## Magnetic Induction At A Point

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# 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

= (μ

_{0}/ 4π) i dl sin θ / r

_{2}

From B, draw BC perpendicular to PA. Let ∟OPB = Φ, ∟BPA = d Φ.

Then, BC = dl sin θ = r d Φ

dB = (μ

_{0}/ 4π) i rd Φ/ r

_{2}= (μ

_{0 }/ 4π) I d Φ/ r

In Δ OPB cos Φ = a / r or r = a / cos Φ

. : dB = (μ

_{0}/ 4π) i cos Φ d Φ / a

The direction of d

**B**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 Φ

_{1}ane Φ

_{2}be the angles made by the ends of the wire at P. Then, magnetic induction at P due to the whole conductor is

B = ∫ (μ

_{0}/ 4π) i cos Φ dΦ / a = (μ

_{0}/ 4π) i / a [sin Φ]

= (μ

_{0}/ 4π) i / a [sin Φ

_{2}– sin (- Φ

_{1})]

= (μ

_{0}/ 4π) i / a [sin Φ

_{2}– sin Φ

_{1}]

If the conductor is indefinitely long, Φ

_{1}= Φ

_{2}= 90.

. : B = (μ

_{0}/ 4π) i / a [1 + 1] = u

_{0}i / 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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