A wire placed along north-south direction carries a current of `5A` from south to north. Find the magnetic field due to a `1cm` piece of wire at a point `200cm` north east from the piece.

Explanation

To find the magnetic field due to a 1 cm piece of wire carrying a current of 5 A at a point 200 cm northeast from the piece, we can use the Biot-Savart Law, which states that the magnetic field \( dB \) at a point due to a small segment of current-carrying wire is given by:

\[ dB = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot dl \cdot \sin(\theta)}{r^2} \]

Where:
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, T \cdot m/A \))
- \( I \) is the current (5 A)
- \( dl \) is the length of the wire segment (1 cm = 0.01 m)
- \( r \) is the distance from the wire segment to the point where the field is being calculated
- \( \theta \) is the angle between the wire segment and the line connecting the wire to the point

### Step 1: Determine the distance \( r \)
The point is located 200 cm (or 2 m) northeast from the wire. Since the wire is oriented north-south and the point is northeast, we can find the distance \( r \) using the Pythagorean theorem.

- The distance in the north direction is \( 2 \cos(45^\circ) = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \, m \)
- The distance in the east direction is \( 2 \sin(45^\circ) = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \, m \)

Thus, the total distance \( r \) from the wire to the point is:

\[ r = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \, m \]

### Step 2: Determine the angle \( \theta \)
Since the wire is vertical (north-south) and the point is northeast, the angle \( \theta \) between the wire and the line connecting the wire to the point is \( 90^\circ - 45^\circ = 45^\circ \).

### Step 3: Substitute values into the Biot-Savart Law
Now we can substitute the values into the Biot-Savart Law:

\[ dB = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot dl \cdot \sin(45^\circ)}{r^2} \]

Substituting the known values:

- \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \)
- \( I = 5 \, A \)
- \( dl = 0.01 \, m \)
- \( r = 2 \, m \)
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)

\[ dB = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{5 \cdot 0.01 \cdot \frac{1}{\sqrt{2}}}{(2)^2} \]

This simplifies to:

\[ dB = 10^{-7} \cdot \frac{5 \cdot 0.01 \cdot \frac{1}{\sqrt{2}}}{4} \]

Calculating this gives:

\[ dB = 10^{-7} \cdot \frac{0.05}{4\sqrt{2}} = 10^{-7} \cdot \frac{0.05}{4 \cdot 1.414} \approx 10^{-7} \cdot \frac{0.05}{5.656} \approx 10^{-7} \cdot 8.84 \times 10^{-3} \approx 8.84 \times 10^{-10} \, T \]

### Final Answer
The magnetic field at the point 200 cm northeast from the piece of wire is approximately:

\[ B \approx 8.84 \times 10^{-10} \, T \]