MCQ
A wire having mass $m$ and length $1$ can freely slide on a pair or parallel smooth horizontal rails placed in vertical magnetic field $B$. The rails are connected by a capacitor of capacitance $C$. The electric resistance of the rails and the wire is zero. If a constant force $F$ acts on the wire as shown in the figure. Then, the acceleration of the wire can be given as
  • $a=\frac{F}{m+C^2 F ^2}$
  • B
    $a=\frac{F}{m+C B 1}$
  • C
    $a=\frac{F^2 B^2 1}{m}$
  • D
    $a=\frac{F}{m+C^2 B^2 1}$

Answer

Correct option: A.
$a=\frac{F}{m+C^2 F ^2}$
a
(a)

Due to the movement of rail with instantaneous velocity $v$, the e.m.f produced across the capacitor, $V = Bv l$

Charge stored in the capacitor $=q= CV = CBvl$

Therefore the current through the capacitor, $i =\frac{ dq }{ dt }= CBl \frac{ dv }{ dt }= CBla$

This same current passes through the wire

Hence the magnetic force on it, $F = Bil = CB ^2 I ^2 a$

Hence, $ma = F - F$

$\Rightarrow ma = F - CB ^2 l ^2 a$

$\Rightarrow a=\frac{F}{m+C^2 F ^2}$

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