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Figure shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a. |
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Answer» Solution : (a) Consider the case R > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop, `L = 2pi r ` `I_e`=Current enclosed by the loop = I The result is the familiar expression for a long straight wire `B(2pi r) = mu_0 I` `B = (mu_0 I)/(2pi r)` `B prop 1/r(r gt a)` (b) Consider the case r < a. The Amperian loop is a circle labelled 1. For this loop, taking the radius of the circle to be r, `L = 2pi r ` Now the current enclosed Ie is not I, but is less than this VALUE. Since the current DISTRIBUTION is uniform, the current enclosed is, `I_e = I ( (pi r^2)/(pi a^2)) = (Ir^2)/(a^2)` Using ampere law `B (2pi r) = mu_0 (Ir^2)/(a^2)` `B = ((mu_0 I)/(2a^2)) r` `B prop r ""(r lt a)`  Figure shows a plot of the magnitude of B with distance r from the centre of the wire. The direction of the field is tangential to the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section. This EXAMPLE possesses the required symmetry so that Ampere’s law can be applied readily. |
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