Answer:
Part a)
Induced EMF when length vector is along Z direction is 0.72 V
Part b)
Induced EMF when length vector is along Y direction is ZERO
Explanation:
As we know that the motional EMF induced in the wire is given as
1)
As we know that
now we have
so we have
2)
If the length vector is along Y direction then we have
so again we have
so we have
EMF = 0
Answer:
W = -510.98J
Explanation:
Force = 43N, 61° SW
Displacement = 12m, 22° NE
Work done is given as:
W = F*d*cosA
where A = angle between force and displacement.
Angle between force and displacement, A = 61 + 90 + 22 = 172°
W = 43 * 12 * cos172
W = -510.98J
The negative sign shows that the work done is in the opposite direction of the force applied to it.
Answer:
Explanation:
You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:
where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.
by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:
(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.
By doing the same for the electron you obtain:
we can equals these expressions for both proton and electron, because the forces qE are the same:
