A circular coil of radius `5 pi` cm having a total of the magnetic meridian plane (i.e. the vertical plane in the north-south direction) . It is rotated about its vertical diameter by `45^(@)` and a current of `sqrt(2A)` is passed through it. A magnetic needle placed at the centre of this coil points west to east. The horizontalcopmonent of the earth's magnetic field is

Options

A. `20 xx 10^(-5)T`
B. `4 xx 10^(-5)T`
C. `4 xx 10^(-7)T`
D. `(4)/(pi) xx 10^(-5)T`

Answer

To solve the problem, we need to calculate the horizontal component of the Earth's magnetic field (B_H) when a circular coil is rotated and a current is passed through it. Here’s a step-by-step solution:

### Step 1: Understand the Setup
- We have a circular coil of radius \( 5\pi \) cm (which is \( 0.05\pi \) m).
- The coil is rotated about its vertical diameter by \( 45^\circ \).
- A current of \( \sqrt{2} \) A is passed through the coil.
- A magnetic needle placed at the center of the coil points from west to east.

### Step 2: Magnetic Field at the Center of the Coil
The magnetic field (B_0) at the center of a circular coil carrying current is given by the formula:
\[ B_0 = \frac{\mu_0 I}{2R} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space),
- \( I = \sqrt{2} \, \text{A} \) (current),
- \( R = 0.05\pi \, \text{m} \) (radius in meters).

### Step 3: Substitute Values into the Formula
Substituting the values into the formula:
\[ B_0 = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{2 \times 0.05\pi} \]
We can simplify this:
\[ B_0 = \frac{4 \times 10^{-7} \times \sqrt{2}}{0.1} = 4 \times 10^{-6} \sqrt{2} \, \text{T} \]

### Step 4: Calculate the Horizontal Component of the Magnetic Field
Since the coil is tilted at \( 45^\circ \), the horizontal component of the magnetic field (B_H) can be found using:
\[ B_H = B_0 \cos(45^\circ) \]
Substituting \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ B_H = 4 \times 10^{-6} \sqrt{2} \times \frac{1}{\sqrt{2}} = 4 \times 10^{-6} \, \text{T} \]

### Step 5: Final Result
Thus, the horizontal component of the Earth's magnetic field is:
\[ B_H = 4 \times 10^{-6} \, \text{T} \]