Two coils each of 100 turns are held such that one lies in vertical plane and the other in the horizontal plane with their centres coinciding. The radius of the vetical coil is 0.20m and that of the horizontal coil is 0.30m. How will you neutralize the magnetic field of the earth at their common centre? What is the current to be passed through each coil? Horizontal component of earth's magnetic field `=0.35xx10^-4T` and angle of dip =`30^@`.

Explanation

To solve the problem of neutralizing the magnetic field of the Earth at the common center of the two coils, we will follow these steps:

### Step 1: Understand the Magnetic Field Components
The Earth's magnetic field has two components at the location of the coils:
- A horizontal component \( B_H = 0.35 \times 10^{-4} \, T \)
- A vertical component \( B_V \), which we need to calculate using the angle of dip \( \theta = 30^\circ \).

### Step 2: Calculate the Vertical Component of the Earth's Magnetic Field
Using the angle of dip, we can find the vertical component \( B_V \) using the formula:
\[ \tan(\theta) = \frac{B_V}{B_H} \]
From this, we can rearrange to find \( B_V \):
\[ B_V = B_H \cdot \tan(\theta) \]
Substituting the values:
\[ B_V = 0.35 \times 10^{-4} \cdot \tan(30^\circ) = 0.35 \times 10^{-4} \cdot \frac{1}{\sqrt{3}} = 0.203 \times 10^{-4} \, T \]

### Step 3: Determine the Required Magnetic Fields
To neutralize the Earth's magnetic field:
- The vertical coil must produce a magnetic field \( B_V \) in the upward direction.
- The horizontal coil must produce a magnetic field \( B_H \) in the opposite direction to cancel the horizontal component of the Earth's magnetic field.

### Step 4: Calculate the Current for the Vertical Coil
The magnetic field at the center of a coil is given by:
\[ B = \frac{\mu_0 n I}{2R} \]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \) (permeability of free space)
- \( n = 100 \) turns
- \( R = 0.2 \, m \) (radius of the vertical coil)

Setting \( B = B_V \):
\[ B_V = \frac{\mu_0 n I_1}{2R} \]
Rearranging for \( I_1 \):
\[ I_1 = \frac{2R B_V}{\mu_0 n} \]
Substituting the values:
\[ I_1 = \frac{2 \cdot 0.2 \cdot 0.203 \times 10^{-4}}{4\pi \times 10^{-7} \cdot 100} \]
Calculating \( I_1 \):
\[ I_1 \approx 0.111 \, A \]

### Step 5: Calculate the Current for the Horizontal Coil
Using the same formula for the horizontal coil:
\[ B_H = \frac{\mu_0 n I_2}{2R} \]
Where \( R = 0.3 \, m \):
\[ I_2 = \frac{2R B_H}{\mu_0 n} \]
Substituting the values:
\[ I_2 = \frac{2 \cdot 0.3 \cdot 0.35 \times 10^{-4}}{4\pi \times 10^{-7} \cdot 100} \]
Calculating \( I_2 \):
\[ I_2 \approx 0.0964 \, A \]

### Final Answer
- The current to be passed through the vertical coil is approximately \( 0.111 \, A \).
- The current to be passed through the horizontal coil is approximately \( 0.0964 \, A \).