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A 0.5 m long metal rod PQ completes the circuit as shown in the Fig. 6.47. The area of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the total circuit is 3Omega, calculate the force needed to move the rod in the direction as indicated with a constant speed of 2 m s^(-1). |
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Answer» Solution :When a conductor of length l moves with a speed v in a uniform magnetic field B such that all the three terms are mutually perpendicular to each other, then induced emf `|varepsilon| = B l v` and induced current for a closed circuit `I = varepsilon/R = (B l v )/R`, where R = RESISTANCE of the circuit. Now the force needed to move the ROD in the direction indicated in FIG. 6.47 against the induced BACK current is `F = B I l = (B^(2)l^(2)v)))/R` In present problem, `I = 0.5 m, B = 0.15 T, v = 2ms^(-1) and R = 3Omega`, hence `F = ((0.15)^(2) xx (0.5)^(2) xx 2)/3 = 3.75 xx 10^(-3)N.` |
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