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 hte orbit remains constant. The application of hte 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 si known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a porportionality constant `lambda`.
The charge in the magnetic dipole moment associated with the orbit. at the end of the time interval of hte magnetic field charge, is

Options

A. `-gammaBQR^(2)`
B. `-gamma(BQR^(2))/2`
C. `gamma(BQR^(2))/2`
D. `gammaBQR^(2)`

Explanation

To solve the problem, we need to find the change in the magnetic dipole moment associated with a point charge \( Q \) moving in a circular orbit under the influence of a changing magnetic field. Here's a step-by-step breakdown of the solution:

### Step 1: Understanding the Current Equivalent
The point charge \( Q \) moving in a circular orbit of radius \( R \) with angular velocity \( \omega \) can be treated as a loop carrying a steady current. The current \( I \) can be calculated as:
\[ I = \frac{Q \omega}{2\pi} \]

### Step 2: Rate of Change of Magnetic Field
The magnetic field \( B \) is switched on and increases at a constant rate from \( 0 \) to \( B \) in \( 1 \) second. Thus, the rate of change of the magnetic field is:
\[ \frac{dB}{dt} = \frac{B - 0}{1 - 0} = B \]

### Step 3: Induced EMF Calculation
The induced EMF (\( \mathcal{E} \)) in the loop can be calculated using Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux. The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A = B \cdot \pi R^2 \]
Thus, the induced EMF becomes:
\[ \mathcal{E} = -\frac{d}{dt}(B \cdot \pi R^2) = -\pi R^2 \frac{dB}{dt} \]
Substituting \( \frac{dB}{dt} = B \):
\[ \mathcal{E} = -\pi R^2 B \]

### Step 4: Electric Field Calculation
The induced electric field \( E \) around the loop can be related to the EMF:
\[ \mathcal{E} = E \cdot (2\pi R) \]
Thus, we can express the electric field as:
\[ E = \frac{\mathcal{E}}{2\pi R} = \frac{-\pi R^2 B}{2\pi R} = -\frac{R B}{2} \]

### Step 5: Torque Calculation
The torque \( \tau \) acting on the charge due to the electric field is given by:
\[ \tau = F \cdot R \]
where \( F \) is the force acting on the charge, which is:
\[ F = Q \cdot E = Q \left(-\frac{R B}{2}\right) = -\frac{Q R B}{2} \]
Thus, the torque becomes:
\[ \tau = -\frac{Q R B}{2} \cdot R = -\frac{Q R^2 B}{2} \]

### Step 6: Change in Angular Momentum
The change in angular momentum \( \Delta L \) can be expressed as:
\[ \Delta L = \tau \cdot dt \]
Over the time interval of \( 1 \) second:
\[ \Delta L = -\frac{Q R^2 B}{2} \cdot 1 = -\frac{Q R^2 B}{2} \]

### Step 7: Change in Magnetic Dipole Moment
The magnetic dipole moment \( m \) is related to angular momentum \( L \) by:
\[ m = \lambda L \]
Thus, the change in magnetic dipole moment is:
\[ \Delta m = \lambda \Delta L = \lambda \left(-\frac{Q R^2 B}{2}\right) = -\frac{\lambda Q R^2 B}{2} \]

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

---