A and B are two identical rods of internal resistance r -1 - If R -2 - V A -4 m - s - V B -3 m - s- length of rods is 1 m- then current in R is Magnetic fieA- 1 AB- 0-4 A- 0-6 AD- 0-8 A

The correct option is B 0.4A The voltage in induced in the left rod is BVL = 2×4×1 = 8V. The bottom part of the rod is at higher potential. The voltage in ind ...

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Standard XII

Physics

Motional EMF

A and B are t...

Question

A and B are two identical rods of internal resistance $r = 1 Ω .$ If $R = 2 Ω, V_{A} = 4 m / s, V_{B} = 3 m / s$ , length of rods is 1 m, then current in R is (Magnetic field is 2T)

1 A

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0.4 A

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0.6 A

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0.8 A

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Solution

The correct option is B $0.4 A$

The voltage in induced in the left rod is $B V L = 2 \times 4 \times 1 = 8 V$ . The bottom part of the rod is at higher potential.

The voltage in induced in the rightside rod is $B V L = 2 \times 3 \times 1 = 6 V$ . The top part of the rod is at higher potential.

By K.V.L,
$8 = i + 2 i^{1}$ ...(1)
$6 = (1 - i^{1}) - 2 i^{1}$
$\Rightarrow 6 = i - 3 i^{1}$ ...(2)
By (1) & (2)
$i^{1} = 0.4 A$

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A and B are two identical rods of internal resistance r=1Ω. If R=2Ω,VA=4m/s,VB=3m/s, length of rods is 1 m, then current in R is (Magnetic field is 2T)

A and B are two identical rods of internal resistance $r = 1 Ω .$ If $R = 2 Ω, V_{A} = 4 m / s, V_{B} = 3 m / s$ , length of rods is 1 m, then current in R is (Magnetic field is 2T)