A 0.360-m-long metal bar is pulled to the left by an applied force F. The bar rides

on parallel metal rails connected through a 45.0$\mathbf{\Omega }$resistor, as shown in Fig., so the apparatus makes a complete circuit. You can ignore the resistance of the bar and rails. The circuit is in a uniform 0.650-T magnetic field that is directed out of the plane of the figure. At the instant when the bar is moving to the left at 5.90 m/s.

(a) is the induced current in the circuit clockwise or counterclockwise?

(b) what is the rate at which the applied force is doing work on the bar?

We have a conducting rod ab, which makes contact with metal rails ca and db where the parallel metal rails are connected through a R = 45.0resistor, the whole device is placed perpendicularity in a magnetic field of B = 0.650 T, as shown in the following figure.

We need to find the direction of induced current in the circuit when the rod is moving under influence of an applied force F toward the left at the instant when the speed is v= 5.90 m/s as shown in the figure. Let x be the length of the expanding side db, and L = 0.360 m is the length of the constant length side ab. The area of the loop abcd decreases as the bar moves to the left, hence the magnetic flux, and the external magnetic field points out of the page, so the induced magnetic field must point out of the page (according to the Lenz's law), so the induced current must circulate counterclockwise in the circuit

The applied force must equal the magnetic force,

$F_{applied}=F_B=lLB$ (1)

where I is the induced current, which is given by,

$I=\frac{\epsilon }{R}$

where $\mathrm{\epsilon }$is the induced emf $\mathrm{\epsilon }=\mathrm{vBL}$, so

$I=\frac{vBL}{R}$

substitute into (1) we get,

$F_{applied}=\frac{v{B}^2{L}^2}{R}$

the rate at which this force does work is the applied force multiplied by the speed, that is,

$$\begin{gathered} F_{applied}=F_{applied}v \\ =\frac{{\left(vBL\right)}^2}{R} \\ =\frac{{\left[\left(5.90m/s\right)\left(0.650T\right)\left(0.360m\right)\right]}^2}{45.0\Omega } \\ {F}_{applied}=0.042W \end{gathered}$$