The correct option is
C 0.2 mA, if both wires move towards left
Each wire can be replaced by a battery whose emf is equal to
Blv=1×4×10−2×5×10−2
=20×10−4V.
The polarity of the battery can be given by Fleming's right hand rule.
When both wire move in opposite direction, the circuit diagram looks like as shown in figure.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1046015/original_original_10S_1.png)
Effective emf of two batteries shown in
E′(=0 V)
Hence , in this case current in
9 Ω resistor is zero.
When both wires move in same direction, the circuit diagram looks like as shown in figure.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1046001/original_original_4Q.png)
Effective emf of two batteries shown is
E=20×10−4V
and internal resistance is
1 Ω.
Hence, current in the circuit is
i=20×10−410=0.2 mA
Hence, choice (c) is correct and choice (a) is wrong.