Step 1: Understanding the Concept:
We are applying Bohr's quantization condition to a charged particle in a magnetic field.
In this scenario, the magnetic Lorentz force acts as the required centripetal force for circular motion.
Bohr's quantization postulates that angular momentum must be an integral multiple of $h/(2\pi)$.
Step 2: Key Formula or Approach:
Equating centripetal force and magnetic force: $\frac{mv^2}{r} = qvB \implies v = \frac{qBr}{m}$.
Bohr's angular momentum quantization: $mvr = \frac{nh}{2\pi}$.
The total energy of a particle moving purely in a magnetic field is entirely kinetic: $E = \frac{1}{2}mv^2$.
Step 3: Detailed Explanation:
From the force equation, we find the expression for velocity:
\[ v = \frac{qBr}{m} \]
Substitute this velocity into the angular momentum quantization equation:
\[ m \left(\frac{qBr}{m}\right) r = \frac{nh}{2\pi} \]
Simplify the left side to isolate $r^2$:
\[ qB r^2 = \frac{nh}{2\pi} \]
\[ r^2 = \frac{nh}{2\pi qB} \]
The energy of the particle is solely its kinetic energy:
\[ E = \frac{1}{2}mv^2 \]
Square the velocity expression derived earlier: $v^2 = \frac{q^2 B^2 r^2}{m^2}$.
Substitute this into the energy equation:
\[ E = \frac{1}{2} m \left(\frac{q^2 B^2 r^2}{m^2}\right) = \frac{q^2 B^2 r^2}{2m} \]
Now, substitute the expression for $r^2$ into this energy equation:
\[ E = \frac{q^2 B^2}{2m} \left(\frac{nh}{2\pi qB}\right) \]
Simplify by canceling $q$ and $B$:
\[ E = \frac{n q B h}{4\pi m} \]
The question asks for the energy in the second level, which means $n = 2$.
Substitute $n=2$ into the energy formula:
\[ E_2 = \frac{2 \cdot q B h}{4\pi m} \]
Simplify the fraction:
\[ E_2 = \frac{q B h}{2\pi m} \]
Step 4: Final Answer:
The energy in the second level is $\frac{\text{qBh}}{2\pi\text{m}}$.