Quizz #4 solutions
PROBLEM 1
a. Let’s assume wires are in the plane of the board, oriented vertically. They are separated by horizontal distance
d. Let’s assume the left current is I1 directed up and the right is I2 directed down. Magnetic field created by
0 I1
current #1 distance d from it is B1 = µ2πd
and due to the right hand rule it’s pointing into the plane.
~ 1 , so the force per length f~1→2 = I~2 × B
~ 1 . The
The force due to magnetic field on the wire is F~1→2 = LI~2 × B
µ0 I1 I2
current # 2 and magnetic field are perpendicular, so magnitude of the force f1→2 = I2 B1 = 2πd and the
direction is to the right, the second wire is repelled from the first one.
0 I2
, and the direction is again into the board (the direction of the current changed, but the
b. Analogously, B2 = µ2πd
I1 I2
point where we are calculating magnetic field is now to the left of the wire). So, f2→1 = I1 B2 = µ02πd
and the
direction is to the left. The first wire is also repelled from the second one.
c. Due to the third Newton’s law, we could expect the force due to interaction between two wires be equal in
magnitude and opposite in direction, as it is.
PROBLEM 2
a. We will assume the left wire is going out of the board and the right one is going into the board. Distance from
0I
.
the wires to point P is d/2, so the magnetic fields produced at the point P by the wires are both B1 = B2 = µπd
The direction of the fields, using right hand rule, is up for both of them. The result is, then, their sum B = 2µπd0 I .
~ 2 will be the same – second current is inversed, but now we are looking at
b. We may guess that the direction of B
the opposite side of the circle.
µ0 I
.
c. Point Q is distance d away from both wires, so magnitudes of the magnetic fields due to the wires B1 = B2 = 2πd
◦
The direction of B1 is perpendicular to the line connecting the left corner with Q and pointing 60 from vertical
line to the right. The direction of B2 is perpendicular to the line connecting the right corner with Q and pointing
60◦ from vertical line to the left. As magnitudes of B1 and B2 are the same, the net horizontal component of
the resulting magnetic field will be zero, so the magnitude of the resulting magnetic field will be equal to its
µ0 I 1
µ0 I
vertical component: B = By = B1 cos 60◦ + B2 cos 60◦ = 2 2πd
2 = 2πd .
PROBLEM 3
~ and pointing to the right as e < 0. So, this is linear motion with constant
a. Force due to electric field is F~ = eE
acceleration to the right. Electron increases its speed. a = F/m = |e|E
m .
b. Change in the kinetic energy equals to the total q
work done by the field: ∆(mv 2 /2) = |e|EL, initial kinetic
p2
2
energy is p2 /(2m), so final velocity is vboundary = m
2 + m |e|EL.
c. In the region of the constant uniform magnetic field, the charged particle moves in a spiral. In our case velocity
component along magnetic field is zero, so the motion is circular in the plane of the drawing. Magnetic field
doesn’t make work, so can’t change the kinetic energy of the particle, thus the magnitude of the velocity is
constant and equal to vboundary . The radius of the circle can be found by equating centripetal force to the
mv
magnetic force: mv 2 /R = |e|vB, so R = |e|B
.
d. Electron continues its circular motion until it leaves the area of magnetic field. This happens after making half
~ and is pointing down at the first moment, so the trajectory of
a circle. Force on the electron is F~ = −|e|~v × B
the electron in magnetic field will be half a circle below the point of entry.
2
e. As the electron moves along the half of the circle, going below its starting point,
and magnitude of the velocity is
r
√ 2
p2
2
m m2 + m |e|EL
p +2m|e|EL
mv
=
−2
.
constant (see above), the point of exit is below: ∆y = −2R = −2 |e|B = −2
|e|B
|e|B
q
p2
2
Magnitude of the velocity is the same, vexit = vboundary = m
2 + m |e|EL and the direction is to the left, or
opposite to the initial direction of motion of the electron.
PROBLEM 4
+i2x
~ =∇
~ ×A
~ = ( −i2y
a. B
x2 +y 2 , x2 +y 2 , 0).
p
b. B = x22i
y 2 + x2 = √ 2i
.
+y 2 ·
2
2
x +y
~ =
c. ~r · B
2i
x2 +y 2
· (x · (−y) + y · x) = 0.
~ is ∇
~ ·A
~ = ∂x Ax + ∂y Ay + ∂z Az = 0.
d. Divergence of A
e. Magnetic field is inversely proportional to the distance from z axis (part b), perpendicular to the radius-vector
from zero and perpendicular to the vector from z axis as well (part c). It resembles magnetic field produced by
infinite wire.
f. Let’s say x = rxy cos φ, y = rxy sin φ, where rxy is the distance from axis z to the point of interest. Nothing
will depend on the z coordinate and Bz = 0 as well. We may rewrite magnetic field using substitution above
~ = 2i( −rxy2 sin φ , rxy 2cos φ , 0) = 2i (cos(φ + π/2), sin(φ + π/2), 0). Any vector in x-y plane may be written
as B
rxy
rxy
rxy
~
~ and x-axis. Using that we can tell that angle
as Cxy = |C|(cos α, sin α), where α is an angle between vector C
between the vector B in x-y plane and axis x is φ + π/2, where radius vector to the point of interest has angle φ
with x axis. That means that in x-y plane magnetic field is tangential to radius-vector, pointing anti-clockwise
in the same way as magnetic field due to the infinite wire with current moving along z axis in the positive
direction.