A vertical circular coil of one turn and radius 9.42 cm is placed with its plane in the magnetic meridian and short magnetic needle is pivoted at the centre of the coil so that it can freely rotate in the horizontal plane. If a current of 6A is passed through the coil, then the needle deflects by `(B_(H) = 4 xx 10^(-5)T)`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to determine the angle of deflection (θ) of the magnetic needle when a current of 6 A is passed through a vertical circular coil of radius 9.42 cm. The horizontal magnetic field (B_H) is given as 4 × 10^(-5) T.

### Step-by-step Solution:

1. **Identify the given values:**
- Radius of the coil (r) = 9.42 cm = 9.42 × 10^(-2) m
- Current (I) = 6 A
- Horizontal magnetic field (B_H) = 4 × 10^(-5) T
- Number of turns (n) = 1 (since it is a one-turn coil)

2. **Use the formula for the magnetic field (B) at the center of a circular coil:**
The magnetic field at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 n I}{2r} \]
where:
- \(\mu_0\) (permeability of free space) = 4π × 10^(-7) T·m/A

3. **Substituting the values into the formula:**
\[ B = \frac{(4\pi \times 10^{-7}) \cdot (1) \cdot (6)}{2 \cdot (9.42 \times 10^{-2})} \]

4. **Calculate the magnetic field (B):**
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 6}{2 \cdot 9.42 \times 10^{-2}} \]
\[ B = \frac{(24\pi \times 10^{-7})}{(18.84 \times 10^{-2})} \]
\[ B = \frac{24\pi \times 10^{-5}}{18.84} \]
\[ B \approx 4.0 \times 10^{-5} \text{ T} \]

5. **Relate the magnetic field to the deflection angle:**
The tangent of the angle of deflection (θ) can be expressed as:
\[ \tan \theta = \frac{B}{B_H} \]

6. **Substituting the values to find θ:**
\[ \tan \theta = \frac{B}{B_H} = \frac{4.0 \times 10^{-5}}{4.0 \times 10^{-5}} = 1 \]

7. **Finding the angle θ:**
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Final Answer:
The angle of deflection (θ) of the magnetic needle is **45 degrees**.