnetic force provides the necessary centripetal force for circular motion:

$\dfrac{mv^2}{r} = qvB$

$\Rightarrow v = \dfrac{qBr}{m}$

According to Bohr's quantization postulate, the angular momentum of the particle is quantized:

$mvr = \dfrac{nh}{2\pi}$

Substituting the expression for $v$ into the quantization condition, we get:

$m \left(\dfrac{qBr}{m}\right) r = \dfrac{nh}{2\pi}$

$\Rightarrow qBr^2 = \dfrac{nh}{2\pi}$

$\Rightarrow r^2 = \dfrac{nh}{2\pi qB}$

The total energy of the particle is purely kinetic (since the magnetic force does no work):

$E = \dfrac{1}{2}mv^2$

Substituting $v = \dfrac{qBr}{m}$:

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

Now, substituting the value of $r^2$ into the energy equation:

$E = \dfrac{q^2B^2}{2m} \left(\dfrac{nh}{2\pi qB}\right)$

$E = \dfrac{nhqB}{4\pi m}$

Answer: $\dfrac{nhqB}{4\pi m}$