In the figure shown , 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 :

Options

A. (a) zero if both wires slide toward left
B. (b) zero if both wires slide in opposite directions
C. (c ) `0.2 mA` if both wires move toward left
D. (d) `0.2 mA` if both wires move in opposite directions

Answer

B, C

Explanation

(b,c): Each wire can be replaced by a battery whose emf is equal to
`Blv = 1 xx 4 xx 10^(-2) xx 5 xx 10^(-2)`
`= 20 xx 10^(-4) V`
The polarity of the battery can be given Fleming's right hand rule. When both wire move in oppsite direction, the circuit diagrame looks as shown in Fig. 3S3.64(a).
The effective emf of the two batteries shown in the diaqgram is zero.
So, coice (b) is correct and coice (D) is wrong.
When both wires move towards left, the circuit diagrame looks like as shown in Fig. s3.64(b)

Effective emf two batteries shown is `E (= 20 xx 10^(-4)V)` and internal resistance is `1 Omega`.
Hence, current in the circuit is
`i = (20 xx 10^(-4))/(10) = 0.2 mA`
Hence, choice ( c) is correct and choice (a) is wrong .