A tangent galvanometer has a coil of 50 turns and a radius of 20cm. The horizontal component of the earth's magnetic field is `B_H = 3 xx 10^(-5) T`. Find the current which gives a diflection of `45^@)`.

Options

A. 0.39A
B. 0.29A
C. 0.19A
D. 0.09A

Explanation

To solve the problem of finding the current that gives a deflection of \(45^\circ\) in a tangent galvanometer, we can follow these steps:

### Step 1: Understand the relationship in a tangent galvanometer
The current \(I\) through the galvanometer is related to the angle of deflection \(\theta\) by the formula:
\[ I = \frac{2 \pi R B_H}{\mu_0 N} \tan(\theta) \]
where:
- \(R\) = radius of the coil (in meters)
- \(B_H\) = horizontal component of the earth's magnetic field (in tesla)
- \(\mu_0\) = permeability of free space \((4 \pi \times 10^{-7} \, \text{T m/A})\)
- \(N\) = number of turns in the coil
- \(\theta\) = angle of deflection

### Step 2: Convert the radius to meters
Given the radius \(R = 20 \, \text{cm}\), we convert it to meters:
\[ R = 20 \, \text{cm} = 0.2 \, \text{m} \]

### Step 3: Substitute the known values
We know:
- \(N = 50\) (number of turns)
- \(B_H = 3 \times 10^{-5} \, \text{T}\)
- \(\theta = 45^\circ\) (for which \(\tan(45^\circ) = 1\))

Substituting these values into the formula:
\[ I = \frac{2 \pi (0.2) (3 \times 10^{-5})}{4 \pi \times 10^{-7} \times 50} \tan(45^\circ) \]

### Step 4: Simplify the equation
Since \(\tan(45^\circ) = 1\), the equation simplifies to:
\[ I = \frac{2 \pi (0.2) (3 \times 10^{-5})}{4 \pi \times 10^{-7} \times 50} \]

### Step 5: Cancel out \(\pi\)
The \(\pi\) in the numerator and denominator cancels out:
\[ I = \frac{2 (0.2) (3 \times 10^{-5})}{4 \times 10^{-7} \times 50} \]

### Step 6: Calculate the numerator and denominator
Calculating the numerator:
\[ 2 \times 0.2 \times 3 \times 10^{-5} = 1.2 \times 10^{-5} \]

Calculating the denominator:
\[ 4 \times 10^{-7} \times 50 = 2 \times 10^{-5} \]

### Step 7: Final calculation
Now substituting back into the equation:
\[ I = \frac{1.2 \times 10^{-5}}{2 \times 10^{-5}} = 0.6 \, \text{A} \]

### Conclusion
The current that gives a deflection of \(45^\circ\) is:
\[ I = 0.6 \, \text{A} \]