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Moving Charges and Magnetism - Biot-Savart Law

Grade 12CBSEPhysics

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

🔑Concepts

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The Biot-Savart Law is a fundamental law of electromagnetism that describes the magnetic field dB⃗d\vec{B} produced by a small current-carrying element Idl⃗I d\vec{l}.

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The magnetic field dB⃗d\vec{B} is proportional to the current II, the length of the element dldl, the sine of the angle θ\theta between the element and the separation vector, and inversely proportional to the square of the distance rr.

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The direction of dB⃗d\vec{B} is perpendicular to both dl⃗d\vec{l} and the displacement vector r⃗\vec{r}, following the Right-Hand Thumb Rule or the vector cross product rule: dl⃗×r⃗d\vec{l} \times \vec{r}.

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The law is analogous to Coulomb's Law in electrostatics, but while Coulomb's Law deals with scalar charges, Biot-Savart Law deals with vector current elements Idl⃗I d\vec{l}.

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Magnetic field at the center of a circular current-carrying loop is given by summing the contributions from all dl⃗d\vec{l} elements, where θ=90∘\theta = 90^\circ for every point on the circumference.

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The constant of proportionality in SI units is μ04π\frac{\mu_0}{4\pi}, where μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A is the permeability of free space.

📐Formulae

dB⃗=μ04πI(dl⃗×r^)r2d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \hat{r})}{r^2}

dB=μ04πIdlsin⁡θr2dB = \frac{\mu_0}{4\pi} \frac{I dl \sin \theta}{r^2}

Bcenter=μ0I2RB_{center} = \frac{\mu_0 I}{2R}

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

Baxis≈μ0IR22x3 (for x≫R)B_{axis} \approx \frac{\mu_0 I R^2}{2x^3} \text{ (for } x \gg R \text{)}

💡Examples

Problem 1:

A circular coil of wire consisting of 100100 turns, each of radius 8.0 cm8.0 \, cm, carries a current of 0.40 A0.40 \, A. What is the magnitude of the magnetic field BB at the center of the coil?

Solution:

B=μ0NI2R=4π×10−7×100×0.402×0.08=3.14×10−4 TB = \frac{\mu_0 N I}{2R} = \frac{4\pi \times 10^{-7} \times 100 \times 0.40}{2 \times 0.08} = 3.14 \times 10^{-4} \, T

Explanation:

We use the formula for the magnetic field at the center of a circular coil with NN turns. Given N=100N=100, I=0.40 AI=0.40 \, A, and R=0.08 mR=0.08 \, m. Substituting these values into B=μ0NI2RB = \frac{\mu_0 N I}{2R} gives the result.

Problem 2:

Consider a circular loop of radius RR. At what distance xx from the center of the loop on its axis is the magnetic field 18\frac{1}{8} of its value at the center?

Solution:

BaxisBcenter=18  ⟹  μ0IR2/2(R2+x2)3/2μ0I/2R=18\frac{B_{axis}}{B_{center}} = \frac{1}{8} \implies \frac{\mu_0 I R^2 / 2(R^2 + x^2)^{3/2}}{\mu_0 I / 2R} = \frac{1}{8} R3(R2+x2)3/2=18  ⟹  RR2+x2=12\frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{8} \implies \frac{R}{\sqrt{R^2 + x^2}} = \frac{1}{2} 2R=R2+x2  ⟹  4R2=R2+x2  ⟹  x=3R2R = \sqrt{R^2 + x^2} \implies 4R^2 = R^2 + x^2 \implies x = \sqrt{3}R

Explanation:

We set the ratio of the axial field formula to the center field formula equal to 1/81/8. Solving the resulting algebraic equation for xx gives the distance in terms of the radius RR.