Question

Here we'll see that an emf can be induced by the motion of the conductor in a static magnetic field. The U-shaped conductor in (Figure 1) lies perpendicular to a uniform magnetic field B⃗ with magnitude B=0.75T , directed into the page. We lay a metal rod with length L=0.10m across the two arms of the conductor, forming a conducting loop, and move the rod to the right with constant speed v=2.5m/s . What is the magnitude of the resulting emf?
With the given magnetic field and rod length, what must the speed be if the induced emf has magnitude 0.73 V ?

Answer 1

Answer:

a) ΔV = 0.1875 V , b) v = 9.73 m / s

Explanation:

For this exercise we can use Necton's second law, as the speed is constant the forces on the driver are equal

               $F_{E} - F_{B}$ = 0

               $F_{E}$ = F_{B}

               q E = q v B

               E = v B

The electrical force induced in the conductor is

                ΔV = E l

                ΔV = v B l

Let's calculate

               ΔV = 2.5 0.75 0.10

               ΔV = 0.1875 V

b) If ΔV = 0.73 what speed should it have

             v = DV / B l

             v = 0.73 / 0.75 0.1

             v = 9.73 m / s