Question

A long conducting wire `AH` is moved over a conducting triangular wire `CDE` with a constant velocity `v` in a uniform magnetic field `vec(B)` directed into the plane of the paper. Resistance per unit length of each wire is `rho`. Then

A. (a) a constant clockwise induced will flow in the closed loop
B. (b) an increasing anticlockwise induced current will flow in the closed loop
C. (c ) a decreasing anticlockwise induced current will flow in the closed loop
D. (d) an constant anticlockwise induced current will flow in the closed loop

Answer 1

Correct Answer - D
(d) Magnetic flux in `ox` direction through the coil is increasing. Therefore, induced current will produce magnetic field in `o.` direction. Thus, the current in the loop is anticlockwise. Magnitude of induced current at any instant of time is
`I = (e)/(R) = (Bv(FG))/(rho(FG + GD + DF))`
When the wire `AH` moves downwards `FG, GD and DF` all increasing in the same ratio. Therfore, `i` is constant.