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A wire carrying current 12 A is bent into a semicircular arc of radius 2.0 cm. The magnetic field at the centre due to the arc is half that of a full loop. Its magnitude is:
A0.95 × 10⁻⁴ T
B3.8 × 10⁻⁴ T
C1.9 × 10⁻⁴ T
D9.5 × 10⁻⁴ T
Answer & Solution
Correct answer: C. 1.9 × 10⁻⁴ T
1. Full-loop centre field: $B_{loop} = \dfrac{\mu_0 I}{2R}$.
2. A semicircle gives half this: $B = \dfrac{\mu_0 I}{4R}$.
3. Substitute $\mu_0 = 4\pi\times10^{-7}$, $I = 12$, $R = 0.02$ m: $B = \dfrac{4\pi\times10^{-7}\times12}{4\times0.02}$.
4. Numerator $= 1.508\times10^{-5}$; denominator $= 0.08$.
5. $B = 1.9\times10^{-4}$ T (the straight segments contribute nothing since dl ∥ r).
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.147_
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