Magnetism(dbt)

please explain why ...u seem to be correct answer has log

Let inner radius be r,and outer radius be R; and no of turns be N....Now just consider a ring of radius x,
magnetic field at the centre dB=(μ₀i/2x)(dx/R-r)(N)

i m considering spiral to be made up of small rings(continuous).

Spring constant is given by k = Gd4/[8nD3]

Where:

k = constant, pounds of load per inch of deflection

G = modulus of rigidity of spring material, pounds per square inch

d = wire diameter, inches

n = number of active coils, which is the number of coils subjected to flexure (always less than the total number of coils)

D = mean coil diameter, inches = Outer Diameter - Wire Diameter.

But we ignore it for our simplicity.

However, If any third party clears the doubt it will be nicer!

@shubodip please specify the kind of spiral

you told about the nature of spiral only in #10

So,according to #10
as Told by shubodip , its a logarithmic spiral

r=R\left(2^{\frac{\theta}{4\pi}} \right)

If you notice carefully the dl element is not perpendicular to the radius vector

The figure Shows the angle formed by two curves

Had the curves exactly overlapped then we can use dl x r =dl*r

Since angle is formed ,we have dl x r = dl*r*cos(α)

where α =angle between the tangent of curves at that point

But we can make use of one thing

dl X r=r²dθ θ(cap direction)

my attempt :

Consider the element dl as shown in figure

From Biot-Savart Law
\mathrm{dB=\frac{\mu_0i}{4\pi}}\left(\frac{\vec{dl} \times \hat{r}}{r^2} \right) \\ \mathrm{\vec{dl} \times \hat{r}= rd\theta \ \hat{z}}\texttt{ Consider spiral in x-y plane} \\ \mathrm{\int dB =\frac{\mu_0i}{4\pi}\int_{0}^{4\pi}\frac{d\theta}{r}} \\ \texttt{Putting r}=\mathrm{R2^{\frac{\theta}{4\pi}}} \\ \mathrm{\int dB =\frac{\mu_0i}{4\pi*R}\int_{0}^{4\pi}\frac{d\theta}{2^{\frac{\theta}{4\pi}}}} \\ \boxed{\mathrm{B=\frac{\mu_0i\ln{2}}{R}}}

sorry but ln2 is coming in denominator only [2]