In a cylindrical region of radius a, magnetic field exists along its axis but the direction of magnetic field is opposite in the four quadrants of the region as shown in Fig. A or AB rotates with its end A at the centre of magnetic field and other end B slides on a smooth wire at the periphery of the region of magnetic field. At t = 0 the rod was situated along the +X direction.

Find and plot the time dependence of the thermal power in the resistance R when rod rotates with a constant angular acceleration `(alpha)`

Options

A.
B.
C.
D. None of these

Answer

A

Explanation

(a) When rod rotates with constant angular speed `omega`.
`e= +- 1/2 B(a)(aomega)= +- 1/2 Ba^(2)omega`
`:. i=(e)/(R)=+- (Ba^(2)omega)/(2R)`
so, for `t=0 to T/4, T/2 to (3T)/(4)` and so on, `i=-(Ba^(2)omega)/(2R)`
The i-t graph is as shown in fig. the i-t equation can be written as,
`i=1/(2R) (-1)^(n+1) Ba^(2)omega`
where `n=1,2,3 ... and t_(n) =(n pi)/(2 omega)`
Here positive current means current from left to right through the resistance and vice-versa.

(b) `alpha` = constant.
`omega=alpha t`
`epsilon = (bomegal^(2))/(2) = ((Bl^2)/(2)) alpha*t, I=E/R = [(Bl^(2)alpha)/(2R)]*t`
I is negative in quardrant I
I is positive in quadrant II
I is negative in quadrant III
I is positive in quadrant IV

also, `t_(1)//4, t_(1)`
`1/2 alpha (t_(1)^(2))/(4), t_(1//4) = sqrt((pi)/(2 alpha))`
`t_(1//2) = sqrt((2 pi)/(2alpha)), t_(3//4)=sqrt((3pi)/(4alpha) xx 2)= sqrt((3 pi)/(2alpha))`
`sqrt(3)-sqrt(2)-sqrt(1)`
`PR=I^(2)R = [(Bl^(2)alpha)/(2R)]^(2) * R*t^(2) P prop t^(2) , y = ax^(2)`
Except at `sqrt((pi)/(2 alpha)), sqrt((2pi)/(2 alpha)), ... sqrt((n pi)/(2 alpha)) (nepsilon N)`
where `P=0`
.