In order to impatt an angular velocity to an earth satellite the geomagnetic field can be used.Find the maximum possible angular velocity about its own axis gained by the satellite if a storage battery with a capacity of `Q=5 Amp`.hours is discharged suddenly through a coil of `N=20` turns wound around the satellite's surface along the circumference of the largest circle.The satellite has a mass of `m=10^(3) kg` and is a thin walled uniform sphere.The geomagnetic field is parallel to the winding plane and its flux density is `B=0.5 Gauss`.`(1 Gauss =10^(-4)Tesla)`

Explanation

To find the maximum possible angular velocity gained by the satellite, we will follow these steps:

### Step 1: Understand the Parameters
We have the following parameters:
- Charge capacity, \( Q = 5 \) Ampere-hours
- Number of turns in the coil, \( N = 20 \)
- Mass of the satellite, \( m = 10^3 \) kg
- Magnetic flux density, \( B = 0.5 \) Gauss \( = 0.5 \times 10^{-4} \) Tesla

### Step 2: Calculate the Dipole Moment
The dipole moment \( p \) of the coil can be calculated using the formula:
\[ p = N \cdot I \cdot A \]
where \( A \) is the area of the coil and \( I \) is the current. The area \( A \) of the coil can be expressed as:
\[ A = \pi r^2 \]
Since we don't have the radius \( r \), we will keep it in the equation for now.

### Step 3: Calculate the Current
To find the current \( I \) from the charge \( Q \), we convert the charge from Ampere-hours to Coulombs:
\[ Q = 5 \, \text{Amp-hours} = 5 \times 3600 \, \text{Coulombs} = 18000 \, \text{Coulombs} \]
The current \( I \) can be calculated as:
\[ I = \frac{Q}{t} \]
Assuming the discharge happens suddenly, we can consider \( t \) to be very small, leading to a very high current. However, for practical purposes, we will consider the effective current during the discharge.

### Step 4: Calculate the Torque
The torque \( \tau \) on the coil in the magnetic field is given by:
\[ \tau = p \times B \]
Substituting for \( p \):
\[ \tau = N \cdot I \cdot A \cdot B = N \cdot I \cdot \pi r^2 \cdot B \]

### Step 5: Relate Torque to Angular Acceleration
The torque can also be expressed in terms of angular acceleration \( \alpha \):
\[ \tau = I \cdot \alpha \]
where \( I \) is the moment of inertia of the satellite. For a thin-walled uniform sphere:
\[ I = \frac{2}{3} m r^2 \]

### Step 6: Set Up the Equation
Equating the two expressions for torque:
\[ N \cdot I \cdot \pi r^2 \cdot B = \frac{2}{3} m r^2 \cdot \alpha \]
We can cancel \( r^2 \) from both sides:
\[ N \cdot I \cdot \pi B = \frac{2}{3} m \cdot \alpha \]

### Step 7: Solve for Angular Velocity
We know that \( \alpha = \frac{d\omega}{dt} \). If we consider the total charge \( Q \) discharged, we can relate it to the change in angular velocity:
\[ \alpha \cdot dt = d\omega \implies \alpha = \frac{d\omega}{dt} \]
Substituting this into the equation gives:
\[ N \cdot I \cdot \pi B = \frac{2}{3} m \cdot \frac{d\omega}{dt} \]
Integrating gives us:
\[ \omega = \frac{3}{2} \cdot \frac{N \cdot I \cdot \pi B}{m} \cdot t \]
Assuming the entire charge is discharged in a very short time, we can approximate \( t \) as a constant factor.

### Step 8: Substitute Values
Substituting the known values:
- \( N = 20 \)
- \( I = \frac{Q}{t} \) (considering \( t \) as a small constant)
- \( B = 0.5 \times 10^{-4} \) Tesla
- \( m = 10^3 \) kg

We can calculate:
\[ \omega = \frac{3}{2} \cdot \frac{20 \cdot I \cdot \pi \cdot (0.5 \times 10^{-4})}{10^3} \]
Assuming \( I \) is effectively high during the discharge.

### Final Calculation
Assuming \( I \) is effectively high, we can compute:
\[ \omega = \frac{3}{2} \cdot \frac{20 \cdot (18000) \cdot \pi \cdot (0.5 \times 10^{-4})}{10^3} \]
After calculating, we find:
\[ \omega \approx 2.7 \times 10^{-2} \text{ rad/s} \]

### Conclusion
The maximum possible angular velocity gained by the satellite is approximately:
\[ \omega \approx 2.7 \pi \times 10^{-2} \text{ rad/s} \]