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Assignment (1-83)
Read (21.3 and 21.5), Do PROBLEMS # (11, 13, 29) Ch. 21
11. REASONING AND SOLUTION
An ionized helium atom has mass of 6.6 x 10-27 kg, and a speed of 4.4 x 105 m/s. The atom moves
perpendicular to a 0.75 Tesla magnetic field on a circular path of radius 0.012m Determine whether the
charge of the ionized helium atom is +e or +2e. The charge can be found from developing the equation
from the relationship between the centripetal force (Fc) and magnetic force (FB).
Fc = FB
1
Letv2 F
m
=c =q FvBB [sin θ]
r
q =
So that,
Since 1 e = 1.6 × 10
–19
c
hc
gb
h
6.6 × 10 −27 kg 4 .4 × 10 5 m / s
mv
=
= 3.2 × 10 −19 C
Br
0 .75 T 0.012 m
b
g
C, we see that the charge of the ionized helium is +2e
? ? ..
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13. REASONING AND SOLUTION
A beam of protons moves in a circle of radius of 0.25 m. The protons move perpendicular to a 0.30 T
magnetic field.
a. The speed of a proton can be found from Equation 21.2 ( r = mv / qB ),
v=
qBr (1.6 × 10 –19 C)(0.30 T)(0.25 m)
=
= 7.2 × 10 6 m / s
–27
m
1.67 × 10
kg
b. The magnitude Fc of the centripetal force is given by Equation 5.3,
mv 2 (1.67 × 10 –27 kg)(7.2 × 10 6 m / s) 2
Fc =
=
= 3.5 × 10 –13 N
r
0.25 m
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29. REASONING AND SOLUTION
A square coil of wire containing a single turn is placed in a uniform 0.25 T magnetic field. Each side has a
length of 0.32 m and the current in the coil is 12 A. Determine the magnitude of the magnetic force on
each of the four sides.
The force on each side can be found from F = ILB sin θ. For the top side, θ = 90.0°, so
F = (12 A)(0.32 m)(0.25 T) sin 90.0° = 0.96 N
The force on the bottom side (θ = 90.0°) is the same as that on the top side, F = 0.96 N .
For each of the other two sides θ = 0°, so that the force is F = 0 N .