If a rod PQ of length 1.25m is moving with velocity 10ms-1 on parallel tracks placed in uniform magnetic field B0 = 0.2T . If resistance of ABCDA is 100Ω , then current through loop at this instant is

ε=Bvl and i=εRv=10ms−1,l=1m,B=0.2Tε= induced emf ==BlV=0.2×10×1=2 volt induced current in loop i=εR=2100=201000=20mA

If a rod PQ of length 1.25m is moving with velocity 10ms-1 on parallel tracks placed in uniform magnetic field B0 = 0.2T . If resistance of ABCDA is 100Ω , then current through loop at this instant is

ε=Bvl and i=εRv=10ms−1,l=1m,B=0.2Tε= induced emf ==BlV=0.2×10×1=2 volt induced current in loop i=εR=2100=201000=20mA

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If a rod PQ of length 1.25m is moving with velocity 10ms^-1 on parallel tracks placed in uniform magnetic field B₀ = 0.2T . If resistance of ABCDA is 100 $Ω$ , then current through loop at this instant is

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10 mA

1.25 mA

20 mA

40 mA

answer is C.

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Detailed Solution

$\begin{array}{l} ε = Bvl and i = \frac{ε}{R} \\ v = 10 {ms}^{- 1}, l = 1 m, B = 0 .2 T \\ ε = induced emf == BlV = 0 .2 \times 10 \times 1 = 2 volt induced current in loop \\ i = \frac{ε}{R} = \frac{2}{100} = \frac{20}{1000} = 20 mA \end{array}$