Question

Determine the emf of movement of a 20 cm long rod that is on a rail and that moves with a speed of 45.0 cm / s. The rod and the rail are in a field 0.350 M magnetic

Answer 1

The induced electromotive force is equal to the change of magnetic flux Φ through the surface area of the loop per time Δt

Ui=−ΔΦ/Δt ...(1)

For the induction flux it applies

Φ=BScosα,Φ=BScosα,

where α is the angle between the normal vector of the surface of the loop, and the induction lines, in our case α = 0°, i.e. cos α = 1. Because the vector of magnetic induction B is of a constant magnitude, the change of the induction flux ΔΦ can be determined as follows

ΔΦ=B.ΔS,

ΔΦ=BΔS,

where ΔS is a change of the surface area and it is equal to

ΔS=−lvΔt.ΔS=−lvΔt.

The minus sign results from the fact that the area of the loop decreases.

The expression for the change of magnetic flux is substituted into equation (1) for induced electromotive force U_i

Ui=−−BlvΔt/Δt=BlvΔtΔt=Blv.

putting the values we get

Ui=0.3*0.45*.2 (all are in SI unit)

=0.027