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For a circular coil of radius R and N turns 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 IR^2 N)/(2 (x^2 + R^2)^(3//2)) (a) Show that this reduces to the familiar result for field at the centre ofthe coil.(b) Consider two parallel co-axial circular coils 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 NI)/(R ) . approximately [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.] |
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Answer» Solution : (b) In a small region of length 2d about the mid-point between the COILS `B = (mu_0 IR^2 N)/(2) xx [ { (R/2 + d)^2 + R^2 }^(-3//2)+ { (R/2- d)^2 + R^2 }^(-3//2)]` `= (mu_0 IR^2 N)/(2) xx ((5R^2)/(4))^(-3//2) xx [ ( 1 + (4d)/(5R)^(-3//2)) + ( 1 - (4d)/(5R) )^(-3//2) ]` ` = (mu_0IR^2N)/(2R^3)xx (4/5)^(3//2) xx [ 1 - (6d)/(5R) + 1 +(6d)/(5R) ]` where in the second and third steps above, TERMS containing `d^2 // R^2`and HIGHER powers of d/R are neglected since `d/R lt lt 1`.The terms linear in d/R cancel giving a UNIFORM field B in a small region: `B = (4/5)^(3//2) (mu_0 IN)/( R) = 0.72 (mu_0 IN)/(R )` |
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