2. (20 pts) The Biot-Savart Law gives the magnetic field at a point in space due to a piece of current,
d~
B=µ0
4π
Id~
ℓ׈r
r2
Apply this to the point a distance aabove a long, straight wire carrying current Ito the right. Work
out the integral that would need to be performed in order to obtain the net magnetic field at that point.
(You do not need to perform the integration. Just write the integral, complete with appropriate limits.)
Hint: Take the point to be above the middle of the wire...
a
x=0
dl
θ
r
-x
I
It’s a long wire, so it extends off to infinity in both directions. Putting point aabove the middle of the
wire locates it at x= 0 m. That makes the math for the section of current d~
ℓa bit easier.
Working out the integral means that the only remaining mathematical step is doing the actual in-
tegration. If there are things remaining in your solution that can’t be directly integrated, then you
didn’t work out the problem completely. For instance, you’d need to work out any cross-products before
integrating. So if there’s a cross-product in your final answer, it’s not really the final answer.
So we’ll need to work out d~
ℓ,r, and the cro ss-product between d~
ℓand ˆrbefore writing down the integral.
Since the current is to the right, d~
ℓ= ˆıdx. Recall that ˆrpoints from the little piece of current towards
point a. So we can work out the cross-product with the right-hand rule. It gives out of the page for all
parts of the current (even those parts on the x > 0portion of the wire). Thus d~
ℓ׈r=ˆ
ksin θ dx. (Don’t
forget the sin θin the cross-product!) We’ll also need r2=x2+a2, since ris the hypotenuse of a right
triangle with sides ±x(depending on which side of the origin you are on) and a.
Putting it all back together gives
d~
B=µ0
4π
Id~
ℓ׈r
r2=µ0
4π
Isin θ dx
x2+a2ˆ
k
But wait, as we integrate over x, the angle θalso changes! So we need to write sin θin term of x. Using
trigonometry, we get
sin θ=a
(x2+a2)1/2
Substituting this into the integral we get
d~
B=µ0
4π
Ia dx
(x2+a2)3/2ˆ
k
All we need to do now is sum over all pieces of the wire, which extends from x→ −∞ to x→ ∞, giving
~
B=
∞
Z
−∞
µ0
4π
Ia dx
(x2+a2)3/2ˆ
k
