A circular coil of radius `20 cm` and `20` turns of wire is mounted vertically with its plane in magnetic meridian.A small magnetic needle (free to rotate about vertical axis) is placed at the center of the coil.It is deflected through `45^(@)` when a current is passed through the coil and in equilbrium (Horizontal component of earth's field is `0.34 xx 10^(-4) T`).The current in coil is:

Options

A. `17/(10 pi) A`
B. `6 A`
C. `6 xx 10^(-3)A`
D. `3/(50) A`

Answer

To find the current in the circular coil, we can follow these steps:

### Step 1: Understand the relationship between the magnetic field and the deflection of the needle
The magnetic field \( B \) at the center of the coil can be related to the horizontal component of the Earth's magnetic field \( B_H \) and the angle of deflection \( \theta \) using the formula:
\[ B = B_H \tan(\theta) \]
Given that \( B_H = 0.34 \times 10^{-4} \, \text{T} \) and \( \theta = 45^\circ \), we know that \( \tan(45^\circ) = 1 \).

### Step 2: Calculate the magnetic field \( B \)
Substituting the values into the equation:
\[ B = 0.34 \times 10^{-4} \times 1 = 0.34 \times 10^{-4} \, \text{T} \]

### Step 3: Use the formula for the magnetic field at the center of a circular coil
The magnetic field \( B \) at the center of a circular coil with \( n \) turns, carrying a current \( I \), is given by:
\[ B = \frac{n \mu_0 I}{2R} \]
where:
- \( n = 20 \) (number of turns)
- \( R = 0.2 \, \text{m} \) (radius converted from cm to m)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)

### Step 4: Rearrange the formula to solve for current \( I \)
Rearranging the formula gives:
\[ I = \frac{2BR}{n\mu_0} \]

### Step 5: Substitute the known values into the equation
Substituting \( B = 0.34 \times 10^{-4} \, \text{T} \), \( R = 0.2 \, \text{m} \), \( n = 20 \), and \( \mu_0 = 4\pi \times 10^{-7} \):
\[ I = \frac{2 \times (0.34 \times 10^{-4}) \times 0.2}{20 \times (4\pi \times 10^{-7})} \]

### Step 6: Simplify the expression
Calculating the numerator:
\[ 2 \times 0.34 \times 10^{-4} \times 0.2 = 0.136 \times 10^{-4} \]
Calculating the denominator:
\[ 20 \times (4\pi \times 10^{-7}) = 80\pi \times 10^{-7} \]
Now substituting back:
\[ I = \frac{0.136 \times 10^{-4}}{80\pi \times 10^{-7}} = \frac{0.136 \times 10^{-4 + 7}}{80\pi} = \frac{0.136 \times 10^{3}}{80\pi} \]

### Step 7: Final calculation
Now, calculate the final value:
\[ I = \frac{136}{80\pi} \times 10^{3} \]
This simplifies to:
\[ I = \frac{17}{10\pi} \, \text{A} \]

### Conclusion
The current in the coil is:
\[ I = \frac{17}{10\pi} \, \text{A} \]