Download Physics 213 — Problem Set 7 —Solutions Spring 1998

Physics 213 — Problem Set 7 —Solutions
Spring 1998
1. Reading Assignment
Serway Sections 29.1–5. Next week Chapter 30.
2. Serway 29.6A
A metal ball having net charge Q is thrown out of a window horizontally at a speed v. The window is at
a height h above the ground. A uniform horizontal magnetic field of magnitude B is perpendicular to the
plane of the ball’s trajectory. Find the magnetic force acting on the ball just before it hits the ground.
SOLUTION: Though this problem can be done by solving the force equation for the motion of the ball, it
is much simpler to use conservation of energy. The initial energy of the ball is the sum of its kinetic and
potential energies, Einitial = mgh + mv 2 /2. Since the magnetic force is always perpendicular to the motion
of the ball it does no work on the ball, and so cannot change its energy. Thus its final energy (when it hits
the ground) will be the same as its initial energy, but will now all be kinetic, Ef inal = mvf2 /2. This allows
us to solve for the final velocity of the ball as
vf =
p
2gh + v 2 .
Since the magnetic force acting on the ball when it hits is Fmag = Qvf × B, its magnitude is
Fmag = Qvf B = QB
p
2gh + v 2 ,
where we have used the fact that the ball is always moving in a plane perpendicular to the magnetic field.
3. based on Serway 29.20A
A rod of mass m slides frictionlessly on two parallel rails (Fig. P29.20 in your text) that are a distance d
apart and have length L. The rod carries a current I in the direction shown. If the rod starts from rest,
what is its speed as it leaves the rails if there is a uniform magnetic field B directed perpendicular to the
rod and the rails?
~ = I(d~ × B).
~ Since d~ and B
~ are
SOLUTION: The force on a straight current carrying wire is given by F
perpendicular to each other, the magnitude of the force is FB = IdB, and using the right hand rule, the
direction is found to be parallel to the rails, (towards the right in figure P29.20). Hence the acceleration
is a = IdB/m. Since the the rod starts from rest, using the equation vf2 = vi2 + 2ax, where vi and vf
are the
and final velocities, and x is the distance travelled, we get the final speed of the rod is
√ initialp
vf = 2aL = 2IdBL/m.
4. Serway 29.26
A wire is formed into a circle having a diameter of 10 cm and placed in a uniform magnetic field of
3 × 10−3 T. A current of 5 A passes through the wire. Find (a) the maximum torque on the wire and (b)
the range of potential energy the wire possesses for different orientations. (Hint for (b): Read Section
26.6 of your text.)
SOLUTION:
a)The magnitude of the magnetic moment of the loop is given by µ = IA, where A = πr2 is the area of
~ So the magnitude of the torque is
τ =µ
~ × B.
the loop. So µ = (5π(0.05)2 ) = 3.9 × 10−2 Amp-m2 . Now ~
given by τ = µB sin θ. Hence the maximum torque occurs when sin θ = 1. Hence the maximum torque is
τ = (3.9 × 10−2 )(3 × 10−3 ) = 1.18 × 10−4 N-m.
~ = −µB cos θ. The range of U depends on the fact that
b)The energy of the system is given by U = −~
µ·B
cos θ ranges between +1 and −1. Hence the range of the potential energy is −1.18 × 10−4 J to 1.18 × 10−4
J.
5. Serway 29.38
The accelerating voltage that is applied to an electron gun is 15 kV, and the horizontal distance from
the gun to the viewing screen is 35 cm. What is the deflection caused by the vertical component of the
Earth’s magnetic field (4 × 10−5 T), assuming that any change in the horizontal component of the beam
velocity is negligible.
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SOLUTION: The speed of the electron after leavingp
the gun is found from the kinetic energy of the electron,
which is 12 mv 2 = eV . Hence the final speed is v = 2eV /m = 7.3 × 107 m/s, where we used the fact that
e = 1.6 × 10−19 C and m = 9.1 × 10−31 Kg. After coming out of the gun it follows a circular path in
the presence of the vertical field. From mv 2 /R = evB, we find that the radius of this circular path is
R = mv/(eB) = 10 m. Using this information it is now just a matter of geometry to find the deflection.
Instead, we will estimate the deflection. Since R >> d, the distance to the viewing screen, it is a good
approximation to assume to leading order that the electron deflection is negligible. Then the force from the
magnetic field can be taken to be a constant force of magnitude F = evB acting perpendicular to the path
of the electron. In a time t such a constant force will cause a deflection δ of δ = F t2 /(2m). The electron
takes a time t = d/v to reach the screen. Putting it all together gives a deflection
δ=
(1.6 × 10−19 )(4 × 10−5 )(.35)2
eBd2
=
= 6 mm.
2mv
2(9.1 × 10−31 )(7.3 × 107 )
6. Serway 29.62
An electron enters a region traveling perpendicular to the linear boundary of a 0.10 T magnetic field.
The direction of the field is perpendicular to the velocity of the electron. (a) Determine the time it takes
for the electron to leave the “field-filled” region, given that it travels a semicircular path. (b) Find the
kinetic energy of the electron if the maximum depth of penetration in the field is 2 cm.
SOLUTION: Once the electron enters the magnetic field, the force due to the field acts as a centripetal
force. Hence we have mv 2 /R = qvB, where R is the radius of the circular path, and v is the electron’s
speed. Hence the electron’s angular velocity in the circular path is given by ω = v/R = qB/m.
(a) To move in a semicircle, it takes half its period T. Hence the time spent in the field is T /2 = (2π/ω)/2 =
πm/(qB). Using |q| = 1.6 × 10−19 C and m = 9.1 × 10−31 Kg, we get time spent t = 1.8 × 10−10 s.
(b) From the above equation for the force, we get v = qBR/m. Here R = 0.02 m. Hence the kinetic energy
of the electron is K.E. = mv 2 /2 = q 2 B 2 R2 /(2m) = 5.6 × 10−14 J.
7. Serway 29.68
A uniform magnetic field of magnitude 0.15 T is directed along the postive x axis. A positron moving at
5 × 106 m/s enters the field along a direction that makes an angle of 85◦ with the x axis (see Fig. P29.68
in your text). The motion of the particle is expected to be a helix, as described in Section 29.4 of your
text. Calculate (a) the pitch p and (b) the radius r of the trajectory.
SOLUTION:
a)The pitch is the distance the positron moves in the positive x-direction in the time that it completes one
circle around the axis. From the previous problem we know that the period of revolution is T = 2π/ω =
2πm/(qB). Hence the distance it moves parallel to the x-axis in time T is d = v|| T = v cos(85◦ )2πm/(qB) =
1.04 × 10−4 m.
b)The radius is R = mv⊥ /(qB) = 1.89 × 10−4 m, where v⊥ = v sin(85◦ ).
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