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Question
A wire of mass m and length l can slide freely on a pair of smooth, vertical rails (figure). A magnetic field B exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance C. Find the acceleration of the wire neglecting any electric resistance.
Solution
Let the velocity of the rod at an instant be v and the charge on the capacitor be q.
The emf induced in the rod is given by
e = Blv
The potential difference across the terminals of the capacitor and the ends of the rod must be the same, as they are in parallel.
\[\therefore\frac{q}{C} = Blv\]
And,
q = C × Blv = CBlv
Current in the circuit:-
\[i = \frac{dq}{dt} = \frac{d(CBlv)}{dt}\]
\[\Rightarrow i = CBl\frac{dv}{dt} = CBla\] ........(a = acceleration of the rod)
The force on the rod due to the magnetic field and its weight are in opposite directions.
∴ mg − ilB = ma
⇒ mg − cBla × lB = ma
⇒ ma + cB2l2a = mg
⇒ a(m + cB2l2) = mg
\[\Rightarrow a = \frac{mg}{m + c B^2 l^2}\]
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