Magnetic induction at a point due to a Straight Conductor Carrying Current

Consider a straight conductor XY carrying a current i in the direction Y to X. P is a point at a perpendicular distance a from the conductor. Consider an element AB of length dl. Let BP = r and ∟OBP = θ Magnetic induction at P due to the element AB
        = dB
        = (μ₀ / 4π) i dl sin θ / r₂

From B, draw BC perpendicular to PA. Let ∟OPB = Φ, ∟BPA = d Φ.
Then, BC = dl sin θ = r d Φ
    dB = (μ₀ / 4π) i rd Φ/ r₂ = (μ₀ / 4π) I d Φ/ r

In Δ OPB cos Φ = a / r or r = a / cos Φ
. : dB = (μ₀ / 4π) i cos Φ d Φ / a



The direction of dB will be perpendicular to the plane containing dl and r. It will be directed into the page at P as shown by right and rule.
Let Φ₁ ane Φ₂ be the angles made by the ends of the wire at P. Then, magnetic induction at P due to the whole conductor is

B = ∫ (μ₀ / 4π) i cos Φ dΦ / a = (μ₀ / 4π) i / a [sin Φ]
   = (μ₀ / 4π) i / a [sin Φ₂ – sin (- Φ₁)]
   = (μ₀ / 4π) i / a [sin Φ₂ – sin Φ₁]

If the conductor is indefinitely long, Φ₁ = Φ₂ = 90.
. : B = (μ₀ / 4π) i / a [1 + 1] = u₀i / 2π a
Magnitude of B depends on i and a. B ∞ 1 / a.
The lines of B form concentric circles around the wire.



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