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In the given figure, the wires P_(1)Q_(1) and P_(2)Q_(2) are made to slide on the rails with same speed of 5 cm s^(-1). In this region, a magnetic field of 1 T exists. The electric current in the 9 Omega resistance is |
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Answer» zero if both wires slide toward left `BLV = 1xx4xx10^(-2)xx5x10^(-2)=20xx10^(-4)V` The polarity of the battery can be given by Fleming.s right hand rule. When both wire move in opposite direction, the circuit diagram look like as shown in figure (a). The effective emf of the two batteries shown in the diagram is zero. So, choice (b) is correct and choice (d) is wrong. When both wires move towards left, the circuit diagram looks like as shown in figure (b). ![]() Effective emf of two batteries shown is `E=(20xx10^(-4)V)` and RESISTANCE is same `(2Omega)` therefore `i_(1)=i_(2)` Hence, current in the circuit is, `i=(20xx10^(-4))/(10)=0.2 mA` Hence, choice (c ) is correct and choice (a) is wrong. |
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