Refer to the figure. The arm $PQ$ of the rectangular conductor is moved from $x=0$ outwards. The uniform magnetic field is perpendicular to the plane and extends from $x=0$ to $x=b$ and is zero for $x>b$. Only the arm $PQ$ possesses substantial resistance $r$. Consider the situation when the arm $PQ$ is pulled outwards from $x=0$ to $x=2b$ and is then moved back to $x=0$ with constant speed $v$. Obtain expressions for the flux,the induced emf,the force necessary to pull the arm,and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

(N/A) Let us first consider the forward motion from $x=0$ to $x=2b$. The magnetic flux $\Phi_B$ linked with the circuit $SPQR$ is:
$\Phi_B = Blx$ for $0 \leq x < b$
$\Phi_B = Blb$ for $b \leq x < 2b$
The induced emf $\varepsilon$ is given by $\varepsilon = -\frac{d\Phi_B}{dt}$:
$\varepsilon = -Blv$ for $0 \leq x < b$
$\varepsilon = 0$ for $b \leq x < 2b$
When the induced emf is non-zero,the current $I$ in magnitude is $I = \frac{|\varepsilon|}{r} = \frac{Blv}{r}$.
The force $F$ required to keep the arm $PQ$ in constant motion is $F = IlB$. Its direction is to the left (opposing motion). In magnitude:
$F = \frac{B^2l^2v}{r}$ for $0 \leq x < b$
$F = 0$ for $b \leq x < 2b$
The Joule heating loss $P_J$ is $P_J = I^2r$:
$P_J = \frac{B^2l^2v^2}{r}$ for $0 \leq x < b$
$P_J = 0$ for $b \leq x < 2b$
Similar expressions are obtained for the inward motion from $x=2b$ to $x=0$. The variations of these quantities with distance are shown in the provided figure.