Electromagnetic Induction and Alternating Current 2 Question 8
9. A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. The distance between the rails is $L$. A conducting massless rod of resistance $R$ can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. A mass $m$ tied to the other end of the string hangs vertically. A constant magnetic field $B$ exists perpendicular to the table. If the system is released from rest. Calculate :
(1997, 5M)
(a) the terminal velocity achieved by the rod, and
(b) the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity.
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Answer:
Correct Answer: 9. (a) $v _T=\frac{m g R}{B^{2} L^{2}}$ (b) $a=\frac{g}{2}$
Solution:
- (a) Let $v$ be the velocity of the wire (as well as block) at any instant of time $t$.
Motional emf, $e=B v L$
Motional current, $i=\frac{e}{r}=\frac{B v L}{R}$
and magnetic force on the wire
$$ F _m=i L B=\frac{v B^{2} L^{2}}{R} $$
Net force in the system at this moment will be
$$ \begin{aligned} F _{\text {net }} & =m g-F _m=m g-\frac{v B^{2} L^{2}}{R} \\ \text { or } \quad m a & =m g-\frac{v B^{2} L^{2}}{R} \\ a & =g-\frac{v B^{2} L^{2}}{m R} \end{aligned} $$
Velocity will acquire its terminal value i.e. $v=v _T$ when $F _{\text {net }}$ or acceleration ( $a$ ) of the particle becomes zero.
$$ \begin{array}{ll} \text { Thus, } & 0=g-\frac{v _T B^{2} L^{2}}{m R} \\ \text { or } & v _T=\frac{m g R}{B^{2} L^{2}} \\ \text { When } & v=\frac{v _T}{2}=\frac{m g R}{2 B^{2} L^{2}} \end{array} $$
(b) When
Then from Eq. (i), acceleration of the block,
$$ a=g-\frac{m g R}{2 B^{2} L^{2}} \quad \frac{B^{2} L^{2}}{m R}=g-\frac{g}{2} $$
or
$$ a=\frac{g}{2} $$
