A coil having N turns is wound tightly in the form of spiral with inner and outer radii a and b respectively when a current I passes through the coil. The magnetic field at the centre is

Question

A coil having N turns is wound tightly in the form of spiral with inner and outer radii a and b respectively when a current I passes through the coil. The magnetic field at the centre is (A) (μ 0 N I/b - a) (B) (μ0 N I/2 π(b-a)) (C) (μ0 N I/2(b-a)) ln (b/a) (D) (μ 0 N I/b - a)ln(b/a)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ua0mpyxy5nbd1ls2.jpg" /> Let d N be the number of turns in width d r. beginarrayl d N=(N/b-a) d r d B=(μ0 d N i/2 r)=(μ0 N i/2(b-a)) d r B=∫ab (μ0 N i/2(b-a)) (d r/r)=(μ0 N i/2(b-a)) ln (b/a) endarray

Q. A coil having $N$ turns is wound tightly in the form of spiral with inner and outer radii $a$ and $b$ respectively when a current $I$ passes through the coil. The magnetic field at the centre is

$\frac{μ _{0} N I}{b - a}$

$\frac{μ _{0} N I}{2 π ( b - a )}$

$\frac{μ _{0} N I}{2 ( b - a )} ln \frac{b}{a}$

$\frac{μ _{0} N I}{b - a} l n \frac{b}{a}$

Let $d N$ be the number of turns in width $d r$ .
$d N = \frac{N}{b - a} d r d B = \frac{μ _{0} d N i}{2 r} = \frac{μ _{0} N i}{2 ( b - a )} d r B = \int_{a}^{b} \frac{μ _{0} N i}{2 ( b - a )} \frac{d r}{r} = \frac{μ _{0} N i}{2 ( b - a )} ln \frac{b}{a}$

Q. A coil having N turns is wound tightly in the form of spiral with inner and outer radii a and b respectively when a current I passes through the coil. The magnetic field at the centre is

b−aμ0​NI​

2π(b−a)μ0​NI​

2(b−a)μ0​NI​lnab​

b−aμ0​NI​lnab​