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A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be |
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Answer» Current is induced in the short - circuited coil due to the imposed time-varying magnetic field. Power `P = (e^(2))/(R )` , Here `e = -(d phi)/(dt)` where `phi = NBA` and `R = (rho l)/(pi r^(2))` where `l` and `r` length and radius of the wire. `:. P = (pi r^(2))/(rho l)[(d)/(dt)NBA]^(2)` or `P = (pi r^(2))/(rho l) N^(2)A^(2)((dB)/(dt))^(2)` or `P =("constant") (N^(2)r^(2))/(l)` when `r_(2) = (r_(1))/(2)` then `t_(2) = 4l_(1)` `:. (P_(2))/(P_(1)) = ((4N)^(2))/(N^(2)) xx ((r)/(2r))^(2) xx ((l)/(4l))` `:. (P_(2))/(P_(1)) = (16 N^(2) xx r^(2) xx l)/(N^(2) xx 4r^(2) xx 4l)` or `(P_(2))/(P_(1)) = (1)/(1)` `:.` Power dissipated is the same. |
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