A straight conductor of length 32 cm carries a current of 30A. Magnetic induction at a point which is in air at a perpendicular distance of 12cm from the mid point of the conductor is

Options

A. 0.2 gauss
B. 0.3 gauss
C. 0.4 gauss
D. 0.5 gauss

Answer

To solve the problem, we will use the formula for the magnetic field (magnetic induction) produced by a straight current-carrying conductor at a point perpendicular to it. The formula is given by:

\[ B = \frac{\mu_0 I}{4 \pi r} \left( \sin \theta_1 + \sin \theta_2 \right) \]

Where:
- \(B\) is the magnetic induction,
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T m/A}\)),
- \(I\) is the current in amperes,
- \(r\) is the perpendicular distance from the midpoint of the conductor to the point where the magnetic field is being calculated,
- \(\theta_1\) and \(\theta_2\) are the angles made by the line connecting the ends of the conductor to the point with the perpendicular line.

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of the conductor, \(L = 32 \, \text{cm} = 0.32 \, \text{m}\)
- Current, \(I = 30 \, \text{A}\)
- Perpendicular distance from the midpoint, \(r = 12 \, \text{cm} = 0.12 \, \text{m}\)

2. **Calculate the half-length of the conductor:**
\[ \frac{L}{2} = \frac{0.32}{2} = 0.16 \, \text{m} \]

3. **Determine the angles \(\theta_1\) and \(\theta_2\):**
Since the point is at the midpoint and the conductor is straight, \(\theta_1 = \theta_2\). We can use the right triangle formed by the conductor and the point to find \(\sin \theta_1\):
\[ \sin \theta_1 = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{0.16}{\sqrt{(0.16)^2 + (0.12)^2}} \]
Calculate the hypotenuse:
\[ \sqrt{(0.16)^2 + (0.12)^2} = \sqrt{0.0256 + 0.0144} = \sqrt{0.04} = 0.2 \, \text{m} \]
Therefore,
\[ \sin \theta_1 = \frac{0.16}{0.2} = 0.8 \]

4. **Substitute the values into the magnetic induction formula:**
Since \(\sin \theta_1 = \sin \theta_2\),
\[ B = \frac{\mu_0 I}{4 \pi r} (2 \sin \theta_1) = \frac{(4\pi \times 10^{-7}) \times 30}{4 \pi \times 0.12} (2 \times 0.8) \]
Simplifying,
\[ B = \frac{(4 \times 10^{-7}) \times 30}{0.12} \times 1.6 \]

5. **Calculate the value:**
\[ B = \frac{1.2 \times 10^{-5}}{0.12} \times 1.6 = 1.0 \times 10^{-5} \times 1.6 = 1.6 \times 10^{-5} \, \text{T} \]
Converting to Gauss (1 T = 10^4 Gauss),
\[ B = 1.6 \times 10^{-5} \times 10^4 = 0.16 \, \text{Gauss} \]

6. **Final answer:**
The magnetic induction at the point is approximately \(0.16 \, \text{Gauss}\).