A pair of parallel conducting rails lie at right angle to a uniform magnetic field of `2.0T` as shown in the fig. two resistor `10Omega` and `5Omega` are to slide without friction along the rail. The distance between the conducting rails is `0.1m`. Then

Options

A. Induced current `=(1)/(150)A` directed clockwise if `10Omega`
resistor is pulled to the right with speed `0.5ms^(-1)` and resistor is held fixed
B. Induced current `=(1)/(300)A` directed anti-clockwise if
`10Omega` resistor is pulled to the right with speed `0.5ms^(-1)` and `10Omega` resistor is held fixed
C. Induced current `=(1)/(300)A` directed clockwise if `5Omega`
resistor is pulled to the left at `0.5ms^(-1)` and `10Omega` resistor is held at res
D. Induced current `=(1)/(150)A` directed anti-clockwise if`5Omega`
resistor is pulled to the left at `0.5ms^(-1)` and `10Omega` resistor is held at rest

Answer

D

Explanation

Induced current `=(10)/(150)A` directed anti-clockwise `5Omega` resistor is pulled to the left at `0.5ms^(-1)` and `10Omega` resistor is held at rest(d) when `5Omega` resistor is pulled left at `0.5 m//sec` induced emf., in the said resistor `=e=vBl=0.5xx2xx0.1=0.1V`.
Resistor `10Omega` is at rest so induced emf in it `(e=vBl)` be zero.
Now net emf in the circuit`=0.1V`
and equivalent resistance of the circuit
`R=15Omega`
Hence current `i=(0.1)/(15)` amp `=(1)/(150)` amp
And its direction will be anti-clockwise (according to Lenz's law)