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Figure shows a square loop 100 turns as area of 2.5 xx 10^(-3) m^(2) and a resistance of 100 Omega. The magnetic field has a magnitude B = 0.40 T . The work done in pulling the loop out of the field slowly and uniformly in 1.0 "s is n" xx 10^(-6) then n is |
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Answer» SOLUTION :The side of the square is 1 = `sqrt(2.5xx10^(-3)m^(2))= 0.05` m As it is uniformly PULLED out in 1.0 s the SPEED of the LOOP is V = 0.05 m/s The emf induced in the left arm of the loop is `epsilon Nv B1 = 100 xx(0.05 m//s)xx (0.40T ) xx (0.05m) = 0.1 v ` The current in the loop is `i=(0.1V) /(100 Omega) = 1.0 xx10^(-3)` A The FORCE on the left arm due to the magnetic field is `F=i//B =(1.0 xx10^(3)A) (0.05 m ) (0.40t) = 2.0xx 10^(-5)N` This force is towards left in the figure. To pull the loop uniformly an external force of `2.0xx10^(-5) N` `W = (2.0 xx10^(-5) N) xx(0.05 m) = 1.0 xx10^(-6) J` |
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