A `16cm^(2)` coil has 20 turns. Its suspended by a phosphor bronze wire of `10^(-6) Nm` per degree as couple per unit twist. When 'i' ampere is passing through it, in a 0.2T field, the wire twisted by 45°. The value of 'i' is?

Options

A. `7xx10^(-3)A`
B. `8xx10^(-3)A`
C. `7xx10^(-2)A`
D. 4A

Answer

To solve the problem step by step, we will use the given data and apply the relevant formulas.

### Step 1: Convert Area to SI Units
The area of the coil is given as \( 16 \, \text{cm}^2 \). We need to convert this to square meters.

\[ A = 16 \, \text{cm}^2 = 16 \times 10^{-4} \, \text{m}^2 \]

**Hint for Step 1:** Remember that \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \).

### Step 2: Identify Given Values
From the problem, we have:
- Number of turns, \( N = 20 \)
- Torque per unit twist, \( \tau = 10^{-6} \, \text{Nm/degree} \)
- Magnetic field, \( B = 0.2 \, \text{T} \)
- Angle of twist, \( \theta = 45^\circ \)

**Hint for Step 2:** Write down all the given values clearly to avoid confusion later.

### Step 3: Calculate Total Torque
The total torque \( \tau \) experienced by the coil when twisted by an angle \( \theta \) can be calculated using the formula:

\[ \tau = \text{(Torque per unit twist)} \times \theta \]

Converting \( \theta \) from degrees to radians (since \( 1 \, \text{degree} = \frac{\pi}{180} \, \text{radians} \)):

\[ \theta = 45^\circ = \frac{45 \times \pi}{180} = \frac{\pi}{4} \, \text{radians} \]

Now, substituting the values:

\[ \tau = 10^{-6} \, \text{Nm/degree} \times 45 \, \text{degrees} = 10^{-6} \times 45 = 45 \times 10^{-6} \, \text{Nm} \]

**Hint for Step 3:** Make sure to convert degrees to radians if necessary, but in this case, we can keep it in degrees since the torque is given per degree.

### Step 4: Use the Torque Formula for a Coil
The torque on a coil in a magnetic field is given by:

\[ \tau = N \cdot i \cdot B \cdot A \]

Where:
- \( N \) = number of turns
- \( i \) = current in amperes
- \( B \) = magnetic field in tesla
- \( A \) = area in square meters

Substituting the known values:

\[ 45 \times 10^{-6} = 20 \cdot i \cdot 0.2 \cdot (16 \times 10^{-4}) \]

### Step 5: Solve for Current \( i \)
Rearranging the equation to solve for \( i \):

\[ i = \frac{45 \times 10^{-6}}{20 \cdot 0.2 \cdot (16 \times 10^{-4})} \]

Calculating the denominator:

\[ 20 \cdot 0.2 = 4 \]

\[ 4 \cdot (16 \times 10^{-4}) = 64 \times 10^{-4} = 6.4 \times 10^{-3} \]

Now substituting back:

\[ i = \frac{45 \times 10^{-6}}{6.4 \times 10^{-3}} = \frac{45}{6.4} \times 10^{-3} \, \text{A} \]

Calculating \( \frac{45}{6.4} \):

\[ i \approx 7.03125 \times 10^{-3} \, \text{A} \]

Thus, the current \( i \) is approximately:

\[ i \approx 7.03 \times 10^{-3} \, \text{A} \]

**Final Answer:** \( i \approx 7.03 \, \text{mA} \)