Using Biot-Savart’s law, derive an expression for magnetic field at any point on axial line of a current carrying circular loop. Hence, find magnitude of magnetic field intensity at the centre of circular coil.
Answers
The magnitude of magnetic field intensity at the centre of circular coil is ∣B0| = μ0(l/2a)
Explanation:
According to Biot-savart’s law magnetic field due to a current element is given by
dB→ = μ0 / 4π x IdlXx¯ r^2
Where r = √x ^2+a^2
dB = μ0 / 4π x Idlsin(90∘) / x^2 + e^2
And direction of dB−→ is ⊥ to the plane containing Idl→ and r→.
Resolving dB−→ along the x-axis and y-axis.
- dBx = dB sinθ
- dBy = dB cosθ
Taking the contribution of whole current loop we get
Bx∮Bx = ∮B.sinθ = ∫μ0 / 4π x 1dl / x^2+a^2 a / √x&^2+a^2
Bx = μ0 / 4π x 1n(x^2+a^2)^E/Z ∮dl = μ0 / 4π x ln×2πa / (x^2+a^2)^E/Z
And By=∮ dBy = ∮dBcosθ = 0
Bp = √ B^2x+B^2y = Bx = μ0 / 4π x 2IA / (x^2+a^2)^E/Z
Bp−→ = μ0 / 4π x 2m / (x^2+a^2)(∵m→ = IA)
For centre x = 0
∣B0|−→ = μ0 / 4π x 2Iπ / a^2a^2 = μ0(l/2a) in the direction of m→
Thus the magnitude of magnetic field intensity at the centre of circular coil is ∣B0| = μ0(l/2a)
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