Question

A copper wire $ab$ of length $l$, resistance $r$ and mass $m$ starts sliding at $t=0$ down a smooth, vertical, thick pair of connected conducting rails as shown in figure. A uniform magnetic field $B$ exists in the space in a direction perpendicular to the plane of the rails. Choose the correct option(s).

• A. The magnitude and direction of the induced current in the wire when speed of the wire $v$ is $\frac{vBl}{r}$, $a$ to $b$.
• B. The download acceleration of the wire at this instant $g-\frac{B^2{l}^2}{mr}v$
• C. The velocity of the wire as a function of time $v_m(1-e^{-gt/v_m}),(\text{where}{v}_m=\frac{mgr}{B^2{l}^2})$
• D. The displacement of the wire as a function of time $V_{m}t-\frac{v{{}_m}^2}{g}(1-e^{-gt/v_m}),(\text{where}{v}_m=\frac{mgr}{B^2{l}^2})$

Answer 1

Answer: (A), (B), (C), (D)

The displacement of the wire as a function of time $V_{m}t-\frac{v{{}_m}^2}{g}(1-e^{-gt/v_m}),(\text{where}{v}_m=\frac{mgr}{B^2{l}^2})$

$\left(A\right)$ current in the wire is $i=\frac{Bvl}{r}\left(atob\right)$

$\left(B\right)$ Net force on the wire is

$mg-\frac{B^2{l}^{2}v}{r}=m\frac{dv}{dt}$

$\therefore \frac{dv}{dt}=g-\frac{B^2{l}^{2}v}{mr}$



$\left(C\right)$ Integrating above equation
${\int }_0^v\frac{dv}{g-\frac{B^2{l}^{2}v}{mr}}={\int }_0^{t}dt$

$\therefore v=v_m(1-e^{\frac{-gt}{v_m}})$
Here $v_m=\frac{mgr}{B^2{l}^2}$

$\left(D\right)$ Integrating the above equation ${\int }_0^{s}ds={\int }_0^t{v}_m(1-e^{\frac{-gt}{v_m}})dt$
$s=v_{m}t-\frac{v_m^2}{g}(1-e^{\frac{-gt}{v_m}})$