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For a circular coil of radius R and N turns carrying current. Prove that the magnitude of the magnetic field at a point on its axis at a distance X from its centre is given by B=(mu_0IR^2N)/(2(x^2+R^2)^(3//2)) |
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Answer» Solution :(a) We have, `B=(mu_(0)NIR^(2))/(2(x^(2)+R^(2))^(3/2))` For finding MAGNETIC field on the centre of above circular coil we should take x = 0 and then, `B_("centre")=(mu_(0)NIR^(2))/(2R^(3))=(mu_(0)NI)/(2R)` Above is also the magnetic field at the centre of current carrying extremely thin circular ring. (b)  (1) Above is the figure as per the statement. (2) Here, `PP_(1)=PP_(2)=DeltaxltltltltR` (3) At point `P_(1)`, MAGNITUDE field due to coil-1 is, `B_(1)=(mu_(0)NIR^(2))/(2(x_(1)^(2)+R^(2))^(3/2))` `thereforeB_(1)=(mu_(0)NIR^(2))/(2{(R/2-Deltax)^(2)+R^(2)}^(3/2))""...(1)` `B_(1)=(mu_(0)NIR^(2))/(2{R^(2)/4+(Deltax)R+(Deltax)^(2)+R^(2)}^(3/2))` (4) Now, since `Deltax` is very small, neglecting `(Deltax)Rand(Deltax)^(2)` we get, `B_(1)~~(mu_(0)NIR^(2))/(2(5/4R^(2))^(3/2))=(mu_(0)NI)/(2(5/4)^(3/2)R)""...(2)` (5) At point `P_(1)`, magnetic field due to coil-2 is, `B_(2)=(mu_(0)NIR^(2))/(2(x_(2)^(2)+R^(2))^(3/2))` `thereforeB_(2)=(mu_(0)NIR^(2))/(2{(R/2+Deltax)^(2)+R^(2)}^(3/2))""...(3)` `thereforeB_(2)=(mu_(0)NIR^(2))/(2{R^(2)/4+(Deltax)R+(Deltax)^(2)+R^(2)}^(3/2))` (6) Now since `Deltax` is very small, neglecting `(Deltax)Rand(Deltax)^(2)` we get, `B_(2)~~(mu_(0)NIR^(2))/(2(5/4R^(2))^(3/2))=(mu_(0)NI)/(2(5/4)^(3/2)R)""...(4)` (7) Now, since `vecB_(1)||vecB_(2)`, resultant magnetic field at point `P_(1)` will be, `B=B_(1)+B_(2)` = `(mu_(0)NI)/((5/4)^(3/2)R)` = `(4/5)^(3/2)(mu_(0)NI)/R` = `=0.715(mu_(0)NI)/R` `thereforeB~~0.72((mu_(0)NI)/R)""...(5)` (8) Similarly if we find out resultant magnetic field at point `P_(2)`, in above figure, we get it same as given be equation (5). THUS in above situation in the length `P_(1)P_(2)=2(Deltax)` along the common axis around centre point P, magnetic field remains uniform whose constant magnitude is equal to `0.72((mu_(0)NI)/R)`. Such a PAIR of coil (shown in the diagram) is called "Helmholtz coils". |
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