Problem #1
A conducting bar with length L and resistance R moves at constant speed v connected to
vertical rails with zero resistance and negligible friction (see figure). An unknown,
uniform, constant magnetic field perpendicular to the loop formed by the bar and the
rail generates an induced current I running clockwise in the loop.
a) Find magnitude and direction of the magnetic field .
b) Calculate the thermal power dissipated into the bar.
c) The fact that the bar moves at constant speed means that the acceleration is 0.
Find the magnitude of the gravitational force pulling on the bar.
d) Calculate the rate of work (work/time) done by the gravitational force on the bar.
Compare your result with what you found in part b). Explain.
Write your results in terms of L, R, v, and I.
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a) Φ
∙
coordinate.
, with
directed into the board, and y being the vertical
1 Φ
where the negative sign is removed because the bar moves in the negative y
direction (the area is reduced!). Therefore:
Into the board (it is positive, therefore parallel to ).
b)
c) If the acceleration is zero, the gravitational force is balanced by the magnetic one.
̂
̂
̂
d)
∙
∙
,
The rate of work done by the gravitational force is equal to the power dissipated
into the circuit.
2
Problem #2
A square loop is made of four identical rods with length L, cross section A, and resistivity
 The loop is placed on the x-y plane with one edge in the origin and the sides parallel to
the x and y axes (see figure). One negligible section of the loop is cut and the two ends
are connected to a battery with e.m.f. , using wires with negligible resistance. A
magnetic field with magnitude
fills the whole space, with C positive constant.
a) Calculate the current running through the loop.
b) Calculate the force exerted by the magnetic field on side a-b of the loop.
c) Calculate the force exerted by the magnetic field on side b-c of the loop.
d) Calculate the force exerted by the magnetic field on side c-d of the loop.
e) Calculate the force exerted by the magnetic field on side d-a of the loop.
Notice that the magnetic field is not uniform in the x direction!
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a)
,
b) Along a-b
c) Along b-c,
, clockwise.
0, therefore
4
d) Along c-d
0, therefore
̂
̂
0
̂
̂
4
16
̂
̂
16
is constant, therefore
̂
̂
4
4
e) Along d-a the force is just equal but opposite to that in part c:
16
̂
̂
3
Problem #3
In the circuit shown, switch S is closed at time t=0. Calculate:
a) The energy stored in the inductor a long time after the switch is closed.
b) The energy stored in capacitor C1 a long time after the switch is closed.
c) The power dissipated in resistor R2 a long time after the switch is closed.
d) The voltage across the inductor L right after the switch is closed.
e) The power provided by the battery right after the switch is closed.
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a) A long time after the switch is closed the inductors act as short circuits and the
capacitors as open circuits.
and the energy
/8 .
The current in the inductor is:
b) There is no current through , therefore there is no voltage across is and the
voltage across the capacitor is just that of the battery, therefore
.
0.
c) There is no current and there no power is dissipated in the resistor:
d) Right after the switch is closed the inductors act as open circuits, the capacitors as
short circuits.
and
.
There potential across the inductor is the same as that across the left resistor
.
e) The current through the battery is
provided by the battery is
,
and the power
.
4