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A copper wire of length 50.0 cm and total resistance of 1.1 xx 10^(-2) Omegais formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 10.0 mT/s. At what rate is thermal energy generated in the loop? |
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Answer» `1.32 xx 10^(-8) W` `R = 1.1 xx 10^(-2) Omega ` `(DB)/(DT) = 10.0 mT//s= 10 xx 10^(-3) T//s` Let r the radius of a circular loop `therefore2pir = I. r = (L)/(2pi) = (50 xx 10^(-2))/(2pi) m = (1)/(4PI) m` Magnetic flux linked with the loop is `phi= BA cos theta = Bpi r^2 cos 0^@= Bpi r^2` Emf induced in the loop is `|epsi| = (dphi)/(dt) = pi r^2(dB)/(dt)` Rate at which thermal energy generated in the loop is `P = (|epsi|^2)/( R) = ((pi r^2)^2 ((dB)/(dt))^2)/( R)` Substituting the GIVEN values ,we GET `P = (pi^2 xx ((1)/(4pi))^4 xx (10 xx 10^(-3))^2)/(1.1 xx 10^(-2) ) = 3.62 xx 10^(-6) W` |
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