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Obtain an expression for magnetic flux d...

Obtain an expression for magnetic flux density B at the centre of a circular coil of radius R, having N turns and carrying a current I.

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Magnetic field at O due to small element `deltal`.
`deltaB=(mu_(0))/(4pi) ( I deltalsin theta)/(R^(2))`

But `theta=90^(@)`
`:. deltaB=(mu_(0))/(4pi)(I delta l)/(R^(2))`
`vecB` at O due to the whole circular coil is given by
`B int dB= int (mu_(0))/(4pi (i dl)/(R^(2))`
`=(mu_(0))/(4pi)(i)/(R^(2))*2piR=(mu_(0)i)/(2)`
If there are N turns in the coil, then
`B=(mu_(0)Ni)/(2R)`
Alternate method : Consider a circular loop of wire of radius .a. carrying current i. Let P be a point on the axis of the coil, at a distance x from its centre O

The resultant magnetic field at P is given by
`B= int d B sin phi`
But `"sin" phi=(a)/(r)`
`B= int (mu_(0))/(4pi) (i)/(r^(2))xx(a)/(r)dl`
`= int (mu_(0))/(4pi) (ia)/(r^(3))dl`
Now, since the terms `mu_(0)//4pi`. i and a are constants and r has the same value for all the elements of loop. Thus, the term `mu_(0)ia//4pir^(3)` can be factored out of integration.
`:.B=(mu_(0))/(4pi)=(ia)/(r^(3)) int dl`
But `int dl =2pi a` (circumference of the loop) and `r=(a^(2)+x^(2))^(1//2)`.
` :. B=(mu_(0))/(4pi)(2pi ia^(2))/((a^(2)+x^(2))^(3//2))`
`=(mu_(0)ia^(2))/(2(a^(2)+x^(2))^(3//2))`
It is a coil of N turns, then each turn will contribute equally to B. Thus,
`B=(mu_(0)Nia^(2))/(2(a^(2)+x^(2))^(3//2))`
The direction of magnetic field `vecB` is along the axis of the loop and of the coil.
At the centre of the coil,
`:. B=(mu_(0)Nia^(2))/(2(a^(2)+0)^(3//2))=(mu_(0)Ni)/(2a)`
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