A thin straight vertical conductor has 10 amp current, vertically upwards. It is present at a place where `B_(H) = 4 xx 10^(-6) T`. Arrange the net magnetic inducitons at the following points in ascending order
a) at 0.5m on south of conductor
b) at 0.5m on west of conductor
c) at 0.5m on east of conductor
d) at 0.5m on north-east of conductor

Options

A. a,b,c,d
B. b,a,c,d
C. a,d,c,b
D. b,a,d,c

Answer

To solve the problem of determining the net magnetic inductions at various points around a vertical conductor carrying a 10 A current, we will follow these steps:

### Step 1: Understand the magnetic field due to a straight conductor
The magnetic field \( B \) created by a long straight conductor carrying current \( I \) at a distance \( d \) from the wire is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi d} \]
where \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \).

### Step 2: Calculate the magnetic field at 0.5 m from the conductor
Given:
- \( I = 10 \, \text{A} \)
- \( d = 0.5 \, \text{m} \)

Substituting the values into the formula:
\[ B = \frac{(4 \pi \times 10^{-7}) \times 10}{2 \pi \times 0.5} = \frac{4 \times 10^{-6}}{1} = 4 \times 10^{-6} \, \text{T} \]

### Step 3: Determine the magnetic field direction at each point
Using the right-hand rule:
- **South (Point A)**: The magnetic field will be directed towards the east.
- **West (Point B)**: The magnetic field will be directed towards the south.
- **East (Point C)**: The magnetic field will be directed towards the west.
- **North-East (Point D)**: The magnetic field will be at an angle of 45 degrees to both the north and east.

### Step 4: Calculate the net magnetic induction at each point
1. **At Point A (0.5 m South)**:
- \( B_A = B + B_H \)
- \( B_H = 4 \times 10^{-6} \, \text{T} \) (horizontal component)
- Resultant \( B_A = \sqrt{(4 \times 10^{-6})^2 + (4 \times 10^{-6})^2} = \sqrt{2 \times (4 \times 10^{-6})^2} = 4\sqrt{2} \times 10^{-6} \, \text{T} \)

2. **At Point B (0.5 m West)**:
- \( B_B = 2B_H = 2 \times 4 \times 10^{-6} = 8 \times 10^{-6} \, \text{T} \)

3. **At Point C (0.5 m East)**:
- The magnetic field is in the opposite direction to \( B_H \), thus:
- \( B_C = 0 \, \text{T} \)

4. **At Point D (0.5 m North-East)**:
- The angle between the two components is 135 degrees.
- Using the cosine rule:
\[ B_D = \sqrt{(4 \times 10^{-6})^2 + (4 \times 10^{-6})^2 + 2(4 \times 10^{-6})(4 \times 10^{-6}) \cos(135^\circ)} \]
- \( \cos(135^\circ) = -\frac{1}{\sqrt{2}} \)
- Thus, \( B_D = \sqrt{2(4 \times 10^{-6})^2 - 2(4 \times 10^{-6})^2/\sqrt{2}} \)
- This simplifies to \( B_D = 4 \times 10^{-6} \sqrt{0.59} \, \text{T} \)

### Step 5: Arrange the net magnetic inductions in ascending order
Now we have:
- \( B_A = 4\sqrt{2} \times 10^{-6} \, \text{T} \)
- \( B_B = 8 \times 10^{-6} \, \text{T} \)
- \( B_C = 0 \, \text{T} \)
- \( B_D = 4 \times 10^{-6} \sqrt{0.59} \, \text{T} \)

Calculating approximate values:
- \( B_A \approx 5.66 \times 10^{-6} \, \text{T} \)
- \( B_B = 8 \times 10^{-6} \, \text{T} \)
- \( B_C = 0 \, \text{T} \)
- \( B_D \approx 4 \times 10^{-6} \times 0.77 \approx 3.08 \times 10^{-6} \, \text{T} \)

### Final Order
The order of magnetic inductions in ascending order is:
1. \( B_C = 0 \, \text{T} \) (Point C)
2. \( B_D \approx 3.08 \times 10^{-6} \, \text{T} \) (Point D)
3. \( B_A \approx 5.66 \times 10^{-6} \, \text{T} \) (Point A)
4. \( B_B = 8 \times 10^{-6} \, \text{T} \) (Point B)