A straight long wire carrying a current of 30 A is placed in an external uniform magnetic field of $4 × 10^{-4} T$ parallel to the current. The magnitude of the resultant magnetic field at a point 2 cm away from the wire is

$5 × 10^{-4} T$

The correct answer is Option (1) → $5 × 10^{-4} T$

Given:

$I = 30~\text{A},~ B_\text{ext} = 4 \times 10^{-4}~\text{T},~ r = 0.02~\text{m}$

Magnetic field due to the wire:

$B_\text{wire} = \frac{\mu_0 I}{2 \pi r} = \frac{4 \pi \times 10^{-7} \cdot 30}{2 \pi \cdot 0.02} = 3 \times 10^{-4}~\text{T}$

Since the fields are perpendicular:

$B_\text{res} = \sqrt{B_\text{wire}^2 + B_\text{ext}^2} = \sqrt{(3 \times 10^{-4})^2 + (4 \times 10^{-4})^2} = 5 \times 10^{-4}~\text{T}$

Answer: $B_\text{res} = 5 \times 10^{-4}~\text{T}$