Question

The given figure shows an inductor and a resistor fixed on a conducting wire. A movable conducting wire $\text{PQ}$ starts moving on the fixed rails from $t=0$ with constant velocity $1\text{m/s}$. The work done (in Joule) by the external force on the wire $\text{PQ}$ in $2\text{seconds}$ is _______

Answer 1

Given:
$L=2\text{H};R=2\Omega ;B=2\text{T};l=2\text{m};v=1\text{m/s};t=2\text{sec}$

$\because$ Induced Emf is given by $ϵ=Bvl$

Thus, $ϵ=2\times 1\times 2=4\text{volt}$

So the induced current , $I=\frac{ϵ}{R}=\frac{4}{2}=2\text{A}$

Now the magnetic force , $F=IlB$

$\therefore F=2\times 2\times 2=8\text{N}$

Now the work done on the wire $\text{PQ}$ in $2$ seconds is -

$W=\overrightarrow{F}.\overrightarrow{S}$

$W=F\times V\times t$

substituting the data given in the question gives,

$W=8\times 1\times 2=16\text{J}$