Two parallel rails, 1 and 2, having negligible resistance are $l = 10 c m$ apart. They are connected by a $5 Ω$ resistor. The two metal rods ab and cd can slide on the rails. The rods have resistance $10 Ω$ and $15 Ω$ respectively. The rods are moved with constant velocities of 4m/s and 2m/s parallel to the rails, as shown. A uniform magnetic field B = 10 mT exists in the entire space perpendicular to the figure. The current x mA in the $5 Ω$ resistor then calculate 55 x.

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Detailed Solution

EMF induced in rod ab is $E_{1} = B v_{1} L \sin 60^{0}$

Here L = length of ab

$L \sin 60^{0} = l = 10 c m =$ separation between the rails

$∴ E_{1} = B v_{1} l = 10 \times 10^{- 3} \times 4 \times 0.1 = 4 m V$

The equivalent circuit is as shown. Current distribution is also shown. We will apply Kirchhoff’s loop law for the two loops.

For loop abfe : $5 i + 10 i_{1} = 4$

For loop cdef: $- 15 (i - i_{1}) - 2 - 5 i = 0$

Solving (i) and (ii), gives $i = \frac{8}{55} m A$

[In equations (i) and (ii), the voltages in mV]

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