MarksRiser
MarksRiser
Back to Practice
Physics-Biot-Savart Law

NEET Physics Biot-Savart Law MCQ Question

Type: MCQ-diagram based-Medium-Class 12

Consider a diagram showing a current-carrying straight wire and a point at a perpendicular distance 'r' from the wire where the magnetic field is measured. Which of the following expressions correctly represents the magnitude of the magnetic field 'dB' produced at that point as per the Biot-Savart Law?

Question diagram
A

dB=μ04π⋅I⋅dl⋅sin⁡θr2dB = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot dl \cdot \sin \theta}{r^2}

B

dB=μ02π⋅IrdB = \frac{\mu_0}{2\pi} \cdot \frac{I}{r}

C

dB=μ04π⋅I⋅dlrdB = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot dl}{r}

D

dB=μ02π⋅I⋅dlr2dB = \frac{\mu_0}{2\pi} \cdot \frac{I \cdot dl}{r^2}

Correct Answer

Option A

Detailed Explanation

The Biot-Savart Law states that the magnetic field 'dB' at a point due to a current element 'I dl' is given by dB=μ04π⋅I⋅dl⋅sin⁡θr2dB = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot dl \cdot \sin \theta}{r^2}, where θ\theta is the angle between 'dl' and 'r'.

Found an issue with this question?

Related Questions

Explore More Questions

All Physics QuestionsMoving Charges and Magnetism Questions