91Ó°ÊÓ

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 point 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 induced current in the resistor equals the induced emf divided by the resistance, that is,

$\mathrm{I}=\frac{\mathrm{Î\mu }}{\mathrm{R}}$

where,

$\mathrm{Î\mu }=\mathrm{vBL}$

so,

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

the rate at which the resistor dissipates electrical energy is,

$P_R=I^{2}R$

Substitute with we get,

${\mathrm{P}}_{\mathrm{R}}=\frac{{\mathrm{v}}^2{\mathrm{B}}^2\mathrm{L}}{\mathrm{R}}$

we need to find the speed of the bar at an instant when the resistor is dissipating electrical energy at a rate of ${\mathrm{P}}_{\mathrm{R}}=0.840\mathrm{J}/\mathrm{s}$= 0.840J /s substitute with the givens we get

$$\begin{gathered} v=\sqrt{\frac{P_{R}R}{B^{2}L}} \\ =\sqrt{\frac{\left(.840J/s\right)\left(45.0\mathrm{Î\copyright }\right)}{{\left(0.650T\right)}^2{\left(0.360m\right)}^2}} \\ =26.3m/s \\ v=26.3m/s \end{gathered}$$