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Derive an expression for the magnetic field at a point on the axis of a current carrying circulat loop. |
Answer» Solution :Expression for the magnetic field at a point on the axis of a current carrying circular loop :  (1) Consider 'O' is the centre of a circular coil of one turn and radius 'a'. (2) Let P is a point at a distance r from the centre, along the axis of coil. (3) The plane of the coil is `_|_^(r)` to the plane of paper. (4) Consider two ELEMENTS AB and A'B' each of length dl which are diameterically OPPOSITE. (5) Then, the magnetic fields at P due to these two elements will be DB and dB in the direction PM and PN respectively. (6) These directions are `_|_^(r)` to the lines joining the mid-points of the elements with the point P. (7) Resolve these fields into two components parallel `(dB sin theta)` and perpendicular `(dB cos theta)` to the axis of the coil. (8) The `dB cos theta` components cancel one another and `dB sin theta` components are in the same direction and add up due to the symmetric elements of the circular coil. (9) Therefore, the total magnetic field along the axis `= B = int dB sin theta` of the circular coil along PC - (I) (10) The magnetic field at 'P' due to current element of length 'dl' is `'dB' = (mu_(0))/(4pi)(I dl sin phi)/((a^(2)+ x^(2))) = (mu_(0))/(4pi) (I dl)/((a^(2) + r^(2))) - (II) [because phi = 90^(@)]` (11) From equations (I) and (II), `B = int(mu_(0))/(4pi) (I dl)/((a^(2) + r^(2))) sin theta` From `Delta^(l E)` OPE, `sin theta = (a)/(sqrt(a^(2)+r^(2)))` `RARR B = int(mu_(0))/(4pi) (I dl a)/((a^(2)+r^(2))^(3//2)) = (mu_(0) I a)/(4pi(a^(2)+r^(2))^(3//2)) int dl` But `int dl` = circumference of the coil `= 2pi a` `:. B = (mu_(0) I a)/(4pi(a^(2) + r^(2))^(3//2)) xx 2pi a = (mu_(0) I a^(2))/(2(a^(2) + r^(2))^(3//2))` (12) If the coil contains N turns, then `B = (mu_(0) NI a^(2))/(2(a^(2)+r^(2))^(3//2))` (13) AT the centre of the coil `r = 0, B = (mu_(0) NIa^(2))/(2a^(3)) = (mu_(0) NI_(A))/(2pi r^(3))` |
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