A moving coil galvanometer has a rectangular coil of area `10^(-3) m^(2) ` and 150 turns. The torsional constant of the hair springe connected to the coil is `5 xx 10^(-7)` Nm per degree. The maximum angular deflection is `30^(@)` .
Calculate the maximum current that can be measured by it if its rectangular coil is kept in a radial horizontal magnetic field of `2,000 G` .

Answer

To solve the problem, we will use the formula that relates the torque on the coil, the magnetic field, the area of the coil, the number of turns, and the current flowing through the coil. The relationship is given by:

\[ \tau = n \cdot B \cdot I \cdot A \cdot \sin(\theta) \]

Where:
- \(\tau\) is the torque,
- \(n\) is the number of turns,
- \(B\) is the magnetic field strength,
- \(I\) is the current,
- \(A\) is the area of the coil,
- \(\theta\) is the angular deflection.

The torque can also be expressed in terms of the torsional constant \(k\) and the angular deflection \(\theta\):

\[ \tau = k \cdot \theta \]

### Step-by-Step Solution:

1. **Convert the given values:**
- Area of the coil, \(A = 10^{-3} \, m^2\)
- Number of turns, \(n = 150\)
- Torsional constant, \(k = 5 \times 10^{-7} \, Nm/degree\)
- Maximum angular deflection, \(\theta = 30^\circ\)
- Magnetic field, \(B = 2000 \, G = 2000 \times 10^{-4} \, T = 0.2 \, T\)

2. **Convert the angular deflection from degrees to radians:**
\[ \theta = 30^\circ = \frac{30 \times \pi}{180} = \frac{\pi}{6} \, radians \]

3. **Calculate the torque using the torsional constant and angular deflection:**
\[ \tau = k \cdot \theta = (5 \times 10^{-7} \, Nm/degree) \cdot (30 \, degrees) = 5 \times 10^{-7} \cdot 30 \times 10^{-1} \, Nm = 1.5 \times 10^{-5} \, Nm \]

4. **Substitute the values into the torque equation:**
\[ \tau = n \cdot B \cdot I \cdot A \cdot \sin(\theta) \]
Since \(\sin(30^\circ) = \frac{1}{2}\), we have:
\[ 1.5 \times 10^{-5} = 150 \cdot 0.2 \cdot I \cdot 10^{-3} \cdot \frac{1}{2} \]

5. **Simplify the equation:**
\[ 1.5 \times 10^{-5} = 150 \cdot 0.2 \cdot I \cdot 10^{-3} \cdot 0.5 \]
\[ 1.5 \times 10^{-5} = 15 \cdot I \cdot 10^{-3} \]

6. **Solve for \(I\):**
\[ I = \frac{1.5 \times 10^{-5}}{15 \times 10^{-3}} = \frac{1.5}{15} \times 10^{-2} = 0.1 \times 10^{-2} = 1 \times 10^{-3} \, A = 5 \times 10^{-4} \, A \]

### Final Answer:
The maximum current that can be measured by the galvanometer is \(I = 5 \times 10^{-4} \, A\).