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(A) Let us consider an elementary ring of radius $r$ and thickness $dr$ in which current $I$ is flowing.The number of turns in this elementary ring is

Vedclass · Accepted Answer

$\frac{\mu_0 N I}{2(b-a)} \ln \left(\frac{b}{a}\right)$

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(A) Let us consider an elementary ring of radius $r$ and thickness $dr$ in which current $I$ is flowing.The number of turns in this elementary ring is $dN = \frac{N}{b-a} dr$.The magnetic field at the center $O$ due to this ring is $dB = \frac{\mu_0 I dN}{2r}$.Substituting $dN$,we get $dB = \frac{\mu_0 I N dr}{2(b-a)r}$.The net magnetic field at the center of the spiral is $B = \int_a^b \frac{\mu_0 I N dr}{2(b-a)r}$.Therefore,$B = \frac{\mu_0 I N}{2(b-a)} \int_a^b \frac{dr}{r}$.Evaluating the integral,$B = \frac{\mu_0 I N}{2(b-a)} [\ln r]_a^b$.Thus,$B = \frac{\mu_0 I N}{2(b-a)} \ln \left(\frac{b}{a}\right)$.

$A$ long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in the $X-Y$ plane and a steady current $I$ flows through the wire. The $Z$-component of the magnetic field at the center of the spiral is

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