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1 Positive ions with charge q and mass m are, in the absence of a magnetic field, accelerated by an electric
! = E î, from rest, along the x-axis, through a potential difference V . At this time the electric field is
field, E
! = B k̂ is turn on.
turned off and a magnetic field B
(a) Determine, in terms of q, m and V , the velocity !v the ions have at the instant when the electric field is
turned off.
The change qV in potential energy must be equal to the change (1/2)mv 2 , i.e.,
1
mv 2 = qV
2
so
v=
!
2qV
m
(b) Determine in terms of B, q, m and V the radius r of the circular path the ion executes when the magnetic
! is turned on.
field B
The radial acceleration ar = v 2 /r must obey mar = Fr , where Fr is the radial component of the magnetic
! force, i.e.,
! B,
force F! = q ×
v2
m = qvB
r
so that
mv
r=
qB
(c) For short times for which y " x, show
y = CBx2
where C is a constant independent of the magnetic field B.
The motion is a circle of radius r however the small part of the circle involved is a parabola. A circle at the
origin is x2 + y 2 = r2 . We need to displace the circle so that it passes through the point x = r, y = 0 this
corresponds to x2 + (y − r)2 = r2 . Write this as x2 + r2 − 2yr + y 2 = r2 which, if y is small simplifies to
y=
1 2
qB 2
x =
x
2r
mv
i.e., is of the form y = CBx2 .
(d) Determine the constant C in terms of q, m and V .
From above
q
C=
mv
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2 (a) Calculate the magnitude B of the magnetic field at point P of the figure in terms of R, I1 and I2 .
Using
µ0 I d!# × r̂
4π r2
one first notes that the straight pieces contribute nothing to the field at point P , which is good since we do
not have enough information to calculate them otherwise.
Consider the contribution to the semi-circle carrying the current I1 . For a given d!#, it is the case that r̂,
! points into the
i.e., the direction of the vector !r which points towards the field point P . It follows that dB
paper. The direction is always the same so
"
"
!
!
B = dB
rather than
B = dB
! =
dB
so
µ0 I1
B1 =
4π
"
d#
µ0 I1
=
2
R
4πR2
"
d# =
µ0 I1
µ0 I1
πR =
2
4πR
4R
#
where it was noted that d# = πR is just the length of a semi-circle.
Examination of an arbitrary d!# for the I2 semi-circle shows the field now points out of the paper so the total
B = B1 − B2
into the paper. The result for the second semi-circle is evidently I1 → I2 :
B2 =
and so for the magnitude:
B=
µ0 I2
4R
µ0
|I1 − I2 |
4R
(b) What does your expression give when I1 = I2 ? Why does this make sense?
The field is zero in this case. This makes sense because of the symmetry between the two legs of the circuit
in this case.
(b) If R = 23cm, I1 = 0.25A and I2 = 1.25A, what is the magnitude B at P ?
B=
µ0
4π × 10−7 N/A2
|I1 − I2 | =
|0.25A − 1.25A| = 1.37 × 10−6 T
4R
4 × 0.23m
The field is in fact out of the paper.
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3 A vertically orientated square loop of copper wire falls into a region where B = 0 from a region where the
! = B î is horizontal, uniform and perpendicular to the loop which lies in the z − y plane, where y is
field B
the vertical axis. The sides have a length s. The mass of the loop is m and its resistance R.
y
B
z
s
v
(a) If the loop starts from rest, show the speed v is given by
dv
= −A(v − v0 ).
dt
Assume the time is such that the top edge of the loop remains in the magnetic field region.
There is a motional emf induced in the wire. This is determined from
E =−
dΦB
dt
Here the flux is equal to
ΦB = ysB
where y the co-ordinate system has been shifted down to the bottom of the magnetic region and where y is
the co-ordinate of the top left corner. This gives
E = −vsB
and for the current
i=
E
vsB
=−
R
R
!
where the sign tells about Lenz’s law. The forces on the sides cancel so the magnetic force dF! = Id!# × B
reduces to FB = isB or, in magnitude,
s2 B 2
v
FB =
R
which is proportional to the velocity v.
Now there are two forces acting on the wire, (i) Fg = mg downwards and (ii) upwards. Newton’s second law
then states F! = m!a which for the (negative) y direction becomes
ma = Fg − FB = mg −
s2 B 2
v.
R
But a = dv/dt so this can be written as
m
dv
s2 B 2
mgR
=−
(v − 2 2 )
dt
R
s B
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which is indeed of the correct form dv/dt = −A(v − v0 ).
(b) Determine A and v0 in in terms of B, R s, and m.
From the pervious expression:
s2 B 2
A=
R
and
mgR
v0 = 2 2 .
s B
(c) What is the terminal velocity of the loop again assuming the top edge remains in the field?
The terminal velocity is when dv/dt = 0 whence it must be that (v − v0 ) = 0 and so the terminal velocity is
v0 =
mgR
.
s2 B 2
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