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A metallic square loop ABCD of size 15 cm and resistance 1.0Omega is moved at a uniform velocity of v m/s in a uniform magnetic field of 2 T, the field lines being normal to the plane of paper. The loop is connected to an electrical network of resistors, each of resistance 2Omega. Calculate the speed of the loop, for which 2 mA current flows in the loop. |
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Answer» Solution :The network of resistors shown in Fig. 6.48 is equivalent to 3 resistors of `2 + 2 = 4Omega,2Omega and 2 + 2 = 4 omega` respectively joined in parallel. Hence, their combined RESISTANCE `R_(1)` is given by `1/R_(1) = 1/4+1/4+1/2=1/1 impliesR_(1) = 1 Omega` As resistance of loop `R_(2) = 1Omega`, hence total resistance `R = R_(1) + R_(2) = 1 + 1 = 2 Omega`. As length of side AB of loop l = 15 cm = 0.15 m and B = 2 T and direction of motion v is perpendicular to B as WELL as I, hence induced EMF `varepsilon = B l v` and induced current `implies v = (IR)/(Bl) = (2m Axx2Omega)/(2Txx0.15m)=(2 xx10^(-3)xx2)/(2xx0.15)m//sor 0.0133cm s^(-1)`. |
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