## Magnetic Induction At Any Point

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# Magnetic Induction at any point on the axis of a Solenoid

Let l represent the length of the solenoid and N the total number of turns in its winding. The number of turns per unit length is then N/L. a is the radius of the solenoid. A current i is flowing in the solenoid. The solenoid contains air in its core. Let us find the magnetic induction B at a point P on the axis of the solenoid.

Consider an elementary length dx of the solenoid, at a distance x from P. We can regard this element AB as a circular coil of radius a containing N dx / L turns.

Magnetic induction at P due to the element dx is

dB =

__μ0 i a2__.

__Ndx__. _____

__1________

____2 __L (a

^{2}+ x

^{2})

^{3/2 }

Let us use the angle Φ instead of x as the independent variable. Then,

x = a cot Φ; dx = - a cosec

^{2}Φ dΦ

Substituting these values of x and dx in Eq. (1), we get

dB =

__μ0 i a__.

^{2}__N__.

__a cosec2Φ dΦ__= -

__μ__sin Φ dΦ

_{0}iN___2_ L [a

^{2}+ a

^{2}cot

^{2}Φ]

^{3/2 }2L

The magnetic induction at P due to the entire length of the solenoid is

B = - μ

_{0 }iN ∫ sin Φ dΦ

2L

=

__μ0 iN__[cos α – cos β]

2L

The direction of B is parallel to the axis of the solenoid.

B = μ iN [cos α – cos β]

2L

Where μ = μ

_{0}μ

_{r}and μ = B/H.

## Special cases

(i) At a point well inside a very long solenoid: α = 0, β = 180.. : B = μ

_{0}iN / L

(ii) At an axial point at one end of a long solenoid: α = 0, β = 90.

B = μ

_{0}iN / 2L

Hence the magnetic induction at either end is one-half its magnitude at points well inside the solenoid.

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