A rectangular coil of a moving coil galvanometer containing 100 turns and 5 cm long and 3 cm broad is suspended in a radial magnetic field of induction 0.025 `Wb//m^(2)` by a fibre of torque constant `1.5 xx 10^(-9)` Nm per degree. The current for which coil will deflect through an angle of `10^(@)` is

Options

A. `4 mu A`
B. `3 mu A`
C. `5 mu A`
D. `6 mu A`

Answer

To find the current for which the coil will deflect through an angle of \(10^\circ\), we can follow these steps:

### Step 1: Understand the Torque Balance
The torque (\(\tau\)) on the coil due to the magnetic field is given by:
\[ \tau = mB \sin \theta \]
where:
- \(m\) is the magnetic moment,
- \(B\) is the magnetic field induction,
- \(\theta\) is the angle of deflection.

The torque due to the fiber is given by:
\[ \tau = c \theta \]
where:
- \(c\) is the torque constant,
- \(\theta\) is the angle in radians.

### Step 2: Calculate the Magnetic Moment
The magnetic moment (\(m\)) is given by:
\[ m = n \cdot i \cdot A \]
where:
- \(n\) is the number of turns,
- \(i\) is the current,
- \(A\) is the area of the coil.

### Step 3: Find the Area of the Coil
The area (\(A\)) of the rectangular coil can be calculated as:
\[ A = \text{length} \times \text{breadth} = 5 \, \text{cm} \times 3 \, \text{cm} = 15 \, \text{cm}^2 = 15 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Convert the Angle to Radians
Convert \(10^\circ\) to radians:
\[ \theta = 10^\circ \times \frac{\pi}{180} \approx 0.1745 \, \text{radians} \]

### Step 5: Set Up the Torque Balance Equation
Equating the two expressions for torque:
\[ mB \sin(10^\circ) = c \theta \]
Since \(\sin(10^\circ) \approx 0.1736\), we have:
\[ mB \cdot 0.1736 = c \cdot 0.1745 \]

### Step 6: Substitute for Magnetic Moment
Substituting \(m = n \cdot i \cdot A\):
\[ (n \cdot i \cdot A)B \cdot 0.1736 = c \cdot 0.1745 \]

### Step 7: Solve for Current \(i\)
Rearranging the equation to solve for \(i\):
\[ i = \frac{c \cdot 0.1745}{n \cdot A \cdot B \cdot 0.1736} \]

### Step 8: Substitute Known Values
Substituting the known values:
- \(n = 100\)
- \(A = 15 \times 10^{-4} \, \text{m}^2\)
- \(B = 0.025 \, \text{Wb/m}^2\)
- \(c = 1.5 \times 10^{-9} \, \text{Nm/degree}\)

Calculating:
\[ i = \frac{(1.5 \times 10^{-9}) \cdot 0.1745}{100 \cdot (15 \times 10^{-4}) \cdot (0.025) \cdot 0.1736} \]

### Step 9: Calculate the Current
Performing the calculations:
\[ i \approx \frac{(1.5 \times 10^{-9}) \cdot 0.1745}{100 \cdot (15 \times 10^{-4}) \cdot (0.025) \cdot 0.1736} \approx 4 \times 10^{-6} \, \text{A} = 4 \, \mu A \]

### Final Answer
The current for which the coil will deflect through an angle of \(10^\circ\) is approximately \(4 \, \mu A\).

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