Question

A conducting rod $MN$ of mass ${}^{\prime }m$ and length ${}^{\prime }{\ell }^{\prime }$ is placed on parallel smooth conducting rails connected to an uncharged capacitor of capacitance ${}^{\prime }{C}^{\prime }$ and a battery of emf $\epsilon$ as shown. A uniform magnetic field ${}^{\prime }{B}^{\prime }$ is existing perpendicular to the plane of the rails.The steady state velocity acquired by the conducting rod $MN$ after closing switch $S$ is (neglect the resistance of the parallel rails and the conducting rod )

• A. $\frac{2CB\ell \epsilon }{\left(m+C{B}^2{\ell }^2\right)}$
• B. $\frac{CB\ell \epsilon }{\left(m+C{B}^2{\ell }^2\right)}$
• C. $\frac{CB\ell \epsilon }{2\left(m+C{B}^2{\ell }^2\right)}$
• D. $\frac{CB\ell \epsilon }{4\left(m+C{B}^2{\ell }^2\right)}$

Answer 1

The correct option is B $\frac{CB\ell \epsilon }{\left(m+C{B}^2{\ell }^2\right)}$

Given :Mass=m ,Length =l, Capacitance=C, Emf=$\epsilon$,Magnetic field =B

Solution: Let the potential difference across conducting rod=$V_1$

Let the potential difference across capacitor=$V_2$

So,$\epsilon =V_1+V_2$

$\epsilon =Bvl+\frac{q}{C}$

$q=(\epsilon -Bvl)C$......(1)

Force required to pull the loop with constant velocity:

$F=ma=m\frac{dv}{dt}$

$Bil=m\frac{dv}{dt}$

Integrating the euation:

$m{\int }_0^{v}dv=Bl{\int }_0^{q}dq\times dt$

$m(v-0)=Blq$

$q=\frac{mv}{Bl}$.....(2)

From equation (1) & (2):

$(\epsilon -Bvl)C=\frac{mv}{Bl}$

The steady state velocity acquired by the conducting rod MN after closing switch S is:

$v=\frac{C\epsilon Bl}{m+B^2{l}^2{C}^2}$

Hence the correct option is B