A rectangular loop with a sliding connector of length l=10m` is the situated in a uniform magnetic for B=2T perpendicular to the plane of the loobs. Resistance of conductor is `r=2Omega`. Two resistances at a`6Omega and 3Omega` are connected as shown in the figure. The external force required to keep the conductor moving with a constant velocity v=2m/s

Options

A. (a) `6n`
B. (b) `4n`
C. ( c) `2n`
D. (d) `1n`

Answer

C

Explanation

( c) Motion emf
`e = Bvl`
`e = (2)(2)(1) = 4 V`
This acts as a cell of emf `E = 4 V` and internal resistance `r = 2 Omega`. The simple circuit can be drawn as follows:

Therefore, current through the connector
`i = (4)/(2+2) = 1 A`
Magnetic force on connector
`F_(m) = ilB = (1)(1)(2) = 2 N` (towards left)
Therefore, to leep the connector moving with a constant velocity, a force of `2 n` will have to be appiled towards right.