A current I flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R. The magnitude of

Q: A current $I$ flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius $R$. The magnitude of the magnetic induction along its axis is:

d Current in a small element, $d I=\frac{d \theta}{\pi} I$ Magnetic field due to the element $d B=\frac{\mu_{0} 2 d I}{4 \pi R}$ The component $d B \cos \theta,$ of the field is cancelled by another opposite component. Therefore, $B_{n e t}=\int d B \sin \theta=\frac{\mu_{0} I}{2 \pi^{2} R_{0}} \int_{0}^{\pi} \sin \theta d \theta=\frac{\mu_{0} I}{\pi^{2} R}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A current $I$ flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius $R$. The magnitude of the magnetic induction along its axis is:

A$\frac{{{\mu _0}I}}{{2{\pi ^2}R}}$
B$\frac{{{\mu _0}I}}{{2\pi R}}$
C$\frac{{{\mu _0}I}}{{4{\pi ^2}R}}$
D$\frac{{{\mu _0}I}}{{{\pi ^2}R}}$

AIEEE 2011, Diffcult

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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