A current of `10A` is flowing east to west in a long wire kept in the east-west direction. Find magnetic field in a horizontal plane at a distance of (i) `10cm`. North (ii) `20cm` south from the wire, and in a vertical plane at a distance of (iii) `40cm` downwards, (iv) `50cm` upwards.

Explanation

To solve the problem, we will use the formula for the magnetic field \( B \) due to a long straight current-carrying wire, which is given by:

\[ B = \frac{\mu_0}{2\pi} \cdot \frac{I}{r} \]

where:
- \( B \) is the magnetic field,
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space),
- \( I \) is the current in amperes,
- \( r \) is the distance from the wire in meters.

### Step-by-Step Solution

**Given:**
- Current \( I = 10 \, \text{A} \)
- Wire is oriented east to west.

#### (i) Magnetic field at a distance of 10 cm North from the wire

1. **Convert distance to meters:**
\[ r_1 = 10 \, \text{cm} = 0.1 \, \text{m} \]

2. **Apply the formula:**
\[ B_1 = \frac{\mu_0}{2\pi} \cdot \frac{I}{r_1} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{10}{0.1} \]

3. **Simplify:**
\[ B_1 = \frac{4 \times 10^{-7}}{2} \cdot 100 = 2 \times 10^{-5} \, \text{T} \]

4. **Determine the direction:**
- Using the right-hand rule, if the current flows from east to west, the magnetic field at a point north of the wire will be directed downwards.

**Result:**
\[ B_1 = 2 \times 10^{-5} \, \text{T} \, \text{(downwards)} \]

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#### (ii) Magnetic field at a distance of 20 cm South from the wire

1. **Convert distance to meters:**
\[ r_2 = 20 \, \text{cm} = 0.2 \, \text{m} \]

2. **Apply the formula:**
\[ B_2 = \frac{\mu_0}{2\pi} \cdot \frac{I}{r_2} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{10}{0.2} \]

3. **Simplify:**
\[ B_2 = \frac{4 \times 10^{-7}}{2} \cdot 50 = 1 \times 10^{-5} \, \text{T} \]

4. **Determine the direction:**
- Using the right-hand rule, the magnetic field at a point south of the wire will be directed upwards.

**Result:**
\[ B_2 = 1 \times 10^{-5} \, \text{T} \, \text{(upwards)} \]

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#### (iii) Magnetic field at a distance of 40 cm Downwards from the wire

1. **Convert distance to meters:**
\[ r_3 = 40 \, \text{cm} = 0.4 \, \text{m} \]

2. **Apply the formula:**
\[ B_3 = \frac{\mu_0}{2\pi} \cdot \frac{I}{r_3} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{10}{0.4} \]

3. **Simplify:**
\[ B_3 = \frac{4 \times 10^{-7}}{2} \cdot 25 = 5 \times 10^{-6} \, \text{T} \]

4. **Determine the direction:**
- The magnetic field at a point directly below the wire will be directed towards the south.

**Result:**
\[ B_3 = 5 \times 10^{-6} \, \text{T} \, \text{(southwards)} \]

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#### (iv) Magnetic field at a distance of 50 cm Upwards from the wire

1. **Convert distance to meters:**
\[ r_4 = 50 \, \text{cm} = 0.5 \, \text{m} \]

2. **Apply the formula:**
\[ B_4 = \frac{\mu_0}{2\pi} \cdot \frac{I}{r_4} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{10}{0.5} \]

3. **Simplify:**
\[ B_4 = \frac{4 \times 10^{-7}}{2} \cdot 20 = 4 \times 10^{-6} \, \text{T} \]

4. **Determine the direction:**
- The magnetic field at a point directly above the wire will be directed towards the south.

**Result:**
\[ B_4 = 4 \times 10^{-6} \, \text{T} \, \text{(southwards)} \]

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### Summary of Results
- \( B_1 = 2 \times 10^{-5} \, \text{T} \) (downwards)
- \( B_2 = 1 \times 10^{-5} \, \text{T} \) (upwards)
- \( B_3 = 5 \times 10^{-6} \, \text{T} \) (southwards)
- \( B_4 = 4 \times 10^{-6} \, \text{T} \) (southwards)