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For a circular coil of radius R and N carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by B = (mu_0 I R^2 N)/(2(x^2 + R^2)^(3//2)) (a) Show that this reduces to the familiar result for field at the centre of the coil. (b) Consider two parallel co-axial circular coil of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by , B = 0.72 (mu_0 N I)/(R) , approximately. [Such an arrangment to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.] |
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Answer» Solution :(a) for the centre of the coil putting x = 0 in the given relation, we have `B = (mu_0 I R^2 N)/(2 R^3) = mu_0 (NI)/(2R)` (b) In a small region length 2d about the mid-point between the coils, the total magnetic field due to both current carrying coils has a value `B = (mu_0 I R^2 N)/(2) xx [{(R/2 + d)^2 + R^2}^(-3//2) + {(R/2 - d)^(2) + R^2}^(-3//2)]` As d is very small, neglecting the term CONTAINING `d^2`, we have `B = (mu_0 I R^2 N)/(2) xx ((5R^2)/(4))^(-3//2) xx [(1 + (4d)/(5R))^(-3//2) + (1 - (4d)/(5R))^(-3//2)]` On EXPANDING by binominal theorem and neglecting higher powers of `((4d)/(5R))`, we get `B = (mu_0 I R^2 N)/(2 R^3) xx (4/5)^(+3//2) xx [1 - (6d)/(5R) + 1+ (6d)/(5R)]` `implies B = (4/5)^(3//2) (mu_0 I N)/(R) = 0.72 (mu_0 I N)/(R)`. |
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