Biot-savart Law Form

Hardphysics

Differential magnetic field dB due to current element Idl at position vector r (angle θ)?

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dB = μ0 I dl sinθ /(4π r²)

dB = μ0 I dl cosθ /(4π r)

dB = μ0 I dl /(2π r²)

dB = μ0 I dl sinθ /(4π r)

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Subject: physics
Chapter: magnetic effects and magnetism
Topic: biot-savart law form
Difficulty: Hard
Year: 2025

Solution

Biot-Savart Law (Vector Form): $d B = \frac{μ _{0}}{4 π} \frac{I ( d l \times r ^ )}{r ^{2}}$ .
Magnitude Extraction: Using the definition of a cross product ( $∣ A \times B ∣ = A B sin θ$ ):
- $∣ d l \times \overset{r}{^} ∣ = d l \times 1 \times sin θ$ .
Resulting Magnitude: $d B = \frac{μ _{0}}{4 π} \frac{I d l s i n θ}{r ^{2}}$ .
Conclusion: The differential magnetic field follows an inverse square law and depends on the sine of the angle between the current element and the position vector.

This hard difficulty physics question is from the chapter magnetic effects and magnetism, covering the topic of biot-savart law form. It appeared in the 2025 exam.

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Biot-savart Law Form

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