Question

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A circular coil of radius R carrying current I at what distance from the centre of the coil the Magnetic field at a point on its axis is equal to the 1/64th times the Magnetic field at the centre of coil ???

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Answer 1

Given:

Magnetic field at distance x from centre is equal to $\frac{1}{64}^{th}$ time of magnetic field at centre .

To Find:

Find the value of distance that is x .

Solution:

We know that Magnetic fied intensity is given by  :

B1=$\frac{(u0/4\pi )\times(2\pi nIa^2)}{(a^2+x^2)^{\frac{3}{2}} }$

Where u0 is permeability of free space= 4$\pi$ x $10^{-7} N/A^{2}$     (Newton per square  Amperes)

I=magnitude of electric current

x=Distance from centre

n=loops per unit length

At centre Magnetic Field Intensity is

B2=  $\frac{(u0/4\pi )\times(2\pi nI)}{a }$

It is given that :

B1=  $\frac{1}{64} \times B2$

so  we have:

$\frac{(u0/4\pi )\times(2\pi nIa^2)}{(a^2+x^2)^{\frac{3}{2}} }$ =$\frac{1}{64} \times$ $\frac{(u0/4\pi )\times(2\pi nI)}{a }$

$\dfrac{a^{2} }{(a^2+x^2)}\times \dfrac{3}{2}$= $\frac{1}{64} \times$ a

Cross multiplying we get :

$a^2\times(64a)=(a^2+x^2)^{3/2}$

$16a^2=a^2+x^2$

We need x so  :

$x=a\times \sqrt(15)$