An `alpha` is moving along a circle of radius R with a constant angular velocity `omega`. Point A lies in the same plane at a distance 2R from the centre. Point A records magnetic field produced by the `alpha`-particle. If the minimum time interval between two successive time at which A records zero magnetic field is t, the angular speed `omega`, in terms of t, is

Options

A. `(2pi)/t`
B. `(2pi)/(3t)`
C. `pi/(3t)`
D. `pi/t`

Explanation

To solve the problem, we need to find the angular speed \(\omega\) in terms of the time interval \(t\) between two successive times at which point A records zero magnetic field.

### Step-by-step Solution:

1. **Understanding the Setup**:
- An alpha particle is moving in a circular path of radius \(R\) with a constant angular velocity \(\omega\).
- Point A is located at a distance \(2R\) from the center of the circle where the alpha particle is moving.

2. **Identifying Conditions for Zero Magnetic Field**:
- The magnetic field at point A will be zero when the velocity vector of the alpha particle is either pointing directly towards point A or directly away from it.
- This occurs at two specific positions of the alpha particle as it moves around the circle.

3. **Geometry of the Problem**:
- The distance from the center of the circle to point A is \(2R\).
- The radius of the circular path of the alpha particle is \(R\).
- When the alpha particle is at an angle \(\theta\) from the line connecting the center of the circle to point A, the angle between the radius \(R\) and the line to point A can be determined using trigonometry.

4. **Finding the Angle**:
- The cosine of the angle \(\theta\) can be expressed as:
\[ \cos(\theta) = \frac{R}{2R} = \frac{1}{2} \]
- This implies that \(\theta = 60^\circ\).
- Therefore, the angle between the radius to point A and the line connecting the center to the alpha particle when the magnetic field is zero is \(120^\circ\) (since there are two positions).

5. **Calculating the Time Period**:
- The time taken for the alpha particle to move from one position where the magnetic field is zero to the next is \(t\).
- Since the total angular displacement for one full circle is \(2\pi\), and the alpha particle takes \(T\) time to complete one full circle, the time taken to move through \(120^\circ\) (which corresponds to the zero magnetic field positions) is:
\[ \frac{120^\circ}{360^\circ} T = \frac{1}{3} T \]
- This means that \(t = \frac{1}{3} T\).

6. **Relating Time Period and Angular Velocity**:
- The time period \(T\) is related to angular velocity \(\omega\) by the formula:
\[ T = \frac{2\pi}{\omega} \]
- Substituting this into the equation for \(t\):
\[ t = \frac{1}{3} \cdot \frac{2\pi}{\omega} \]

7. **Solving for Angular Velocity \(\omega\)**:
- Rearranging the equation gives:
\[ \omega = \frac{2\pi}{3t} \]

### Final Result:
Thus, the angular speed \(\omega\) in terms of \(t\) is:
\[ \omega = \frac{2\pi}{3t} \]