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An electric current is flowing in a circular coil of radius a. at what distance from the centre of the axis of the coil will the magnetic field be 1/8th of

its value at the centre.



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ΜThe point P is on the axis of the circular coil, I suppose.  Let the center of the coil be O.  Let OP be equal to x.  Radius is a.  Current is I.  Let  dL be a small element of the coil.  So  r = distance between dL and P = √(a²+x²).

Let the angle made by the line joining dL and P with the axis  be θ.  Here the angle θ is same for all elements on the coil.  Also r is also same.  The angle between the vectors dL and r is 90 deg. always.  Let us say that the axis of the coil is x axis.

Cos θ = a / √(a² + x²) = a/r

Using Biot Savarts law: 
\vec{B}=\frac{\mu_0\ I}{4 \pi} \int\limits^{}_{} {\frac{\vec{dL} \times \vec{r}}{r^3}} \, {} \\\\B_x=\frac{\mu_0 I}{4 \pi}  \int\limits^{2\pi a}_{0} { \frac{\ Sin \frac{\pi}{2}\ Cos\theta}{r^2}} \, dL \\\\=\frac{\mu_0 I}{4 \pi} [2 \pi a\ cos \theta/r^2]\\\\B_x=\frac{\mu_0 I\ a^2}{2 (a^2+x^2)^{\frac{3}{2}}}\\\\B_x=\frac{\mu_0 I}{2a}cos^3\theta

The y and z components of the magnetic field induction By and Bz are zero, as the components along these axes will be cancelled by diametrically opposite elements.

B₀ at the center of the coil when x = 0, is:    μ₀ I / (2 a)
Bx at a distance x on the axis will be  B₀ / 8 when Cos θ = 1/2

ie.,  Cos² θ = 1/4 = a² / (a² + x²)
   =>  x² = 3 a²       
   =>  x = √3 a

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