A magnetic dipole of magnetic moment `6Am^(2)` is lying in a horizontal plane with its north pole pointing toward `60^(@)` East of North. Find the net horizontal magnetic field at a point· on the axis of the magnet 0.2m away from it. Horizontal Component of earth field at this place is `30 mu T`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a magnetic dipole with a magnetic moment \( m = 6 \, \text{Am}^2 \) lying in a horizontal plane. The north pole of the dipole points \( 60^\circ \) East of North. We need to find the net horizontal magnetic field at a point on the axis of the magnet, located \( 0.2 \, \text{m} \) away from it. The horizontal component of the Earth's magnetic field at this location is \( 30 \, \mu T \).

### Step 2: Calculate the Magnetic Field Due to the Dipole
The magnetic field \( B_m \) at a distance \( r \) along the axis of a magnetic dipole is given by the formula:
\[ B_m = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 3: Substitute the Values
Given:
- \( m = 6 \, \text{Am}^2 \)
- \( r = 0.2 \, \text{m} \)

Substituting the values into the equation:
\[ B_m = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 6}{(0.2)^3} \]
This simplifies to:
\[ B_m = 10^{-7} \cdot \frac{12}{0.008} = 10^{-7} \cdot 1500 = 1.5 \times 10^{-4} \, \text{T} = 150 \, \mu T \]

### Step 4: Calculate the Net Magnetic Field
The net magnetic field \( B \) at the point is the vector sum of the magnetic field due to the dipole \( B_m \) and the horizontal component of the Earth's magnetic field \( B_e = 30 \, \mu T \).

Since the dipole is oriented at \( 60^\circ \) East of North, we can resolve the Earth's magnetic field into components:
- \( B_{e_x} = B_e \cos(60^\circ) = 30 \cos(60^\circ) = 30 \times 0.5 = 15 \, \mu T \)
- \( B_{e_y} = B_e \sin(60^\circ) = 30 \sin(60^\circ) = 30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \mu T \)

The magnetic field due to the dipole is directed along the axis of the dipole, which we can consider as the x-direction.

### Step 5: Calculate the Resultant Magnetic Field
Now, we can find the resultant magnetic field using the Pythagorean theorem:
\[ B_{net} = \sqrt{(B_m + B_{e_x})^2 + (B_{e_y})^2} \]
Substituting the values:
\[ B_{net} = \sqrt{(150 + 15)^2 + (15\sqrt{3})^2} \]
Calculating:
\[ B_{net} = \sqrt{165^2 + (15\sqrt{3})^2} = \sqrt{27225 + 675} = \sqrt{27900} \]
Calculating the square root:
\[ B_{net} \approx 167.1 \, \mu T \]

### Final Answer
The net horizontal magnetic field at the point \( 0.2 \, \text{m} \) away from the dipole is approximately \( 167.1 \, \mu T \).

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