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1. None on the lower wire.  The upper
left wire has force ILBsin60 into page.
The upper right wire has force ILBsin60
out of page.  Thus total is zero.

2. It is a current loop.  All current
loops in a uniform B field have a net
force of zero.

3. Use Lorentz mag force.  Pos charges
in bar would experience a leftward force,
and neg charges a rightward force, so
diagram 4 is the correct one.

4. Use Ampere's Law.  For a long,
straight wire, it yields B=mu0I/(2pir),
in the tangential dir as given by RHR.
5 mm from either wire, means over 3 mm
(to the central point in between the
wires) and then perpendicularly (normal
to the plane of the wires) by 4 mm.
Draw a good picture. Add the two vector
fields; the components normal to the
plane of the wires cancel.  So the
total B=2*B0*sin(53) where
B0=mu0(40 A)/(2pi*0.005 m)=.0016 T.
So B=.00256 T.

5. Use BiotSavart.  Do the integral
for the field at center due to a
quarter circle of current.  The integral
is of (mu0/4pi)*Ids/R^2 over a quarter
circle or just (mu0/4pi)I(pi/2R)=
(mu0I/8R).  All contributions to the
total field are the in the same DIR
(into page by RHR) and there are 3
quarter circles of radius 4 cm but
only 1 quarter circle of radius 2 cm,
so B=(mu0I/8)((1/a)+(3/b))=39.3 microT.

6. Use Ampere's Law for the long wire.
The field due to current I is mu0I/(2pir)
tangential to I with DIR by RHR.  Add
vector fields to find the total field
at a selected wire, then do ILcrossB for
the mag. force on the current in that wire.
The fields due to the other two wires at
the location of our wire of interest are
the same size (0.0200 mT) and differ in DIR
by 60 degrees (draw a good picture).  Let
the bisector of this angle be the x axis.
Then the y components cancel, so the mag of
Btotal is twice the x component, or
2(0.0200 mT)cos(30)=0.0346 mT.  For
ILcrossB the angle in the sin fuction
is 90 degrees, thus F=ILB=0.0693 mN=
69.3 microN.

7. Use Ampere's Law for the long wire.
The field due to current I is mu0I/(2pir)
tangential to I with DIR by RHR.  Add
vector fields to find the total field
at wire A, then do ILcrossB for the mag.
force on the current in wire A.  The
fields due to B and C (at A) are the same
size (0.800 mT) and differ in DIR by 120
degrees (draw a good picture).  Let +x be
to the right in the figure.  The y
components cancel, so the mag of Btotal
is twice the x component, or
2(0.800 mT)cos(60)=0.800 mT.  For
ILcrossB the angle in the sin fuction
is 90 degrees, thus F=ILB=32.0 mN.

8. Use Ampere's Law for the long wire.
The location is inside the wire, not the
more common outside-the-wire case.  The
circulation of B around a circular Amperean
loop of radius r is 2pir*B.  The current
enclosed by such a loop is 25 A*r^2/R^2,
where R is the radius of the wire.  Solving
for B yields mu0(25 A)r/(2piR^2), or
2.50 mT.

9. Add vector fields.  From Ampere's
Law for the solenoid, B is mu0*n*I and
parallel to the solenoid axis, or
0.362 mT parallel to the solenoid axis.
From Ampere's Law for the long wire,
the field is mu0I/(2pir) tangential,
which is perpendicular to the solenoid
axis; so 0.120 mT normal to solenoid axis.
Use Pythagoras to add the perpendicular
fields, thus B=0.381 mT.

10. Add vector fields.  From Ampere's
Law for the solenoid, B is mu0*n*I and
parallel to the solenoid axis, or
0.452 mT parallel to the solenoid axis.
From Ampere's Law for the long wire,
the field is mu0I/(2pir) tangential,
which is perpendicular to the solenoid
axis; so 0.400 mT normal to solenoid axis.
Use Pythagoras to add the perpendicular
fields, thus B=0.604 mT.

11. Use Faraday's Law.  The change in
flux per turn is 0.0050 m^2*4 T, so
the total change in flux is 20*0.020 Wb
=0.4 Wb.  So the rate of change of flux
is 0.2 Wb/s; that is also the induced
EMF so the induced current is 0.2 V/.4 Ohms
or 0.5 A.

12. Use Motional EMF (Lorentz mag force
law), and Ampere's Law result for B due
to long straight current, i.e. mu0I/(2pir).
So B at the bar is 2.5 mT into page.
Integrate vcrossBdotdL from B to A.  v is
towards top of page and B is into page
so vcrossB is leftwards.  vcrossB and dL
thus have same direction so that the dot
is cos0=1.  So the integral is trivial,
just vBdL from zero to L or just vBL=.015 V.
Double check the sign with RHR for
hypothetical POS charges in bar.  Thumb is
towards top of page and fingers are into page,
so palm (force) is leftwards, so POS charges
gather at A, and A is the high voltage end
of the bar.

13. Use Motional EMF.  Integrate
vcrossBdotdL from P to A.  Choose rotation
CW then, at time shown, v is towards
bottom of page.  Then with B into page,
the vcrossB vector is rightwards.  So
vcrossB has the same direction as dL and
the dot gives cos0degrees or 1.  Write
v as omega*x and dL becomes dx, then the
integral is of B*omega*xdx from 0 to L/2,
which yields 0.5*B*omega*L^2/4=0.008 V.
The chosen CW rotation gives a positive
EMF from P to A, so a CCW rotation (the
plus sign on the 2.0 rad/s) would give a
negative EMF from P to A.  Practically,
almost everyone chooses to do the integral
to get a positive result, then gets the
sign by RHR.  With rotation CCW, thumb
towards top of page, fingers into page,
so force on a hypothetical POS charge is
leftwards, towards P.  The POS charges
thus gather at P, and P is the high voltage
end of the rotating bar.

14. Use Faraday's Law and the result
from Ampere's law for the field in a
solenoid, namely B=mu0In.  The circulation
of the tangential E around an imaginary
The rate of change of flux through such a
loop is piR^2 times the time derivative of
B, where R is the solenoid radius.  The
time derivative of B is mu0*n*the time
derivative of I.  So solve Faraday's Law
for E=(mu0*n*R^2/(2r))*(dI/dt)
=(mu0*(2500/m)*(.03m)^2/.08m)*((.3*200)A/s)*
=0.00186 V/m

15. Need a CCW induced E field for the
left to right current requested.  So
the B due to such a current will be out
of page.  Nature would create such a B
to fight an increase in the flux.  No
rotation can increase the flux.  Either
increase the B field or expand the loop.

16. Use Faraday's Law.  The changing
flux in the metallic loop is due only to
the changing current in the straight wire,
so the greater the rate of change of
current, the greater the rate of change
of flux through the metallic loop.
C has the greatest slope, then E, then
B; D and A are both zero.

150 T/s*.06m*.12m=1.08 V.  This 1.08 V
is due to a CW induced E field by Lenz's
Law, as nature resists the change in flux.
The battery EMF opposes this 1.08 V.  So
the net EMF is 2.92 V and the current
is 2.92 V/10 Ohms = .292 A.

18. Use Faraday's Law.  EMF=(amount of
flux change)/deltaT= .25 T*pi*.15m^2/.3 s.
So EMF=0.0589 V.

19. Use Lorentz mag force on hypothetical
POS charges in the metal bar; those must
be driven in the bar towards top of page.
Use RHR for these POS charges; thumb right
and palm towards page top.  Fingers are
then into page.

20. Use Faraday's Law.  A changing mag
flux is required passing through the wire
loop.  In 1, the flux is rightward and
decreasing.  In 2, the flux is rightward
and decreasing.  In 3, the flux is not
changing.  In 4, the flux is rightward
and increasing.
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