current i is being carried by an infinity wire passing through origin along ditection i^j^k^.find magnetic field due yo wire at a point (1,0,0)

To find the magnetic field due to an infinitely long current-carrying wire in the direction of a vector î + ĵ + k̂ passing through the origin, we’ll use the vector form of the Biot–Savart Law for a straight infinite wire and a bit of vector geometry.

Step-by-step Approach

1. Direction Vector of the Wire

The wire lies along the direction of the vector:

𝐥̂ = (î + ĵ + k̂) / √3

This is the unit vector of the wire’s direction.

2. Point Where We Need Magnetic Field

The observation point is at (1, 0, 0).

3. Use of Formula for Infinite Wire

The magnetic field **B** at a perpendicular distance **r⊥** from an infinite wire is:

B = (μ₀ I) / (2π r⊥)

But B is a vector

Its direction is given by the cross product:

𝐁 ∝ (𝐥̂ × 𝐫̂)

4. Find Perpendicular Distance (r⊥)

We calculate the perpendicular distance from the point (1,0,0) to the line passing through origin along 𝐥̂.

Let 𝐫 = (1,0,0), and 𝐥̂ = (1,1,1)/√3

Perpendicular distance formula from point to a line:

r⊥ = |𝐥̂ × 𝐫| / |𝐥̂|

𝐥̂ × 𝐫 = (1,1,1) × (1,0,0) = (0,1,–1)

|𝐥̂ × 𝐫| = √(0² + 1² + (–1)²) = √2

|𝐥̂| = √(1² + 1² + 1²) = √3

So, r⊥ = √2 / √3 = √(2/3)

5. Compute Magnitude of Magnetic Field

B = (μ₀ I) / (2π r⊥) = (μ₀ I) / (2π √(2/3)) = (μ₀ I √3) / (2π √2)

6. Direction of Magnetic Field

Direction is given by 𝐥̂ × 𝐫̂. We already found 𝐥̂ × 𝐫 = (0,1,–1), which is in the direction of **ĵ – k̂**

Final Answer:

Magnitude: (μ₀ I √3) / (2π √2)
Direction: Along the vector (ĵ – k̂), or more precisely,

𝐁 = [(μ₀ I √3) / (2π √2)] × (ĵ – k̂) / √2 = (μ₀ I √3) / (4π) × (ĵ – k̂)