A moving coil galvanometer has a circular coil of area `16 cm^(2)` and 30 turns. It is mounted on a torsional spring of spring constant 100 Nm/rad. The coil turns by angle `21^(@)` when a current 40 mA is passed through it. The magnitude of the uniform magnetic field produced by the permanent magnet inside the galvanometer is _________________T.
[use `cos 21^(@)=15/16, 21^(@)~~11/30` rad]

Explanation

To find the magnitude of the uniform magnetic field produced by the permanent magnet inside the galvanometer, we can follow these steps:

### Step 1: Note down the given values
- Area of the coil, \( A = 16 \, \text{cm}^2 = 16 \times 10^{-4} \, \text{m}^2 \)
- Number of turns, \( N = 30 \)
- Spring constant (torsional constant), \( C = 100 \, \text{Nm/rad} \)
- Current, \( I = 40 \, \text{mA} = 0.04 \, \text{A} \)
- Angle of deflection, \( \theta = 21^\circ \approx \frac{11}{30} \, \text{rad} \)
- Cosine of the angle, \( \cos(21^\circ) = \frac{15}{16} \)

### Step 2: Write the torque equation
The torque \( \tau \) acting on the coil is given by:
\[ \tau = C \cdot \theta \]
Also, the torque due to the magnetic field is given by:
\[ \tau = N \cdot I \cdot A \cdot B \cdot \cos(21^\circ) \]
Setting these two expressions for torque equal gives:
\[ C \cdot \theta = N \cdot I \cdot A \cdot B \cdot \cos(21^\circ) \]

### Step 3: Solve for the magnetic field \( B \)
Rearranging the equation to solve for \( B \):
\[ B = \frac{C \cdot \theta}{N \cdot I \cdot A \cdot \cos(21^\circ)} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ B = \frac{100 \, \text{Nm/rad} \cdot \frac{11}{30} \, \text{rad}}{30 \cdot 0.04 \, \text{A} \cdot (16 \times 10^{-4} \, \text{m}^2) \cdot \frac{15}{16}} \]

### Step 5: Simplify the expression
Calculating the numerator:
\[ 100 \cdot \frac{11}{30} = \frac{1100}{30} \]
Calculating the denominator:
\[ 30 \cdot 0.04 \cdot (16 \times 10^{-4}) \cdot \frac{15}{16} = 30 \cdot 0.04 \cdot 10^{-4} \cdot 15 = 0.018 \, \text{(after simplification)} \]

### Step 6: Calculate \( B \)
Now substituting back into the equation for \( B \):
\[ B = \frac{\frac{1100}{30}}{0.018} \]
Calculating this gives:
\[ B \approx 1.1 \, \text{T} \]

### Final Answer
The magnitude of the uniform magnetic field produced by the permanent magnet inside the galvanometer is approximately **1.1 T**.

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