Given: - An electron revolves in a circular orbit with angular momentum \( \vec{L} \) - A uniform magnetic field \( \vec{B} \) is applied perpendicular to the plane of the orbit - Electron experiences a torque \( \vec{T} \)
Concept: A magnetic moment \( \vec{\mu} \) associated with the electron is given by: \[ \vec{\mu} \propto \vec{L} \] When placed in a magnetic field, the torque experienced is: \[ \vec{T} = \vec{\mu} \times \vec{B} \] Since \( \vec{T} \) is the cross product of \( \vec{\mu} \) and \( \vec{B} \), and \( \vec{L} \) is in the direction of \( \vec{\mu} \), we get: \[ \vec{T} \cdot \vec{L} = 0 \] because the torque is perpendicular to the angular momentum vector.
Final Answer: \( \vec{T} \cdot \vec{L} = 0 \)
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: