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A long straight wire carries a current of 35 A. The magnitude of the magnetic field at a point 20 cm from the wire is:

A1.75 × 10⁻⁵ T
B7.0 × 10⁻⁵ T
C3.5 × 10⁻⁵ T
D3.5 × 10⁻⁴ T
Answer & Solution
Correct answer: C. 3.5 × 10⁻⁵ T
1. Field of a long straight wire: $B = \dfrac{\mu_0 I}{2\pi r}$. 2. Write $\dfrac{\mu_0}{2\pi} = 2\times10^{-7}$ T m A⁻¹. 3. Substitute $I = 35$ A, $r = 0.20$ m: $B = \dfrac{2\times10^{-7}\times35}{0.20}$. 4. Numerator $= 7\times10^{-6}$; dividing by 0.20 gives $B = 3.5\times10^{-5}$ T. _Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", p.158_
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