Question

A thin straight vertical conductor has 10 amp current flows vertically upwards. It is present at a place where $B_H = 4 \times 10^{-6}T$ . Arrange the net magnetic induction at the following points in ascending order a) at 0.5m on south of conductor b) at 0.5ra on west of conductor c) at 0.5m on east of conductor d) at 0.5m on north-east of conductor

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

Answer 1

The Correct Option is D

Answer (d) b,a,d,c

Answer 2

Given:
Current I = 10A (upwards)
Horizontal magnetic field \(B_H = 4 \times 10^{-6}\)T

Let's calculate the magnetic field due to the current-carrying conductor at a distance r = 0.5m from the wire using the formula:
\(B = \frac{\mu_0 I}{2 \pi r}\)
where:
\(\mu_0 = 4 \pi \times 10^{-7} \, \text{T} \cdot \text{m}/\text{A}\)

\(B = \frac{4 \pi \times 10^{-7} \times 10}{2 \pi \times 0.5} = \frac{4 \times 10^{-6}}{0.5} = 8 \times 10^{-6} \, \text{T}\)

Net Magnetic Induction at Each Point:
1. At 0.5 m South of the Conductor:
Magnetic field due to current \(B_{\text{current}} = 8 \times 10^{-6} \, \text{T}\)(towards the west)
Horizontal magnetic field \(B_H = 4 \times 10^{-6} \, \text{T}\) (towards the north)
Net magnetic field:
\(B_{\text{south}} = \sqrt{(8 \times 10^{-6})^2 + (4 \times 10^{-6})^2} = \sqrt{64 \times 10^{-12} + 16 \times 10^{-12}} = \sqrt{80 \times 10^{-12}} = \sqrt{8 \times 10^{-11}} = 8.94 \times 10^{-6} \, \text{T}\)

2. At 0.5 m West of the Conductor:
Magnetic field due to current \(B_{\text{current}} = 8 \times 10^{-6} \, \text{T}\) (towards the south)
Horizontal magnetic field \(B_H = 4 \times 10^{-6} \, \text{T}\) (towards the east)
Net magnetic field:
\(B_{\text{west}} = 8 \times 10^{-6} - 4 \times 10^{-6} = 4 \times 10^{-6} \, \text{T}\)

3. At 0.5 m East of the Conductor:
Magnetic field due to current \(B_{\text{current}} = 8 \times 10^{-6} \, \text{T}\) (towards the north)
Horizontal magnetic field \(B_H = 4 \times 10^{-6} \, \text{T}\) (towards the west)
Net magnetic field:
\(B_{\text{east}} = 8 \times 10^{-6} + 4 \times 10^{-6} = 12 \times 10^{-6} \, \text{T}\)

4. At 0.5 m North-East of the Conductor:
Magnetic field due to current \(B_{\text{current}} = 8 \times 10^{-6} \, \text{T}\)
Horizontal magnetic field \(B_H = 4 \times 10^{-6} \, \text{T}\)
Net magnetic field:
\(B_{\text{north-east}} = \sqrt{(8 \times 10^{-6})^2 + (4 \times 10^{-6})^2} = \sqrt{64 \times 10^{-12} + 16 \times 10^{-12}} = \sqrt{80 \times 10^{-12}} = \sqrt{8 \times 10^{-11}} = 8.94 \times 10^{-6} \, \text{T}\)

Ascending Order:
1. \(B_{\text{west}} = 4 \times 10^{-6} \, \text{T}\)
2. \(B_{\text{south}} = 8.94 \times 10^{-6} \, \text{T}\)
3. \(B_{\text{north-east}} = 8.94 \times 10^{-6} \, \text{T}\)
4. \(B_{\text{east}} = 12 \times 10^{-6} \, \text{T}\)

So, the correct option is (D): \(b, a, d, c\).