Homework Hint
Assignment (1-85)
Read (21.9), Do PROBLEMS # (35, 51, 53) Ch. 21 + AP 2000 #7 (handout)
35. REASONING
The drawing below shows a side view of the conducting rails and the aluminum rod. Three forces act on
the rod:
(1) its weight (W = mg),
(2) the magnetic force (F = ILB sin θ) and
(3) the normal force (FN = mg cos θ)
! An application of Right-Hand Rule No. 1 shows that the magnetic force is directed to the left, as
shown in the drawing.
! Since the rod slides down the rails at a constant velocity, its acceleration is zero.
! If we choose the x-axis to be along the rails, Newton’s second law states that the net force along the
x-direction is zero:
ΣFx = max = 0.
That is,
! Using the components of mg and FB that are along the x-axis, Newton’s second law becomes
− F cos 30. 0° + mg sin 30. 0° = 0
144444244444
3
∑ Fx
The magnetic force is given by Equation 21.3 as F = ILB sin θ, where θ = 90.0° is the angle between the
magnetic field and the current. We can use these two relations to find the current in the rod.
Current (out
of paper)
FN
Aluminum
rod
F
x
mg
30.0 o
Conducting
rail
SOLUTION
Substituting the expression F = ILB sin 90.0° into Newton’s second law and solving for the current I, we
obtain
I =
b
mg sin 30.0 °
kerchunk…
=
LB sin 90.0 ° cos 30.0 °
g
b 0.20 kg gc 9.80 m / s h sin 30.0° =
.6 m gb 0.050 T g sin 90.kerchunk…
0° cos 30.0 °
b1kerchunk…
2
14 A
51.
REASONING
The two rods attract each other because they each carry a current (I) in the same direction.
The bottom rod floats because it is in equilibrium. The two (2) forces that act on the bottom
rod are the:
• Downward force of gravity (mg) and
I1
• Upward magnetic force (ILB) of attraction
to the upper rod.
I2
If the two rods are a distance (r) apart, the magnetic field (B) generated by the top rod at the
location of the bottom rod is given by the equation B = µ0 I
2πr
According to the FB = ILB sin θ, the magnetic force exerted on the bottom rod is:
FB = ILB sin θ = IL µ0 I sin θ
2πr
So that,
2
FB = µalgebra
?? θ
0 I L sin
2πr
where θ is the angle between the
magnetic field at the location of the
bottom rod and the direction of the
current in the bottom rod.
Since the rods are parallel, the magnetic field is perpendicular to the direction of the current
(RHR-2), and θ = 90.0 ° , so that sin θ = 1.0 .
SOLUTION
Taking upward as the positive direction, the net force on the bottom rod is
substitution
I 2 L sinand
µ
θ
0
mg
algebra−??
2π s
=0
Solving for the current (I), we find
I=
2 π mgs
=
µ0L
2 π (0.073 kg)(9.80 m / s 2 )( 8.2 × 10 –3 m)
= 190 A
(4 π × 10 –7 T ⋅ m / A)(0.85 m)
53.
REASONING AND SOLUTION
Two long straight wires are separated by 0.12 m. The wires carry currents of 8.0 amps in opposite
directions as shown. Find the magnitude of the net magnetic field at points A and B.
Let the current in the left-hand wire be labeled I1
and that in the right-hand wire I2.
a. At point A: B1 is up and B2 is down, so we subtract them to
get the net field.
We have
B1 = µ0I1/(2πd1) = µ0(8.0 A)/[2π(0.030 m)]
B2 = µ0I2/(2πd2) = µ0(8.0 A)/[2π(0.150 m)]
So the net field at point A is
BA = B1 − B2 = 4.3 × 10 −5 T
b. At point B: B1 and B2 are both down so we add the two.
We have
B1 = µ0(8.0 A)/[2π(0.060 m)]
B2 = µ0(8.0 A)/[2π(0.060 m)]
So the net field at point B is
BB = B1 + B2 = 5.3 × 10 −5 T