# Question c27e7

Jul 3, 2016

(3)

#### Explanation:

Consider an element of thickness $\mathrm{dr}$ at a distance $r$ from the centre of the coil.
The number of turns in this element are dN=N/(b−a)dr

Magnetic field* at the centre of the coil due to this element is given by the expression dB=(μ_0dN.I)/(2r)
Inserting values of $\mathrm{dN}$ we obtain
dB=(μ_0N/(b−a)dr.I)/(2r)
=>dB=(μ_0NI)/(2(b−a)) 1/rdr
Total magnetic field is found by Integrating from limits $a \text{ to } b$

B=int_a^b(μ_0NI)/(2(b−a)) 1/rdr
=>(μ_0NI)/(2(b−a))ln(|r|)|_a^b, ignoring constant of integration as it is proper/definite integral
=>(μ_0NI)/(2(b−a))(lnb-lna)
=>(μ_0NI)/(2(b−a))ln(b/a)#

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*Magnetic field at Center of a coiled wire of radius $R$ carrying a current $I$ can be calculated as below:

Let $\mathrm{dv} e c l$ be the infinitesimal length of the coil, along the circumference, carrying current and
$\mathrm{dh} a t r$ be the unit vector along the radius of the coil.
The magnetic field $\mathrm{dv} e c B$ produced by this element is given by the simplified expression for Biot-Savart Law
$\mathrm{dv} e c B = \frac{{\mu}_{0} I \mathrm{dv} e c l \times \mathrm{dh} a t r}{4 \pi {R}^{2}}$
which simplifies as the angle $= {90}^{\circ}$ all elements along the path and the distance to the centre is constant. The scalar part becomes
$\mathrm{dB} = \frac{{\mu}_{0} I \cdot \mathrm{dl}}{4 \pi {R}^{2}}$

To find the total magnetic field we need to find the line integral of the length which is the circumference $2 \pi R$, due to symmetry. $B = \frac{{\mu}_{0} I}{4 \pi {R}^{2}} \oint \text{ } \mathrm{dl}$
$\implies B = \frac{{\mu}_{0} I}{4 \pi {R}^{2}} 2 \pi R$
$\implies B = \frac{{\mu}_{0} I}{2 R}$
If there are $N$ number of turns, the total current is $N I$ and magnetic field becomes
$B = \frac{{\mu}_{0} N I}{2 R}$