Chapter 20
Problem 1
Φ = BA cos θ = (0.3)(π(0.25)2 ) cos(0) = 5.8 · 10−2 T m2
Problem 3
Φ = BA cos θ = (0.3)(2 · 2) cos(50) = 0.771T m2
Problem 7
The only field that is penetrating through the colored surface is the Bx field, since the other two have
cos θ = 0. The flux is therefore Φ = 5 ∗ 0.025 ∗ 0.025 ∗ cos 0 = 3.125 · 10−3 T m2 . The total flux through the
whole cube is zero, since all flux coming in through one side exits the other side. Flux coming into a region
is negative, flux going out is positive, and they cancel exactly.
Problem 9
= − ∆Φ
∆t . The magnetic field and angle are constant, so ∆Φ = B cos θ∆A and
1.8 · 10−3 = B(1)(0.1) which you can solve to find B.
∆A
δt
= 0.1. This gives
Problem 10
The same equation as above, except now we are looking for the EMF. The change in flux is going to be the
final flux (zero) minus the initial flux (π(0.12)2 · 0.15). That number is 6.78 · 10−3 . Divide this by 0.2s to get
the average induced EMF.
Problem 11
The trick here is to remember that B and A are vectors technically. The initial flux is Bi Ai = (0.3T )(π(0.3)2 m2 )
and the final flux is Bf Af = (−0.2T )(π(0.3)2 m2 ). It is important to note that the final flux is negative,
reflective of the fact that the field changed direction. The change in flux is then final flux minus initial flux,
and dividing this by the t = 1.5s you find the average EMF change.
Problem 19
The charges in the wings of the plane moving in the magnetic field experience a lorentz force F = qvB sin(58)
which pushes the positive charges one way and the negative charges the other way (think right hand rule).
As the charges separate, the electric attraction between them pulls them back together, so eventually whwen
enough charges are separated, an equilibrium is reached when the electric force pulls just as hard as the
magnetic force. This happens when qvB sin(58) = qE. Use this to find the electric field, and then multiply
by the wingspan to find the potential (since ∆V = E∆X).
Problem 20
The beam is falling vertically, and the field is pointing horizontally. Just before impact, we can figure out the
velocity of the beam since it is falling under gravity. vf2 = vi2 + 2a∆y = 0 − 2 · 9.8(−9). g is negative because
gravity points down, but Deltay is negative because the height decreases. Solve this to find the velocity.
Now use the same procedure as problem 19. Set E = vB, solve for E, and multiply by the length of the bar
to find the voltage.
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