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)/(2t)`

Answer

To solve the problem, we need to analyze the motion of the alpha particle and how it affects the magnetic field at point A.

### 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.

2. **Magnetic Field Calculation**:
- The magnetic field \( B \) at point A due to a current element \( dl \) of the moving charge can be calculated using the Biot-Savart law:
\[ B = \frac{\mu_0}{4\pi} \int \frac{I \, dl \times r}{r^3} \]
- Here, \( r \) is the position vector from the current element to point A.

3. **Condition for Zero Magnetic Field**:
- The magnetic field at point A will be zero when the angle \( \theta \) between the velocity vector \( v \) of the alpha particle and the position vector \( r \) from the alpha particle to point A is either \( 0^\circ \) (parallel) or \( 180^\circ \) (antiparallel).
- As the alpha particle moves in a circle, it will create a zero magnetic field at point A at specific positions.

4. **Finding the Angles**:
- The distance from the center to point A is \( 2R \), and the radius of the circular path of the alpha particle is \( R \).
- Using the cosine rule, when the angle \( \theta \) is \( 60^\circ \), the cosine of the angle between the position vector and the velocity vector is:
\[ \cos \theta = \frac{R}{2R} = \frac{1}{2} \]
- This indicates that the angle \( \theta \) is \( 60^\circ \).

5. **Time Interval Calculation**:
- The alpha particle moves \( 120^\circ \) (or \( \frac{2\pi}{3} \) radians) between the two positions where the magnetic field at point A is zero.
- The angular displacement \( \Delta \theta \) in terms of angular velocity \( \omega \) and time \( t \) is given by:
\[ \Delta \theta = \omega \cdot t \]
- Since \( \Delta \theta = \frac{2\pi}{3} \), we can write:
\[ \omega \cdot t = \frac{2\pi}{3} \]

6. **Expressing Angular Velocity**:
- Rearranging the equation gives us:
\[ \omega = \frac{2\pi}{3t} \]

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