The magnetic needle of a tangent galvanometer is deflected at an angle `30^(@)` due to a magnet. The hoeizontal component of earth's magnetic field `0.34xx10^(-4)T` is along the plane of the coil. The magnetic intensity is

Options

A. `1.96xx10^(-4)T`
B. `1.96xx10^(-5)T`
C. `1.96xx10^(4)T`
D. `1.96xx10^(5)T`

Explanation

To solve the problem, we will use the formula for the magnetic intensity (B) in a tangent galvanometer, which is given by:

\[ B = B_h \cdot \tan(\theta) \]

where:
- \( B \) is the total magnetic intensity,
- \( B_h \) is the horizontal component of the Earth's magnetic field,
- \( \theta \) is the angle of deflection.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The angle of deflection, \( \theta = 30^\circ \).
- The horizontal component of the Earth's magnetic field, \( B_h = 0.34 \times 10^{-4} \, \text{T} \).

2. **Convert the Angle to Radians (if necessary)**:
- In this case, we can directly use degrees since the tangent function can take degrees as input.

3. **Calculate the Tangent of the Angle**:
- Calculate \( \tan(30^\circ) \):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \]

4. **Substitute the Values into the Formula**:
- Substitute \( B_h \) and \( \tan(30^\circ) \) into the formula:
\[ B = B_h \cdot \tan(30^\circ) = (0.34 \times 10^{-4}) \cdot \left(\frac{1}{\sqrt{3}}\right) \]

5. **Calculate the Magnetic Intensity**:
- Now, calculate the value:
\[ B = 0.34 \times 10^{-4} \cdot 0.577 \approx 0.196 \times 10^{-4} = 1.96 \times 10^{-5} \, \text{T} \]

6. **Final Answer**:
- The magnetic intensity \( B \) is approximately \( 1.96 \times 10^{-5} \, \text{T} \).

### Conclusion:
The magnetic intensity due to which the needle is deflected at an angle of \( 30^\circ \) is \( 1.96 \times 10^{-5} \, \text{T} \).