A long straight wire carrying current of `30 A` is placed in an external uniform magnetic field of induction `4xx10^(-4) T`. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point `2.0 cm` away from the wire is

Options

A. `10^(-4)`
B. `3xx10^(-4)`
C. `5xx10^(-4)`
D. `6xx10^(-4)`

Explanation

To solve the problem, we need to find the resultant magnetic induction at a point 2.0 cm away from a long straight wire carrying a current of 30 A, while also considering the external magnetic field of induction 4 × 10^(-4) T that is parallel to the direction of the current.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current (I) = 30 A
- External magnetic field (B_ext) = 4 × 10^(-4) T
- Distance from the wire (r) = 2.0 cm = 0.02 m

2. **Calculate the Magnetic Field due to the Current-Carrying Wire:**
The magnetic field (B_w) created by a long straight wire at a distance r is given by the formula:
\[ B_w = \frac{\mu_0 I}{2 \pi r} \]
where \(\mu_0\) (the permeability of free space) is approximately \(4\pi \times 10^{-7} \, \text{T m/A}\).

3. **Substitute the Values into the Formula:**
\[ B_w = \frac{(4\pi \times 10^{-7} \, \text{T m/A}) \times 30 \, \text{A}}{2 \pi \times 0.02 \, \text{m}} \]
Simplifying this:
\[ B_w = \frac{(4 \times 30 \times 10^{-7})}{2 \times 0.02} \]
\[ B_w = \frac{120 \times 10^{-7}}{0.04} = 3 \times 10^{-4} \, \text{T} \]

4. **Determine the Resultant Magnetic Field:**
Since the external magnetic field is parallel to the current, the two magnetic fields (B_w and B_ext) are perpendicular to each other. We can find the resultant magnetic field (B_r) using the Pythagorean theorem:
\[ B_r = \sqrt{B_w^2 + B_{ext}^2} \]
Substituting the values:
\[ B_r = \sqrt{(3 \times 10^{-4})^2 + (4 \times 10^{-4})^2} \]
\[ B_r = \sqrt{(9 \times 10^{-8}) + (16 \times 10^{-8})} \]
\[ B_r = \sqrt{25 \times 10^{-8}} = 5 \times 10^{-4} \, \text{T} \]

5. **Final Result:**
The magnitude of the resultant magnetic induction at the point 2.0 cm away from the wire is:
\[ B_r = 5 \times 10^{-4} \, \text{T} \]