Aman Bansal
Last Activity: 13 Years ago
dear ayush
We Will Use Superposition in this problem. You have an outer hollow conductor. Within the center of that you have an inner solid conductor .
Field due to the outer hollow conductor carrying uniform axial current is 0 in the center. To convince yourself look at tangential and radial components within the hollow conductor center:
First tangential component. Integral Htangential dot dL = Ienclosed. Draw the countour of the integral as a circle. There is no Ienclosed (remember we are looking only at field from outer conductor)
Look at radial component:
By symmetry, the radial flux inside the conductor must be uniform (doesn't depend on angle theta). But flux must flow in loops, and the radial symmetry doesn't allow any return path for flux. So radial flux must be 0.
Above we have shown the outer hollow conductor creates no flux in it's center. So the flux that the inner conductor is exposed to is only the flux that it creates.... which is a simple textbook problem...
thus you can easily calculate
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AMAN BANSAL