Quiz 7 Solutions
1. Using Ampere’s law, find the magnetic field (magnitude and direction) a distance d
below a horizontal long straight wire carrying a constant current I to the left. Explicitly
state and explain all the steps in the calculation in detail.
d
•P
Pick a circular contour centered on the wire and lying in a
perpendicular plane. Then since by symmetry B must point circumferentially –
tangent to the circle,
point in the same direction. So
.
But B can only depend on r, the radial distance from the wire, and not on z, the
distance along the wire, or θ the angle around the wire, by symmetry. Therefore
on the circular contour B is a constant, since r = d on the circle, and the integral
becomes simply B(2πd). The enclosed current is just I and so solving for B we
find B = µoI/2πd. The direction is out of the paper, by the right hand rule.
2. Suppose three long straight wires, each carrying the same constant current I, are parallel
to one another as shown with I = 2A and d = 0.5 m. Find the magnetic field (separate
magnitude and direction) at point P.
•
P
X
d
2d
d
X
Using (1), we have
net
= [µoI/2πd] + [µoI/2πd] + [µoI/2π(2d)]
1/2
[µoI/2πd] [ + (3/2) ]; So, Bnet = [µoI/2πd] [1+9/4]
an angle of 56.3o from the x-axis (tan-1 3/2).
=
-6
= 1.44 x 10 T, pointing at
3. The long wire shown carries a constant current I. Find the magnetic field (magnitude and
direction) at point P at the center of the semi-circle of radius R. Can you solve this
problem easily using Ampere’s law?
•P
Can not use Ampere’s law easily because of a lack of symmetry. Using the Biot
Svart law, we see that the straight portions do not contribute to B at P since
is zero for those parts (with along . For the circular part,
=
where the direction is into the paper. Since
R is constant and the integral of ds is simply πR, we have at P that
directed into the paper.
4. Consider the system of linear equations
and let A =
. Then A is the matrix of coefficients of the given system.
a. Compute |A|, the determinant of A.
Expanding down the second column, we obtain:
b. What does your answer to part a tell us about the number of solutions to the given
system?
Since |A| ≠ 0, we know that the given system has a unique solution.
c. Solve for z, using Cramer’s rule.
Let A3 =
.
Then, |A3| =
.
By Cramer’s rule, it follows that
5. Let F(x,y,z) = 4zi + 2xj + 3yk and let C be the be the rectangle in the xy plane with
vertices (0,0,0), (0,3,0), (2,3,0), and (2,0,0), with orientation clockwise when viewed
from above. Use Stoke’s Theorem to compute
.
First, we compute curl(F):
Let S be the filled-in rectangle with boundary C. k and -k are both unit normals
to S. The given orientation to C implies that we must let n = -k. Then,
-2 (area of S) =
-2 (2)(3) =
-12