A rod of length `L` rotates in the form of a conical pendulum with an Angular velocity `omega ` about its axis as shown in Fig. 3.165. The rod makes an angle `theta` with the axis. The magnitude of the motional emf developed across the two emf of the rod is

Options

A. (a) `(1)/(2)BomegaL^(2)`
B. (b) `(1)/(2)BomegaL^(2)tan^(2) theta`
C. ( c) `(1)/(2)BomegaL^(2)cos^(2)theta`
D. (d) `(1)/(2)BomegaL^(2)sin^(2)theta`

Answer

D

Explanation

(d) `epsilon = int(vec(v) xx vec(B)).vec(dl)`

`rarr epsilon = int [(l sin theta)omega hat(k)) xx B(-hat(j))]. [dl sin theta(-hat(i)) + dlcostheta(-gat(j))]`
`rarr epsilon = int [B omega sin theta -hat(i)]. [dl sin theta(-hat(i)) + dl costheta(-hat(j))]`
`= -Bomegasin^(2)theta int_(0)^(L)ldl = -(1)/(2)Bomegasin^(2)thetaL^(2)`
`rarr epsilon = V_(b) - V_(a) = -(1)/(2)BomegaL^(2)sin^(2)theta`
Negative sign indicates that end `b` will be negative `w.r.t.a`.
Alternate Method:
Consider two more conductors `ac` and `bc`. This comletes a closed loop. The net emf inducd in this closed loop should be zero, as net flux through this loop always remains zero.

`e_(ab) + e_(bc) + e_(ca) = 0`
but `e_(bc) = (1)/(2)Bomega(Lsintheta)^(2), e_(ca) = 0` putting the values, we get
`e_(ab) = -(1)/(2)Bomega(L sin theta)^(2)`