A circuit coil of radius 20 cm and 20 turns of wire is mounted vertically withits plane in the magnetic meridian. A small magnetic needle placed at the center of the coil is deflected through `45^(@)` when a current is passed through the coil. What is the value of the current? (horizontal induction of earth's field = `3.6xx10^(-5)Wb//m^(2)`

Options

A. 0.6A
B. 6A
C. `6xx10^(-3)A`
D. 0.06A

Explanation

To solve the problem step by step, we will follow the method outlined in the video transcript while providing a detailed explanation for each step.

### Step 1: Identify Given Values
We are given the following values:
- Radius of the coil, \( R = 20 \, \text{cm} = 20 \times 10^{-2} \, \text{m} \)
- Number of turns, \( N = 20 \)
- Angle of deflection, \( \theta = 45^\circ \)
- Horizontal magnetic induction of the Earth's field, \( B_h = 3.6 \times 10^{-5} \, \text{Wb/m}^2 \)

### Step 2: Calculate the Net Magnetic Field at the Center of the Coil
The net magnetic field \( B \) at the center of the coil when the current is passed can be calculated using the relationship:
\[ B = B_h \tan(\theta) \]
Since \( \tan(45^\circ) = 1 \), we have:
\[ B = B_h \cdot 1 = B_h = 3.6 \times 10^{-5} \, \text{Wb/m}^2 \]

### Step 3: Use the Formula for Magnetic Field of a Circular Coil
The magnetic field \( B \) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 N I}{2R} \]
where:
- \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( N \) is the number of turns
- \( I \) is the current
- \( R \) is the radius of the coil

### Step 4: Rearranging the Formula to Find Current \( I \)
We can rearrange the formula to solve for the current \( I \):
\[ I = \frac{2BR}{\mu_0 N} \]

### Step 5: Substitute the Known Values
Now we substitute the known values into the equation:
- \( B = 3.6 \times 10^{-5} \, \text{Wb/m}^2 \)
- \( R = 20 \times 10^{-2} \, \text{m} \)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( N = 20 \)

Substituting these values:
\[ I = \frac{2 \times (3.6 \times 10^{-5}) \times (20 \times 10^{-2})}{4\pi \times 10^{-7} \times 20} \]

### Step 6: Simplifying the Expression
Calculating the numerator:
\[ 2 \times (3.6 \times 10^{-5}) \times (20 \times 10^{-2}) = 2 \times 3.6 \times 20 \times 10^{-7} = 144 \times 10^{-7} = 1.44 \times 10^{-5} \]

Calculating the denominator:
\[ 4\pi \times 10^{-7} \times 20 = 80\pi \times 10^{-7} \]

Now substituting back into the equation for \( I \):
\[ I = \frac{1.44 \times 10^{-5}}{80\pi \times 10^{-7}} \]

### Step 7: Final Calculation
Calculating \( I \):
\[ I = \frac{1.44 \times 10^{-5}}{80\pi \times 10^{-7}} = \frac{1.44}{80\pi} \times 10^{2} \]

Using \( \pi \approx 3.14 \):
\[ I \approx \frac{1.44}{251.2} \approx 0.00573 \times 10^{2} = 0.573 \, \text{A} \]

### Conclusion
The value of the current \( I \) is approximately \( 0.6 \, \text{A} \).