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Physics-Biot-Savart Law

NEET Physics Biot-Savart Law MCQ Question

Type: MCQ-numerical-Hard-Class 12

Using the Biot-Savart Law, calculate the magnitude of the magnetic field at a point located 0.5 meters away from a straight wire carrying a current of 10 A, if the angle between the current element and the position vector is 90 degrees. Assume the permeability of free space μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}.

A

4×10−8 T4 \times 10^{-8} \text{ T}

B

2×10−7 T2 \times 10^{-7} \text{ T}

C

1×10−6 T1 \times 10^{-6} \text{ T}

D

5×10−8 T5 \times 10^{-8} \text{ T}

Correct Answer

Option A

Detailed Explanation

The Biot-Savart Law gives B=μ04π⋅2Ir⋅sin⁡θB = \frac{\mu_0}{4\pi} \cdot \frac{2I}{r} \cdot \sin \theta. Substituting I=10 A,r=0.5 m,sin⁡θ=1I = 10 \text{ A}, r = 0.5 \text{ m}, \sin \theta = 1, we get B=4×10−8 TB = 4 \times 10^{-8} \text{ T}.

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