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If X is a capacitor C, then the constant...

If `X` is a capacitor C, then the constant accelerationn a of the wire.

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A time `t` suppose velocity of wire is `v`.Then due to motional emf `e=Bvl` capacitor gets charged.

` q=CV=C(Bvl)`
This charge is increasing as `v` will be inscreasing
Hence, there will be a current in the circuit as shown in figure.
`i=(dq)/(dt)=(BlC)=(dv)/(dt)`
or `i=(BlC)a (as (dv)/(dt)=a)`
Due to this current a magnetic force `F_m` will act in the direction shown in figure.
`F_m=ilB=(B^2l^2C)a`
Now, `F_("net")=F-F_m`
or `ma=F-(B^2l^2C)a`
`:. a=f/(m+B^2l^2C)`
now, we can see that this acceleration is constant but less than `F//m`
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