Answer:
5.69 m/s.
Explanation:
The given data is as follows.
Magnetic field strength (B) = 0.65 T
Speed (v) = 4.2 m/s
Induced emf (E) = ?
Formula for emf induced at the ends of the rod of length L which is moving with a speed of v is as follows.
E = BvL
Putting the given values into the formula as follows.
.
[tex]E_{1} = BvL = 0.65 T \times 4,2 m/s \times L\\=2.73L[/tex]
When magnetic field is changed to [tex]B_{2} = 0.48 T[/tex]
Now, we assume that the speed be [tex]v_{2}[/tex] to get the emf[tex]E_{2} = E_{1}.[/tex]
Then,
[tex]0.48 T \times v_{2} \times L = 0.65 T \times 4.2 m/s \times L v_{2} = 5.69 m/s[/tex]
we can conclude that the speed v of the rod be adjusted to reestablish the emf induced between the ends of the rod at its initial value is 5.69 m/s.
