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An alternating sinusoidal current of frequency omega=1000s^(-1) flows in the winding of a straight solenoid whose cross - sectional radius is equal to R=6.0 cm. Find the ratio of peak values of electric and magnetic energies within the solenoid. |
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Answer» Solution :Here `I = I_(m) COS omegat`, then the peak magnetic energy is `W_(m) = (1)/(2)LI_(m)^(2) = (1)/(2)mu_(0)n^(2)I_(m)^(2)piR^(2)d` Changing magntic field induces an electric field which by Fararday's law is given by `E.2pi R =- (d)/(dt) int oversetrarr(B).d oversetrarr(S) = pir^(2) mu_(0)nI_(m)omega SIN omega t` `E = (1)/(2)rm_(0)nI_(m)omega sin omega t` The associated peak electric energy is `W_(e) = int (1)/(2)epsilon_(0)E^(@)d^(3)r = (1)/(8)epsilon_(0)mu_(0)^(2)n^(2)I_(m)^(2)omega^(2)sin^(2)omega xx (piR^(4))/(2)d` Hence `(W_(e))/(W_(m)) = (1)/(8) epsilon_(0)mu_(0) (omegaR)^(2) = (1)/(8)((omegaR)/(c))^(2)` Again we expect the results to be valid if and only if `((omegaR)/(c)) lt lt 1` |
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