12. (a) Here, the enclosed electric flux is found by integrating 320012(0.500 V/m s)(2 )123rrEErdrtrdrtrRRπππ⎛⎞Φ==⋅−=−⎜⎟⎝⎠∫∫with SI units understood. Then (after taking the derivative with respect to time) Eq. 32-3 leads to 321(2)rBrrRπεµ=−. For r= 0.0200 m and R= 0.0300 m, this gives B= 3.09 ×10−20 T. (b) The integral shown above will no longer (since now r> R) have ras the upper limit;
This is the end of the preview. Sign up
access the rest of the document.