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Calculate the axial field of a finite solenoid. |
Answer» Solution :Let the solenoid of figure consists of `n` turns PER unit length.  Let its length be `2l` and radius a. We can EVALUATE the axial field at a point P, at a distance r from the centre O of the solenoid. Consider a circular element of thickness dx of the solenoid at a distance x from the centre. It consists of n dx turns. Let I be the current in the solenoid. According to formula for the magnitude of magnetic field at axis of coil of N turn, number of tum `N= n` dx and taking distance from O to P `= (r-x)`. `B= ( mu_(0) NIa^(2) )/( 2[ (r-x)^(2) + a^(2) ]^(3//2) )` `therefore B= (mu_(0) n dx Ia^(2) )/(2[ (r-x)^(2) +a^(2) ]^(3//2))` Total field is obtained by summing over all the elements that is by INTEGRATING from `x=-l` to `x=+l`. Then the denominator is approximately by, `[(r-x)^(2) +a^(2) ]^(3//2) ~~ r^(3) [ because l` means `x` and a is neglected to compare with r ] `therefore B= (mu_(0) n Ia^(2) )/( 2r^(3) ) int_(-l)^(l) dx` `= (mu_(0) nIa^(2) )/( 2r^(3) ) [l-(-l)]` `= (mu_(0) nIa^(2) )/( 2r^(3) ) [2l]` Arranging the terms appropriately, `B = (mu_(0) )/( 2r^(3) ) [ (n2l) (Ia^(2) )]` multiply and dividing by `pi` to right side term, `B= (mu_0)/( 2r^3) [( (n2l)(I pi a^(2) ))/( pi)]` Here `(n2l)` are the total turns of the solenoid. `pia^(2) = A` is the cross-section area of one side of solenoid. `B = (mu_0)/( 2r^3) [ (NIA)/( pi) ]` Here, NIA = dipole moment of solenoid, `B= (mu_0)/(2pi ) (m)/(r^3)` `therefore B=2 (mu_0)/( 4pi) (m)/( r^3)`. Is the magnetic field at the axis of solenoid. This also similar the far axial magnetic field of a bar magnet. Thus, a bar magnet and a solenoid produce similar magnetic field. |
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