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Derive an expression for the axial field of a solenoid of radius ''a'', containing ''n'' turns per unit length and carrying current ''I''. |
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Answer» Solution :(1) Consider a solenoid of length '2l' and radius 'a' having 'n' turns per unit length. (2) Let 'I' be the current in the solenoid. (3) We have to calculate magnetic field at any point P on the axis of solenoid, where OP = r. (4) Consider a small element of thickness DX of the solenoid, at a distance 'x' from 'O'. (5) Number of turns in the element = ndx. (6) Magnitude of magnetic field at P due to this current element is `dB = (mu_(0) ndx Ia^(2))/(2[(r-x)^(2)+a^(2)]^(3//2))`. (7) If P LIES at a very large distance from 0, i.e., `r gt gt a` and `r gt gt x`, then `[(r-x)^(2)+a^(2)]^(3//2) ~~ r^(3)`. `rArr dB = (mu_(0) n dx I a^(2))/(2r^(3))`  (8) To get total magnetic field, integrating the above equation between the limites from `x = -l` to `x = +l`. i.e., `B = UNDERSET(-l)overset(+l)int dB = underset(-l)overset(+l)int (mu_(0) I a^(2) n dx)/(2r^(3)) = (mu_(0) n I a^(2))/(2r^(3))[x]_(-l)^(+l)` `:. B = (mu_(0) n I 2l)/(2) (a^(2))/(r^(3))(2l) = (mu_(0))/(4pi)(2n(2l) I pi a^(2))/(r^(3))` (9) The magnitude of the magnetic moment of the solenoid is, `m = n(2l) I (pi a^(2))`. `:. B = (mu_(0))/(4pi) (2m)/(r^(3))` (10) Therefore, magnetic moment of a bar magnetic is equal to magnetic moment of an equivalent solenoid that produces the same magnetic field. |
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