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Applications of Biot-Savart's Law > Magnetic Field at the Centre of a Circular Current-carrying Loop

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Estimated time: 8 minutes
CISCE: Class 12

Derivation

Step 1: Consider a circular loop of radius a carrying current I, lying in a plane, with centre C.

Step 2: Take a small current element \[d\vec{l}\] on the loop. The distance from this element to the centre C is r = a, and the angle between \[d\vec{l}\] and r is always 90°.

Step 3: By Biot–Savart's Law, the magnetic field due to this element at C is:

  • dB = \[\frac{\mu_0}{4\pi}\cdot\frac{Idl\sin90°}{a^2}=\frac{\mu_0Idl}{4\pi a^2}\]

Step 4: Every current element contributes a field dB in the same direction (perpendicular to the plane of the loop), so the fields add up directly (no vector resolution needed).

Step 5: Integrating over the entire loop (circumference = 2πa):

  • B = \[\int dB=\frac{\mu_0I}{4\pi a^2}\int dl=\frac{\mu_0I}{4\pi a^2}(2\pi a)\]

Step 6: Simplify:

  • B = \[\frac{\mu_0I}{2a}\]
CISCE: Class 12

Special Cases

Case Formula Notes
Single circular loop (centre) B = \[\frac {μ_0I}{2a}\] Basic result
Coil with N turns (centre) B = \[\frac {μ_0NI}{2a}\] Field scales linearly with turns
Circular arc subtending angle θθ (radians) B = \[\frac {μ_0Iθ}{4πa}\] Reduces to full-loop formula when θ = 2π
On the axis, at distance x from centre B = \[\frac {μ_0Ia^2}{2(a^2+x^2)^{3/2}}\] Maximum at x = 0 (i.e., at the centre)
CISCE: Class 12

Direction of Magnetic Field

  • Determined by the Right-Hand Rule: curl the fingers of the right hand in the direction of current flow; the thumb points in the direction of \[\vec{B}\].
  • The field is always perpendicular to the plane of the loop.
  • Reversing the current direction reverses the field direction.
CISCE: Class 12

Example 1

Given: Along the axis of a circular loop, find the distance x at which the field becomes 1/64 of its value at the centre.

Solution: Set \[\frac {B_{axis}}{B_{centre}}\] = \[\frac {1}{64}\] and solve using the axial-field formula.

Answer: x = a\[\sqrt {15}\].

CISCE: Class 12

Example 2

Given: A circular arc of radius a = 10 cm = 0.1 m subtends an angle of 60° at the centre and carries a current I = 9 A.

Find: Magnetic field at the centre due to the arc.

Solution: Convert angle to radians: θ = 60° = \[\frac {π}{3}\] rad

B = \[\frac {μ_0Iθ}{4πa}\] = \[\frac {(4π×10^{−7})(9)(π/3)}{4π(0.1)}\]
B ≈ 9.42 × 10−6 T (N A−1m−1)

Answer: B ≈ 9.42 × 10−6 T, directed as per the right-hand rule.

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