A wire carrying current 12 A is bent into a semicircular arc of radius…

Question

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:

Accepted 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 $\mu_0 = 4\pi	imes10^{-7}$, $I = 12$, $R = 0.02$ m: $B = \dfrac{4\pi	imes10^{-7}	imes12}{4	imes0.02}$.
4. Numerator $= 1.508	imes10^{-5}$; denominator $= 0.08$.
5. $B = 1.9	imes10^{-4}$ T (the straight segments contribute nothing since dl ∥ r).
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.147_

Answer

0.95 × 10⁻⁴ T

Answer

3.8 × 10⁻⁴ T

Answer

9.5 × 10⁻⁴ T