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(A) Consider a small circular element of thickness $dr$ at a distance $r$ from the centre of the spiral as shown in the figure.Total number of turns i

Vedclass · Accepted Answer

$2 \pi \ln (2) 	ext{ mT}$

Vedclass · Answer

(A) Consider a small circular element of thickness $dr$ at a distance $r$ from the centre of the spiral as shown in the figure.Total number of turns in this element is $dN = \frac{N}{b-a} dr$.The current passing through this element is $di = I \cdot dN = \frac{N I}{b-a} dr$.The magnetic field at the centre of the spiral due to this element is given by $dB = \frac{\mu_0 di}{2r} = \frac{\mu_0 N I}{2(b-a)} \frac{dr}{r}$.Integrating this from $r = a$ to $r = b$,the total magnetic field $B$ is:$B = \int_a^b \frac{\mu_0 N I}{2(b-a)} \frac{dr}{r} = \frac{\mu_0 N I}{2(b-a)} \ln \left( \frac{b}{a} ight)$.Given $N = 100$,$I = 1 	ext{ A}$,$a = 1 	ext{ cm} = 10^{-2} 	ext{ m}$,and $b = 2 	ext{ cm} = 2 	imes 10^{-2} 	ext{ m}$.Substituting these values:$B = \frac{4 \pi 	imes 10^{-7} 	imes 100 	imes 1}{2(2 	imes 10^{-2} - 1 	imes 10^{-2})} \ln \left( \frac{2 	imes 10^{-2}}{1 	imes 10^{-2}} ight)$$B = \frac{4 \pi 	imes 10^{-5}}{2 	imes 10^{-2}} \ln(2) = 2 \pi 	imes 10^{-3} \ln(2) 	ext{ T}$.Since $1 	ext{ T} = 10^3 	ext{ mT}$,we get $B = 2 \pi \ln(2) 	ext{ mT}$.

$A$ coil having $100$ turns is wound tightly in the form of a spiral with inner and outer radii $1 \text{ cm}$ and $2 \text{ cm}$,respectively. When a current $1 \text{ A}$ passes through the coil,the magnetic field at the centre of the coil is

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