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Moving Charges and Magnetism - Magnetic Field on the Axis of a Circular Current Loop

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Consider a circular loop of radius RR carrying a steady current II. We calculate the magnetic field B⃗\vec{B} at a point PP on its axis at a distance xx from the center OO.

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According to the Biot-Savart Law, the magnetic field dB⃗d\vec{B} due to a small current element dldl is given by dB=μ04πIdlsin⁡θr2dB = \frac{\mu_0}{4\pi} \frac{I dl \sin \theta}{r^2}. In this geometry, the angle between dldl and rr is always 90∘90^\circ.

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The magnetic field dB⃗d\vec{B} at point PP can be resolved into two components: dBcos⁡ϕdB \cos \phi (perpendicular to the axis) and dBsin⁡ϕdB \sin \phi (along the axis).

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Due to the axial symmetry of the loop, the perpendicular components dBcos⁡ϕdB \cos \phi from diametrically opposite elements cancel each other out.

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The total magnetic field BB is obtained by integrating the axial components: B=∫dBsin⁡ϕB = \int dB \sin \phi.

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The direction of the magnetic field is along the axis of the loop. If the current is clockwise (as seen from PP), the field is away from the loop; if counter-clockwise, it is towards the loop (Right-Hand Thumb Rule).

📐Formulae

B=μ0IR22(R2+x2)3/2B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}

Bcenter=μ0I2R (At the center, where x=0)B_{center} = \frac{\mu_0 I}{2R} \text{ (At the center, where } x = 0\text{)}

BN=μ0NIR22(R2+x2)3/2 (For a coil with N turns)B_{N} = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} \text{ (For a coil with } N \text{ turns)}

B≈μ0IR22x3 (For points far away on the axis, x≫R)B \approx \frac{\mu_0 I R^2}{2x^3} \text{ (For points far away on the axis, } x \gg R\text{)}

B=μ04π2mx3 (Where m=IA=I(πR2) is the magnetic dipole moment)B = \frac{\mu_0}{4\pi} \frac{2m}{x^3} \text{ (Where } m = I A = I (\pi R^2) \text{ is the magnetic dipole moment)}

💡Examples

Problem 1:

A circular coil of wire consisting of 100100 turns, each of radius 8.08.0 cm carries a current of 0.400.40 A. What is the magnitude of the magnetic field B⃗\vec{B} at the center of the coil?

Solution:

Given: N=100N = 100, R=8.0 cm=0.08 mR = 8.0 \text{ cm} = 0.08 \text{ m}, I=0.40 AI = 0.40 \text{ A}. Using the formula for the center of the coil: B=μ0NI2RB = \frac{\mu_0 N I}{2R} B=4π×10−7×100×0.402×0.08B = \frac{4\pi \times 10^{-7} \times 100 \times 0.40}{2 \times 0.08} B=12.56×10−7×400.16B = \frac{12.56 \times 10^{-7} \times 40}{0.16} B=3.14×10−4 TB = 3.14 \times 10^{-4} \text{ T}

Explanation:

The magnetic field at the center is the maximum value for the axis. We use the modified Biot-Savart result for NN turns at x=0x=0.

Problem 2:

Calculate the magnetic field at a point on the axis of a circular loop of radius RR at a distance x=Rx = R from the center, if the field at the center is B0B_0.

Solution:

Field at center: B0=μ0I2RB_0 = \frac{\mu_0 I}{2R}. Field at x=Rx = R: B=μ0IR22(R2+R2)3/2B = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} B=μ0IR22(2R2)3/2=μ0IR22⋅22R3B = \frac{\mu_0 I R^2}{2(2R^2)^{3/2}} = \frac{\mu_0 I R^2}{2 \cdot 2\sqrt{2} R^3} B=μ0I42RB = \frac{\mu_0 I}{4\sqrt{2} R} Comparing with B0B_0: B=B022B = \frac{B_0}{2\sqrt{2}}

Explanation:

This shows how the magnetic field strength drops as we move along the axis away from the center of the loop.