SATHEE: Magnetics 2 Question 12

Magnetics 2 Question 12

12. A coil having $N$ turns is wound tightly in the form of a 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

(2001, 2M)

(a) $\frac{μ_{0} N I}{b}$

(b) $\frac{2 μ_{0} N I}{a}$

(d) $\frac{μ_{0} I^{N}}{2 (b - a)} \ln \frac{b}{a}$

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Answer:

Correct Answer: 12. (c)

Solution:

Consider an element of thickness $d r$ at a distance $r$ from the centre. The number of turns in this element,

$d N = \frac{N}{b - a} d r$

Magnetic field due to this element at the centre of the coil will be

$\begin{aligned} d B & = \frac{μ_{0} (d N) I}{2 r} = \frac{μ_{0} I}{2} \frac{N}{b - a} \cdot \frac{d r}{r} \\ ∴ B & = \int_{r = a}^{r = b} d B = \frac{μ_{0} N I}{2 (b - a)} \ln \frac{b}{a} \end{aligned}$

NOTE The idea of this question is taken from question number 3.245 of IE Irodov.

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Magnetics 2 Question 12

12. A coil having N turns is wound tightly in the form of a 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

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12. A coil having $N$ turns is wound tightly in the form of a 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