Answers  Magnetic Fields


1.

F_{m}(N) = q (C) v(m/s) B sin v,
B.
The unit of B is then N/(Cm/s) or
N/(C/sm) = N/Am.
So 1 T = 1 N/(C/sm) = 1 N/Am.


2.

With F_{m }= q(v x B), F_{m
}is always perpendicular to v and to B
and to the plane that contains v and B, but
the angle between v, B (v,
B) can have any value.


3.

Let the magnetic field B be out of the page, south
be down and north be up in the plane of the page so that v
is up. The v,
B = 90^{0}.
 F_{m } = qvB sin 90^{o }= 1.6 x 10^{19
}C(3.2 x 10^{7 }m/s)(1.2 x 10^{3}
N/C/sm)(1) = 6.1 x 10^{15 }N.
 If you point the fingers of your right hand up (in the
direction of v) and then curl them toward B
(out of the page), your thumb points to the right or to
the east.




6.

F_{m }=
qvB sin v,B
= (10^{6 }C)(10^{3} m/s)(2 Ns/Cm)(0.50) =
10^{3} N.


7.

 In Fig. for #7a above, with the velocity v parallel
to the magnetic field B,
v,B
= 0 and sin 0^{o} = 0, so there is no force on the
electron. The electron continues to move with velocity v
into the page.
 In Fig. for #7b above, the velocity v is perpendicular
to the magnetic field B, v,B
= 90^{o} and sin 90^{o} = 1. At the
instant shown in the figure, the magnetic force F_{m
}is down.
Note: the force on a negative charge is opposite to that
on a positive charge. Since this force is perpendicular
to the velocity, it does not change the magnitude of the
velocity, but it changes its direction. As the direction
of the velocity changes, the direction of F_{m
}changes so it always remains perpendicular to the
velocity and the path is a circle.
 At some other angle with the magnetic field, the electron
will continue to move with constant speed in the direction
of the component of B which is parallel to velocity
of the electron and proceed in a circle from the force due
to the component of B which is perpendicular to the
velocity of the electron. The resultant path is a spiral.
 There is an acceleration of the electron in parts (b)
and (c).


8.

F_{net }= ma 

qvB sin 90^{o}
= mv^{2}/r or
v = qBr/m 
v_{A }= qBr_{A}/m
and
v_{B }= qBr_{B}/m 
v = 2 πr/T,
where T is the period of the motion. 
v_{A} = 2 πr_{A}/T_{A}
= qBr_{A}/m. T_{A} = 2 πm/qB 
v_{B} = 2 πr_{B}/T_{B}
= qBr_{B}/m. T_{B} = 2 πm/qB
= T_{A}.



9.

From #8, r = mv/qB. r_{2 }= m_{2}v/qB.
r_{1 }= m_{1}v/qB. m_{2}/m_{1
} = r_{2}/r_{1} = 22/20.


10.

 F_{e }= kQq/r^{2} = 9.0 x10^{9}
Nm^{2}/C^{2}(10^{4 }x 2 x 10^{5})C^{2}/10^{4}m^{2} =
18 x 10^{4} N from q to  Q.
 F_{m }= qvB sin 90^{o} = 2 x 10^{5}(4)Ns/m
v
 If the particle moves
 clockwise
the magnetic force is away
from the center of the circle
(Fig. for #10b) and
 counterclockwise
the magnetic force is toward
the center of the circle (Fig. for #10a).

F_{net }= ma
Using F_{m }into the center and positive
as in Fig. for #10a,
kQq/r^{2} + qvB = mv^{2}/r or
mv^{2}/r  qBv  kQq/r^{2} = 0.
v = {(qB) ± [(qB)^{2} + 4mkqQ/r^{3}]^{1/2}}/2m/r
v =
v = {8 ± [784]^{1/2}}m/s/2
= {8 ± 28}m/s/2 = 18 m/s or
10 m/s.
The velocity of 18 m/s corresponds to counterclockwise motion
of the particle and the negative velocity of 10 m/s to clockwise
motion of the particle.


11.

 As the positive particle with charge +q, mass m, and velocity
v_{o} down enters Region 1 with an electric
field E to the right (the positively charged particle would
be urged from the positive plate toward the negative plate),
it experiences an electric force F_{e }to the right
equal to qE. In order for the +q to pass through the electric
field with no deflection, there must be a magnetic force
F_{m }to the left. This magnetic force must be perpendicular
to the magnetic field B and to the velocity v_{o.
}For v_{o }down and F_{m }to
the left, B cannot be up or down or to the
right or left because B is always perpendicular to
F_{m} and in this problem B is said
to be perpendicular to v. B must be
out of the page because only then will rotating v
into B give F_{m }to the left. F_{m
}= q(v_{o }x B). Since
we are told that B is perpendicular to v_{o},
v_{o},
B = 90^{o} and sin 90^{o} = 1. Thus,
F_{m} = qv_{o}B. For no deflection,
F_{net} = ma = 0 because a = 0. Taking to the right
to be positive, qE  qv_{o}B = 0 or
B = E/v_{o} = 10^{6} N/C/(10^{6}
m/s) = 1.0 Ns/Cm = 1.0 N/Am = 1.0 T out of the page.
 If,
 v = 5 x 10^{5 }m/s < 10^{6} m/s,
qE > qvB and the particle is deflected to the
right.
 v = 5 x 10^{6 }m/s > 10^{6} m/s,
qvB > qE and the particle is deflected toward
the left. This arrangement "selects" a velocity
of a particle, which allows it to enter the slit of
Region II. For this reason, the apparatus is called
a "velocity selector."
 As the particle enters region 2 without an electric field,
it experiences a force to the right, which makes it go in
a counterclockwise circle. For a velocity down and a magnetic
force to the right, the new magnetic field B’
must be into the page. Again B’ is perpendicular
to v_{o} and the magnetic force which produces
a centripetal acceleration is F_{m }= qvB’.
Now,
F_{net } = ma, where
a is the centripetal acceleration
qv_{o}B’ = mv_{o}^{2}/R
or
R = mv_{o}/qB’ = 2 x 10^{12 }kg(10^{6
}m/s)/(10^{6 }C)(2 Ns/Cm) = 1 m.



13.

 The Fig. for #13 above shows a side view of the inclined
plane and the bar that carries a current out of the page.
The forces acting on the bar are the magnetic force F_{m
}to the left, the gravitational force mg, and
the normal force N.
 F_{m }= ILB. For the bar to remain at rest,
(F_{net})_{y }= 0 and
(F_{net})_{x }= mg sin 22.5^{o}
 F_{m} cos 22.5^{o }= 0 or
mg tan 22.5^{o} = ILB.
I = mg tan 22.5^{o}/LB = (1.2 kg)(9.8 m/s^{2})(0.414)/(2.0
m)(0.50 N/Am)
= 4.87 A.


14.

I have taken up as north in Fig. for #14. In b of this figure,
you see that the compass points north. When the compass needle
is placed above a long current carrying wire (Fig. for #14c)
the needle points to the west. Point the thumb of your right
hand down. Above the wire your fingers and the magnetic field
lines are counterclockwise. The tangent to the field lines is
to the west, the direction of the magnetic field. The current
in the wire that produces this deflection of the compass needle
is south.


15.

The magnetic field lines
due to a current I into the page are clockwise. The tangent
to the magnetic field line at any point gives the direction
of the field at that point.
At P_{1}, the magnetic field is down.
At P_{2}, the magnetic field is to
the left.




16.

The magnetic field B due to a very long
wire is directly proportional to the current I and inversely
proportional to the distance r from the wire.
 If I is doubled, B is doubled. If r is halved, B
is also doubled.
If I is doubled and r is halved, B goes up
by a factor of 4.
The new B is 4 x 0.50 T = 2.0 T.
 If I is halved, B is halved. If r is doubled, B
is halved.
If I is halved and r is doubled, B goes down
by a factor of 4.
The new B is 0.50 T/4 = 0.125 T.


17.

 The magnetic field lines for I_{a} (labeled a)
and the magnetic field lines for I_{b} (labeled
b) are shown in the Fig. for #17 above. I_{a}
= I_{b} = I. At P_{1}, the magnetic
field_{ }due to I_{a} is B_{a }=
µ_{o}I/2 πr
= B_{b, }the magnetic field due to I_{b}.
B_{b }=  B_{a}. The
resultant magnetic field at P_{1} is zero.
 At P_{2}, the magnetic field B_{a }due
to I_{a} equals µ_{o}I/2 πr,
and the magnetic field B_{b }due to I_{b }equals
µ_{o}I/2 π
(3r), where r is the distance of P_{2} from I_{a}.
Both fields are up. The total field at P_{2}
= µ_{o}I/2 πr
+ µ_{o}I/6 πr
= 2µ_{o}I/3 πr
up.
 A proton out of the page at P_{2} experiences
a force to the left since v is out of the page and
B is up.


18.

The magnetic field lines for very long currentcarrying wires
are circles with the wire at the center. To find the direction
of the field line, point the thumb of your right hand in the
sense of the current and your fingers curl in the direction
of the field.
In the figure for both I_{1 }and I_{2} the direction
of the field lines are counterclockwise. Notice I have drawn
both so they pass through P because that is where I wish to
find the field. The magnetic field at P is tangent to the circle
at P.
For I_{1}, the circle has AP as a radius and the
field due to it B_{1 }is perpendicular to the
radius and tangent to the field line so it is to the left and
up, as shown in the figure above.
Similarly, the radius for the field lines for I_{2}
is CP, and B_{2} is perpendicular to CP and tangent
to its field line, so it is to the left and down.
The angle between B_{1} and B_{2 }is
60^{o }because B_{1} is perpendicular
to AP and B_{2} is perpendicular to CP. Two angles
that have their sides mutually perpendicular are equal.
Since I_{1} = I_{2} = 3 A and
r = 1.0 m, B_{1} = B_{2}
= µ_{o}I/2 πr
= 2 x 10^{7 }Tm/A (3A)/1m = 6 x 10^{7} T.
B_{1y }= B_{2y} = B_{1} sin 30^{o},
but B_{1y }is positive, while B_{2y }is
negative so there is no component of B in the Ydirection.
B = 2B_{1x} = 2B_{2x }= 2B_{1} cos 30^{o}
= 2B_{2} cos 30^{o} =
2(6 x 10^{7 }T)(0.866) =10.4 x 10^{7 }T in
the negative Xdirection.


19.

In Fig. for #19a above, I_{1} sets up a field
B_{1 }at P_{1} up. A magnetic field
line with I_{1} at the center would be counterclockwise.
A tangent to this field line at P_{1} is up. When
I_{2} is placed at P_{1}, it experiences
a force to the left.
In Fig. for #19b above, I_{2} sets up a field
B_{2 }at P_{2} down. A magnetic
field line with I_{2} at the center would be counterclockwise.
A tangent to this field line at P_{2} is down. When
I_{1} is placed at P_{2}, it experiences
a force to the right.
The field approach correctly predicts that two wires carrying
currents in the same direction attract each other.


20.

The magnetic field due to a very long wire is B = µ_{o}I/2 πr.
For the long wire carrying a current I = 10 A at the position
of side ea
of the loop with r = 0.01 m = 10^{2} m,
B = 2 x 10^{7 }N/A^{2}(10A)/10^{2
}m = 2 x 10^{4 }N/Am.
At the position of the side of the loop cd,
with r = 0.03 m = 3 x 10^{2 }m,
B = 2 x 10^{7}N/A^{2}(10A)/(3x10^{2}m)
= 2/3(10^{4)}N/Am.
Along ed
and ac,
the field varies with the distance r from the wire, but is the
same for the same r.
To the right of the long wire, the magnetic field of the wire
is into the plane of the paper.
The force F_{m }on a wire carrying a current
I’ with length L in a magnetic field B is F_{m
}= I’(L x B), where the direction of
L is taken in the sense of I’ in that wire. For
all lengths of the wire, L,
B_{due to the long wire} is 90^{o}
and sin 90^{o} = 1. For the same value of r, B
at that position on the length de
or ac
is the same. The force on an element of length there is the
same magnitude, but opposite in direction because L_{de}
is to the left and L_{ac} to the right. The net
force on L_{de} and L_{ac} is zero.
F_{ea }= (20 A)(3 x 10^{2 }m)(2 x 10^{4}
N/Am) sin 90^{o} = 12 x 10^{5 }N to the right.
F_{cd }= (20 A)(3 x 10^{2 }m)(2/3 x 10^{4
}N/Am) sin 90^{o }= 4 x 10^{5 }N to the
left.
Net force on loop = (12  4) x 10^{5} N = 8 x 10^{5
}N to the right.


21.

There will be equal forces up and down on the upper and lower
parts of the loop, respectively, but they will not produce any
torque about the Yaxis. The force on length cd with I = 1.0
A, L = 0.25mj, and B = 0.5 N/Am i
is
F_{m} = 1.0 A(0.25 m j x 0.5 N/Am i)
= 0.125 N k. The torque τ
= (r x F), where r is drawn from the axis
to the point of application of the force F. The angle
between r and F_{m } is 60^{o}.
The magnitude of r is 0.20 m and the magnitude of F_{m }is
0.125 N. τ =
(0.20 m)(0.125 N) sin 60^{o} = 0.022 Nm. The
direction of the torque from τ
= r x F_{m }is in the negative
y or  j direction.


22.

To find the direction of the field at P, point the thumb
of your right hand at the upper (u) portion of the loop, out
of the page. For this element, you can think of the magnetic
field lines as being a concentric circle about it. At
P, the tangent to the line would be in the direction of B_{u}.
Now go to an element at a lower portion of the loop. Again
point your thumb there in the sense of the current or into
the page. At P, the field due to this current at P is
shown as B_{l}. For every other element of
the loop, the magnetic field vectors will make a cone about
P. The components of the vectors in the Y and Z direction
cancel, leaving the direction of the field in the positive
X direction. When you look at a circular coil carrying a current
in a counterclockwise direction, the magnetic field is along
the axis of the coil, outward.


23.

If the velocity v is parallel
to the axis of the loop, it is parallel to the magnetic
field, v,B
= 0, and the force on it is zero. The electron will continue
merrily on its path.


24.

 For a very long solenoid, the magnetic field inside the
solenoid, far from either end, is a constant with B_{solenoid}
= µ_{o}nI_{solenoid}, where n is the
number of turns per unit length.
B = (4 π x 10^{7
}N/A^{2})(1000/m)(0.20 A)^{ }= 2.5
x 10^{4} N/Am = 2.5 x 10^{4 }T.
When viewed from the right end of the solenoid, the current
is counterclockwise. The magnetic field points to the right,
as shown in the figure above.
 The magnetic field due to the very long wire B_{wire}
= µ_{o}I_{wire}/2 πr.
For a current in the wire to the left, below the axis of
the solenoid at P the magnetic field due to the currentcarrying
wire is out of the page. The resultant field due to both
makes an angle of 45^{o} when
B_{solenoid }= B_{wire } or
µ_{o}nI_{solenoid } = µ_{o}I_{wire}/2 πr
or
r = (I_{wire}/I_{solenoid})/2 πn
= (6.0/0.2)/(2000 π
m^{1})
= 4.8 x 10^{3 }m = 4.8 mm.


25.

In the figure above the dashed circles represents paths around
which you can evaluate Ampere's law. There is also a
path for a < r < b that is solid with arrows to denote
a magnetic field. Notice that B,ds
= 0.
From Ampere's law, (current
enclosed)
By symmetry for all cases, B,ds
= 0, and B is constant everywhere on the surface so,
In general, B2 πr
= µ_{o }(current enclosed)
 For r ≤ a, B2 πr
= µ_{o }(0) = 0
 For a ≤ r ≤ b, the current
enclosed = NI, where N = the number of turns. B2 πr
= µ_{o }NI and
B = µ_{o}NI/2 πr.
 For r ≥ b, the current enclosed = I  I = 0 and
again B = 0.


26.

There are two problems concerning magnetic
fields, just as there were for electric fields.
 Given a distribution of current elements, find the magnetic
field due to these current elements. There are three approaches
to this problem:
 use the field due to a single long wire and then
find the field due to a number of them by vector addition;
 use the BiotSavart law for unsymmetrical situations;
and
 use Ampere's Law for symmetrical situations.
 Given a magnetic field, find the force on
 a moving charged particle, F_{m }=
q(v x B)
for a positively charged particle and
 a currentcarrying wire of length L, F_{m
}= I(L x B).


27.

 The force dF = (6 A)(dx)B sin 90^{o} = (6 A)(dx
m)(1 + 3x^{2} m^{2})N/Am(1).
 Unit of dF is N. The direction of dF is in the
+Zdirection.
 dτ_{y} =
x dF sin 90^{o} = x(m) (6 A)(dx m)(1 + 3x^{2}
m^{2})N/Am
= 6x(1 + 3x^{2}
m^{2})Nm.
 dτ_{z} = 0.
dF is parallel to the Zaxis.
 F = 6N / m(1 + 3x^{2}m^{2})dx = 6(0.5 + 0.125)N = 3.75N.




30.

For r ≤ a, the current density = J_{r≤a
}= I/ πa^{2
} Current enclosed by circle of radius r = ( λ/ πa^{2}) πr^{2}
= Ir^{2}/a^{2.
}For a ≤ r ≤ b, current enclosed
= I
For b ≤ r ≤ c, the current density =
J_{b≤r≤c} =  I/ π(c^{2}b^{2})
Current enclosed = I[1  π(r^{2
} b^{2})/ π(c^{2
} b^{2})] = I[(c^{2 } r^{2})/(c^{2
} b^{2})].
For r ≥ c, current enclosed = I  I = 0.
Only the crosssections of the coaxial cylinders are shown in
the figure above. The dashed circles in the figure are the paths
drawn to evaluate
= µ_{o}I.
By symmetry, B is tangent to the path and constant for a given
r.
For r ≤ a or a ≤ r ≤
b, the magnetic field lines are counterclockwise for the
current of the inner cylinder out of the page. B,^{
}d=
0.
Since B is constant over the path, B (2 πr).
This is true for all values of r.
 For r ≤ a, B2 πr
= µ_{o}(current enclosed) = µ_{o}(Ir^{2}/a^{2}),
and
B
= µ_{o}(Ir/2 πa^{2})
 For a ≤ r ≤ b,
B2 πr = µ_{o}I and
B = µ_{o}I/2 πr
 For b ≤ r ≤ c
B2 πr = µ_{o}I[(c^{2
} r^{2})/(c^{2 } b^{2})]
B = µ_{o}I[(c^{2 }
r^{2})/(c^{2 } b^{2})]/2 πr
 For r ≥ c B2 πr
= 0 and B = 0.


31.

Number of wires/length = N/w. Number of wires in length
dx = (N/w)dx.
 For length dx, dB = Nµ_{o}I dx/2 πrw
= Nµ_{o}I dx/2 πw(x^{2
}+ a^{2})^{1/2}.
 If you point the thumb of your right hand in the direction
of the current at x, the magnetic field line through point
P is counterclockwise. The tangent to the field line at
P is perpendicular to the radius r of the circle in the
direction of dB in the figure above.
For a corresponding dx on the left hand of the origin, the
dB would be perpendicular to the line from dx to P. You
can see from the figure that the components of the dB's
in the Ydirection dB_{y} cancel. The resultant
component is in the negative Xdirection.
dB_{x} = dB cos Θ =
Nµ_{o}(I/w) dx/2 π(x^{2
}+ a^{2})^{1/2} (a/(x^{2 }+
a^{2})^{1/2})
= Nµ_{o}(I/w)a dx/2 π(x^{2
}+ a^{2}).
Let x = a tan Θ.
dx = a sec^{2}ΘdΘ.
dB_{x} = Nµ_{o}(I/w)a^{2} sec^{2}
ΘdΘ/2 π(a^{2
}tan^{2 }Θ^{
}+ a^{2})
= Nµ_{o}(I/w)a^{2}
sec^{2} ΘdΘ/2 πa^{2}sec^{2
}Θ
= (Nµ_{o}I/2w π)
dΘ.
The lower limit on x is  w/2 and the upper limit on
x is + w/2.
The lower limit on Θ
is tan ^{1 }(w/2)/a and the upper limit
on Θ
is tan^{1} (+ w/2)/a.
B =(Nµ_{o}I/2 π)
d Θ
= (Nµ_{o}I/2w π)[tan^{1}
(+w/2)/a  tan^{1} (w/2)/a]
= (Nµ_{o}I/w π)[tan^{1}
w/2a].


32.

 The magnetic moment µ = I πR^{2}.
 B = µ_{o}I πR^{2}/2 π(R^{2}
+ z^{2})^{3/2} becomes
B = µ_{o}µ/2 π(R^{2}
+ z^{2})^{3/2}.
 For z >> R, you can neglect R in the denominator
of B and then you have 2 π(z^{2})^{3/2}
= 2 πz^{3} and
B becomes
μ_{o}μ/2 πz^{3}.
 B = 2 x 10^{7} N/A^{2} (9.27 x 10^{24}
Am^{2})/(10^{10 }m)^{3} = 1.9
N/Am = 1.9 T.
 The spinning electron gives a field and magnetic moment
opposite in direction to a positive charge moving in a loop.
Since both change sign, the net effect for the potential
energy will be the same as for current loops.
That is U
=  μ ^{.} B
=  µB cos μ,B.
 When they are aligned they will have minimum energy.
 Antialigned U = µB
=
9.27 x 10^{24 }Am^{2} x 1.9 N/Am/(1.6
x 10^{19} J/eV)
=
1.1 x 10^{4 }eV.


