a current I is flowing in a conductor placed along the x axis as shown in the figure find the magnitude and direction of the magnetic field due to small current element dl lying at the origin at points (0,d,0) and 0,0,d

To solve this, we apply the **Biot–Savart Law**, which gives the magnetic field **d𝐁** due to a small current element:

Biot–Savart Law:

d𝐁 = (μ₀ / 4π) × (I d𝐥 × 𝐫̂) / r²

• I: current
• d𝐥: vector length of current element
• 𝐫: position vector from the current element to the field point
• r: magnitude of 𝐫
• 𝐫̂: unit vector in direction of 𝐫

Case 1: Point at (0, d, 0)

Let’s define:

• Current element d𝐥 = dx î (since wire lies along x-axis)
• Position vector 𝐫 = 𝐫_P – 𝐫_source = (0, d, 0) – (0, 0, 0) = d ĵ

Now compute d𝐥 × 𝐫:

d𝐥 × 𝐫 = dx î × d ĵ = dx·d (î × ĵ) = dx·d k̂

So the magnetic field at (0, d, 0):

d𝐁 = (μ₀ / 4π) × (I dx·d k̂) / d² = (μ₀ I dx) / (4π d) k̂

Result:

Magnitude: (μ₀ I dx) / (4π d)
Direction: Along the positive z-axis (**k̂**)

Case 2: Point at (0, 0, d)

Now, position vector 𝐫 = d k̂

d𝐥 × 𝐫 = dx î × d k̂ = dx·d (î × k̂) = dx·d (–ĵ)

So the magnetic field at (0, 0, d):

d𝐁 = (μ₀ / 4π) × (–I dx·d ĵ) / d² = –(μ₀ I dx) / (4π d) ĵ

Result:

Magnitude: (μ₀ I dx) / (4π d)
Direction: Along the negative y-axis (**–ĵ**)

Final Summary: