A point charges Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity `omega`. This can be considered as equivalent to a loop carrying a steady current `(Q omega)/(2 pi)`. S uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an `EMF` in the orbit. The induced `EMF` is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant `lambda`. The charge in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field charge, is

Options

A. `lamda BQR^(2)`
B. `-lamda(BQR^(2))/(2)`
C. `lamda(BQR^(2))/(2)`
D. `-lamdaBQR^(2)`

Explanation

To solve the problem, we need to determine the change in the magnetic dipole moment associated with a point charge \( Q \) moving in a circular orbit of radius \( R \) when a uniform magnetic field is switched on and increases at a constant rate.

### Step-by-step Solution:

1. **Identify the Current Equivalent**:
The point charge \( Q \) moving in a circular orbit can be treated as a current loop. The current \( I \) can be expressed as:
\[ I = \frac{Q}{T} = \frac{Q \omega}{2\pi} \]
where \( T \) is the period of one complete revolution, which is \( T = \frac{2\pi}{\omega} \).

2. **Magnetic Dipole Moment**:
The magnetic dipole moment \( m \) for a current loop is given by:
\[ m = I \cdot A \]
where \( A \) is the area of the loop. For a circular loop of radius \( R \):
\[ A = \pi R^2 \]
Therefore, substituting the expression for \( I \):
\[ m = \left(\frac{Q \omega}{2\pi}\right) \cdot \pi R^2 = \frac{Q \omega R^2}{2} \]

3. **Change in Magnetic Field**:
The magnetic field \( B \) is switched on and increases from \( 0 \) to \( B \) in \( 1 \) second. The rate of change of the magnetic field \( \frac{dB}{dt} \) is:
\[ \frac{dB}{dt} = \frac{B}{1 \text{ s}} = B \]

4. **Induced EMF**:
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) in the loop is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux. The change in magnetic flux \( \Phi \) through the loop is:
\[ \Phi = B \cdot A = B \cdot \pi R^2 \]
Therefore, the induced EMF becomes:
\[ \mathcal{E} = -\frac{d}{dt}(B \cdot \pi R^2) = -\pi R^2 \cdot \frac{dB}{dt} = -\pi R^2 B \]

5. **Change in Angular Momentum**:
The angular momentum \( L \) of the charge in the orbit is given by:
\[ L = m R^2 \omega \]
The change in angular momentum due to the induced EMF can be expressed as:
\[ \Delta L = \lambda \Delta m \]
where \( \lambda \) is a proportionality constant.

6. **Change in Magnetic Dipole Moment**:
The change in the magnetic dipole moment \( \Delta m \) can be expressed as:
\[ \Delta m = \lambda \cdot \frac{Q B R^2}{2} \]
Thus, the final expression for the change in the magnetic dipole moment is:
\[ \Delta m = -\frac{\lambda B Q R^2}{2} \]
The negative sign indicates that the induced magnetic dipole moment opposes the change in the magnetic field, in accordance with Lenz's law.

### Final Answer:
The change in the magnetic dipole moment associated with the orbit at the end of the time interval of the magnetic field change is:
\[ \Delta m = -\frac{\lambda B Q R^2}{2} \]