PHYS-340 - Problem Set 10 - SolutionsSolution by: Olivier Trottier

November 26, 2012

In the following, there is an integral that comes up many times. I will state its value for once and invokeit when necessary. ∫

dx

(x2 + z2)3/2 = x

z2√x2 + z2

(1)

11 a)We use the Biot-Savart law to find the total magnetic field on the x-y plane. The 3 contributions to thismagnetic field will come from the left straight wire (1), the parabolic wire (2) and the right straight wire(3).

Btot = B1 +B2 +B3

As a reminder, here is the Biot-Savart Law:

B(x, y, z) = µ0I

4π

∫dl× r|r|2

For the first wire, the calculus is pretty straightforward.

r = (x, y, 0), r′ = (0, y′, 0), dl1 = (0, dy′, 0),r = (x, y − y′, 0), dl1 × r = (0, 0,−xdy′),

B1(r) = µ0I

4π

∫ 0

−∞

−xdy′z(x2 + (y − y′)2)3/2

= −µ0I

4π

∫ −y/x−∞

(1x

)dz′z

(1 + z′2)3/2

= −µ0I

4πxz′√

1 + z′2

∣∣∣∣−y/x−∞

z

= −µ0I

4πx

(−y

x√

1 + (y/x)2−(

limz′→−∞

z′√1 + z′2

))z

= µ0I

4πx

(y√

x2 + y2− 1)z

In the integral above, we have used the change of variable z′ = y′−yx . Also, we have assumed that the

point (x, y, 0) is not on the y-axis (x 6= 0). Otherwise, the B-field vanishes because the vector productis zero in that case. The second wire is a little tricky. We won’t compute the integral as the result is

1

pretty messy.

r = (x, y, 0), r′ = (x′, (x′)2, 0), dl2 = (dx′, 2xdx′, 0), (dy′ = 2x′dx′)r = (x− x′, y − x′2, 0), dl2 × r = (0, 0, (y − x′2 − 2x′(x− x′))dx′)

= (0, 0, (y − x2 + (x′2 − 2xx′ + x2))dx′

= (0, 0, (y − x2 + (x− x′)2))dx′

B2(r) = µ0I

4π

∫ 1

0

(y − x2 + (x− x′)2)dx′z((x− x′)2 + (y − x′2)2)3/2

For the third wire, the calculations are similar to ones for the first wire. One can simply replace x→ x−1in the formula for B1(r).

r = (x, y, 0), r′ = (1, y′, 0), dl1 = (0, dy′, 0),r = (x− 1, y − y′, 0), dl1 × r = (0, 0,−(x− 1)dy′),

B3(r) = µ0I

4π

∫ 0

−∞

−(x− 1)dy′z((x− 1)2 + (y − y′)2)3/2

= . . .

= µ0I

4π(x− 1)

(y√

(x− 1)2 + y2− 1)z

1 b)In this problem, we fix the origin to be at the center of both circles of wire and we assume that thecircular loops have a common center. When I talk about the bigger circle of wire, I refer to the part ofthe longer wire that is not squarish. Also, when I’ll use polar coordinates, the direction of increasingangle will be counterclockwise (opposite to the currents’ direction!).

First, we will find the angle that subtends the missing arcs of the bigger loop. Since the setup is centered,the missing arcs will have the same length on both sides so I only have to focus on one side. Let α bethe angle subtending the missing arc. Then, by doing some simple trigonometry, we find:

sin(α

2

)= a2

2(R+ a1) ⇒ cos(α

2

)=

√1− a2

24(R+ a1)2 =

√4(R+ a1)2 − a2

24(R+ a1)2

We separate the B-field into 3 parts: the inner wire loop (1), the outer wire loop (2) and the squares ofwire (3). For the inner wire, the B-field is easily found using Biot-Savart.

r = (0, 0, 0), r′ = Rr′, dl = −Rdθ′θ′,r = −Rr′, dl× r = R2dθ′z

B1(r) = µ0I

4π

∫ 2π

0

R2dθ′z

R3

= µ0I

2R z

2

The contribution of the outer circle of wire is similarly derived with the difference that the the angularintegral misses an angle of 2α and the radius is now R+ a1. So,

B2(r) = µ0I

4π

(∫ π−α/2

α/2

(R+ a1)2dθ′

(R+ a1)3 +∫ 2π−α/2

π+α/2

(R+ a1)2dθ′

(R+ a1)3

)z

= µ0I(π − α)2π(R+ a1) z

For the square part, we focus on the one on the left hand side and we will multiply by 2 at the end:B3 = 2BSquare)

r′up = (x′, a2

2 , 0), dlup = (dx′, 0, 0),

rup = (−x′,−a2

2 , 0), dlup × rup =(−a2

2

)dx′z

r′vert = (d, y′, 0), (d = R cos(α

2

)+ a2) dlvert = (0,−dy′, 0),

rvert = (−d,−y′, 0), dlvert × rvert = (−d) dy′z

r′down = (x′,−a2

2 , 0), dldown = (−dx′, 0, 0),

rdown = (−x′, a2

2 , 0), dldown × rdown =(−a2

2

)dx′z

In the above, d is the distance of the vertical wire from the center. Also, we see that the upper and lowerwire have an equal contribution.

Bsquare(r) = µ0I

4π

2∫ R cos(α2 )+a2

R cos(α2 )−a2

2 dx′((

a22)2 + x′2

)3/2 +∫ a2

2

− a22

−d dy′

(d2 + y′2)3/2

z

= −µ0I

4π

(a2)

x′(a22)2√

x′2 +(a22)2∣∣∣∣∣∣R cos(α2 )+a2

R cos(α2 )

+ (d)

y′

(d)2√y′2 + d2

∣∣∣∣∣a22

− a22

z

= −µ0I

4π

( 4a2

) d√d2 +

(a22)2 − R cos

(α2)√

(R cos(α2))2 +

(a22)2+

(1d

) a2√(a22)2 + d2

z

= −µ0I

4π

(

4a2d

)√d2 +

(a2

2

)2− Rd√

1 +(

a22R cos(α2 )

)2

z

= − µ0I

πa2d

√d2 +(a2

2

)2− Rd√

1 + a22(R+a1)2

R2(4(R+a1)2−a22)

z

In the case where the the inner loop is surrounded by a square wire of radius R+a1, one replaces B2 +B3by a B-field oriented in the −z direction. The norm is given by the result of question c) after substitutingn = 4 and r = R+ a1.

1 c)This problem is simple once one realizes this fact: the orientation of the segments of the n-gon don’tchange the magnitude of the B-field at the origin. In other words, one just have to find the contributionof one side and multiply the result n-times to get the B-field created by the whole loop.

3

First, let’s remember some facts about polygons. An n-gon can be constructed by gluing n isoscelestriangles together. The angle between the common edges of each isosceles triangle is given by 2π

n . Thecommon edges have a length a = `

2 sin(πn ) where ` is the length of the n edges that form the regular

n-gon. Also, the center of each of the n edges is a distance b = `

2 tan(πn ) away from the center of the

polygon. Using Pythagoras’ theorem, one finds: a2 = b2 +(`2)2.

Since the circular loop is circumscribed by the polygon, one has b = r which implies ` = 2r tan(πn

)and

a = r

cos(πn ) . It remains to find the magnetic field of a rod of length `. We assume that this rod is verticaland that its endpoints are a distance a from the center. We then find the usual quantities required bythe Biot-Savart law. We also assume that the current is flowing counterclockwise in the loop.

r = (0, 0, 0), r′ = (b, y′, 0), dl = (0, dy′, 0),r = (−b,−y′), dl× r = bdy′z

Bside(r) = µ0I

4π

∫ `/2

−`/2

bdy′z

(b2 + y′2)3/2

= µ0Ib

4π

y′

b2√b2 + y′2

∣∣∣∣∣`/2

`/2

z

= µ0I

4πb

`√b2 +

(`2)2 z

= µ0I`

4πab z

= µ0I

2πr sin(πn

)z

⇒ Btot,n = µ0nI

2πr sin(πn

)z

⇒ B� =√

2µ0I

πrz, B∆ = 3

√3µ0I

4πr z

Since limn→∞ n =∞ and limn→∞ tan(πn

)= 0, we can use L’Hospital rule to find the limit.

Btot ≡ limn→∞

Btot,n = limn→∞

µ0nI

2πr sin(πn

)z

= µ0I

2πr limn→∞

sin(πn

)1/n z

= µ0I

2πr limn→∞

(cos(πn

)) (−πn2

)−1/n2 z

= µ0I

2r limn→∞

cos(πn

)z

= µ0I

2r z

This is indeed the magnetic field created by a circular loop of radius r as shown by Griffiths in eq. 5.38(setting z = 0).Another problem related to this one is one where the polygon is circumscribed by the circular loop ofradius r. In this case, one has a = r which implies ` = 2r sin

(πn

)and b = r cos

(πn

). Then, using the

4

formula above with the proper values for a, b and `, one finds:

Bside(r) = µ0I`

4πab z

= µ0I

2πr tan(πn

)z

⇒ Btot,n = µ0nI

2πr tan(πn

)z

⇒ limn→∞

Btot,n = µ0I

2r z

Obviously, the limit is the same.

1 d)First of all, the work that we need to provide in order to rotate the loop comes from the fact that wehave to work against the torque created by the straight wire. In other words:

W =∫ π

0τ · dθ

where dθ is oriented along the axis of rotation. Now let’s put the straight along the z-axis and assumethat the loop is rotated by an angle θ with respect to its axis of rotation (θ = 0 corresponds to the initialsetup). The x-axis is directed to the right and the y-axis is into the page. The origin is on the straightwire and is aligned with the center of the loop. Recall the expressions for the magnetic force and thetorque

dF = Idl×B, dτ = r × dF

r is the vector joining the point where the infinitesimal magnetic force dF is applied to the closest pointon the axis of rotation of the loop. Using a vector identity and the factor that r and dl are perpendicular,we can simplify the expression of the torque.

dτ = I(dl(r ·B)−B(r · dl)) = Idl(r ·B)

As it is usually the case in this kind of setup, the magnetic force on the horizontal segments of the loopwill cancel and the magnetic force on the vertical segments will add. We will identify the left (right)vertical segment with 1 (2).

r1 = a

2 (cos(π + θ), sin(π + θ), 0) = a

2 (− cos(θ),− sin(θ), 0)

r2 = a

2 (cos(θ), sin(θ), 0)

B1 = µ0I

2πd1φ1 = µ0I

2πd1(− sin(φ1), cos(φ1), 0)

B2 = µ0I

2πd2φ2 = µ0I

2πd2(− sin(φ2), cos(φ2), 0)

d1 =√(

c+ a

2

)2+(a

2

)2− 2

(c+ a

2

)(a2

)cos(θ)

d2 =√(

c+ a

2

)2+(a

2

)2+ 2

(c+ a

2

)(a2

)cos(θ)

dl1 = (0, 0, dz′), dl2 = (0, 0,−dz′)

r1 ·B1 = −µ0Ia

4πd1(− sin(φ1) cos(θ) + cos(φ1) sin(θ)) = −µ0Ia

4πd1(sin(θ − φ1))

r2 ·B2 = µ0Ia

4πd1(− sin(φ2) cos(θ) + cos(φ2) sin(θ)) = µ0Ia

4πd2(sin(θ − φ2))

5

To find the value of sin(θ− φ1) and sin(θ− φ1) in terms of θ we use the law of sines (I inserted a sketchof the setup to help with the geometry).

α = θ − φ2, β = −(π + θ − φ1)sin(α)c+ a

2= sin(π − θ)

d2= sin(θ)

d2

sin(β)c+ a

2= sin(θ)

d1

⇒ sin(θ − φ1) = sin(−π − β) = sin(β) =c+ a

2d1

sin(θ)

⇒ sin(θ − φ2) = sin(α) =c+ a

2d2

sin(θ)

ΘΦ1

Φ2

2 Π � Φ1

Α

Β

Θ

a

2

a

2

c �a

2

d2

d1

y

x

Figure 1: Bird-view of the setup of question 1d)

We are then ready to find the total torque τ and the work of a 180◦ rotation.

τ = µ0I2a sin(θ)4π

(c+ a

2

)∫ b/2

−b/2

(−dz

′

d21

+ −dz′

d22

)z

= −µ0I2ab sin(θ)

4π

(c+ a

2

) 2((c+ a

2)2 +

(a2)2)((

c+ a2)2 +

(a2)2)2

−(2(c+ a

2) (

a2))2 cos(θ)2

z

⇒W =∫ π

0(τ z) · (dθz)

= µ0I2ab

2π

(c+ a

2

)((c+ a

2

)2+(a

2

)2)∫ π

0

− sin(θ)dθ((c+ a

2)2 +

(a2)2)2

−(2(c+ a

2) (

a2))2 cos(θ)2

= µ0I

2ab

2π

(c+ a

2

)((c+ a

2

)2+(a

2

)2) 1

2(c+ a

2) (

a2) ((

c+ a2)2 +

(a2)2) arctan

2(c+ a

2) (

a2)x((

c+ a2)2 +

(a2)2)

∣∣∣∣∣∣π

0

= µ0I

2b

2π arctan

(c+ a

2)aπ((

c+ a2)2 +

(a2)2)

6

1 e)By passing through this potential difference, the particle has gained kinetic energy. We can find thenorm of the initial horizontal velocity by using conservation of energy.

QU0 = 12mv

20 ⇒ v0 =

√2QU0

m

For the second part of the problem, let’s assume that the plates are in the x-y plane and are separatedalong the z-direction by a distance `. Without loss of generality, let’s assume that one plate is at z = 0and the other is at z = `. Since there is no charge in between the plates, the electric potential in betweenthe plates solves the 1d Laplace’s equation in vacuum.

d2V (z)dz2 = 0⇒ V (z) = mz + b.

Using the given boundary conditions (V (0) = 0, V (`) = U), we can find the electric field inside theplates.

V (z) = U

`z ⇒ E = −U

`z

Now, let’s assume that the particle initially travels in the +x direction. Since the particle has to travelin a straight line, we have to cancel the electric force by a magnetic force in the −z direction. Twistingyour right hand for a couple of seconds, you can figure out that the B-field must have the form:

B = By

To find the value of B, we impose a vanishing net force.

F = Q(E + v ×B)

= Q(−U`z + v0Bx× y)

= Q(−U`z + v0Bz)

= 0

⇒ B = U

v0`= U

`

√m

2QU0

1 f)First, we work out the case where the wires are perpendicular and lying in the same plane, say the x-yplane. We will assume that the I1 current flows along the x-axis and the I2 current flows along they-axis. Both wires create a magnetic field along the z-direction.

Since the particle is constrained to move on the plane, the only possibility that it feels no magnetic forcealong a particular path is if the magnetic field along that path vanishes. From that fact, one realizesthat there are 2 options: I1 = I1x, I2 = I2y or I1 = −I1x, I2 = −I2y. Without loss of generality, wewill focus on the first option. The total magnetic field is easily found from common formulas:

B(x, y) =(µ0I12πy −

µ0I22πx

)z

So the particle will feel no magnetic force along the path that satisfies B(x, y) = 0.

B(x, y) = 0⇒ y

x= I1I2

or y = I1I2x

This no more and no less than a straight line in the x-y plane with slope I1I2.

7

In the parallel case, let’s put the straight wires along the y direction as shown in the picture below.Also, we will put the point M at the origin. The wires are therefore positioned at z1 = −a2 and z2 = a

2with currents I1 and I2 flowing in them, respectively. Since the magnetic field cancels at M, we deducethat the currents have the same direction and value: I1 = I2. We can also deduce that the currentsmust flow in the −y direction by using the information that the magnetic field points up (+z direction)along the positive side of the x-axis. Then, I1 = I2 ≡ −Iy. Using the Biot-Savart law, we can find themagnetic field in the x-z plane.

x

z

ya

I2

I1

Figure 2: Bird-view of the setup of question 1f)

r = (x, 0, z),

r1′ = (0, y′, a2 ), dl1 = (0,−dy′, 0),

r1 = (x,−y′, z − a

2 ), dl1 × r1 = (−(z − a

2

)dy′, 0, xdy′),

r2′ = (0, y′,−a2 ), dl2 = (0,−dy′, 0),

r2 = (x,−y′, z + a

2 ), dl2 × r2 = (−(z + a

2

)dy′, 0, xdy′),

B1(r) = µ0I

4π

∫ ∞−∞

dy′

(x2 +(z − a

2)2 + y′2)3/2

(−(z − a

2

)x+ xz

)

= µ0I

4π

y′((x2 +

(z − a

2)2)√

x2 +(z − a

2)2 + y′2

∣∣∣∣∣∣∣∞

−∞

(−(z − a

2

)x+ xz

)

= µ0I

2π−(z − a

2)x+ xz(

(x2 +(z − a

2)2)

Sending a→ −a, we can find the magnetic field created by the wire with current I2.

B2(r) = µ0I

2π−(z + a

2)x+ xz(

(x2 +(z + a

2)2)

8

Btot(r) = µ0I

2π

− (z − a2)x+ xz(

(x2 +(z − a

2)2) +

−(z + a

2)x+ xz(

(x2 +(z + a

2)2)

As a check, the total magnetic field is indeed vanishing at the origin. Also, on the negative side of thex-axis, the magnetic field will point in the −z direction. Finally, we find the point on the x-axis wherethe field is maximal.

Btot(z = 0, x) = µ0I

2π

a2 x+ xz − a

2 x+ xz((x2 +

(a2)2)

= µ0I

π

x((x2 +

(a2)2)

z

We now maximize the B-field magnitude with respect to x.

d

dxBtot(z = 0, x)

∣∣∣∣x=x0

= 0

⇒ 1((x2

0 +(a2)2) − 2x2

0((x2

0 +(a2)2)2 = 0

⇒ x20 =

(a2

)2

So the magnetic field is maximized at x0 = ±a2 .

1 g)Let’s first focus on the problem of figure 4. For this problem, we put the origin at the center of the biggercircle of wire. The x and y axes point right and up, respectively. Hence, the center of the of the smallercircle is positioned at y = R1 −R2.

We are interested in finding the magnetic field at r = (0, 0, 0). We will first find contributions of thecircles of wire and we will finish with the straight wires.

r′ = R1r, dl1 = −R1dθ′θ′

r = −R1r, dl1 × r = −R21dθ′z

B1 = µ0I

4π

∫ 3π/2

π/2

−R21dθ′

R31

z

= − µ0I

4R1z

For the second loop, since the center is not at the origin, we shift the origin at the center of the secondloop and find the magnetic field at the point r = (0,−(R1 − R2), 0) which is the position of the trueorigin according to this temporary system of reference.

r′ = R2r, dl2 = R2dθ′(− sin(θ′), cos(θ′), 0)

r = (−R2 cos(θ′),−(R1 −R2)−R2 sin(θ′), 0), dl2 × r =(R2

2 + (R1 −R2)R2 sin(θ′))dθ′z

B2 = µ0I

4π

∫ π/2

−π/2

(R2

2 + (R1 −R2)R2 sin(θ′))dθ′

(R22 + (R1 −R2)2 + 2(R1 −R2)R2 sin(θ))3/2 z

The evaluation of the integral above is pretty messy so I will left it like so.

9

To calculate the contributions of the straight wires, we will first derive general formulas for the magneticfield generated at the origin by a horizontal and vertical wire. In the above formulas, s = ±1 dependingon the direction of the current.

1) Vertical wire which endpoints have coordinates (x0, y1, 0), (x0, y2, 0), y1 < y2.

r′ = (x0, y′, 0), dl = (0, sdy′, 0),

r = (−x0,−y′, 0), dl× r = (0, 0, sx0dy′)

Bvert(r) = µ0I

4π

(∫ y2

y1

sx0dy′

(x20 + y′2)3/2

)z

= µ0Isx0

4π

y′

x20√x2

0 + y′2

∣∣∣∣∣y2

y1

z

= µ0Is

4πx0

(y2√x2

0 + y22− y1√

x20 + y2

1

)z

2) Horizontal wire which endpoints have coordinates (x1, y0, 0), (x2, y0, 0), x1 < x2.

r′ = (x′, y0, 0), dl = (sdx′, 0, 0),r = (−x′,−y0, 0), dl× r = (0, 0,−sy0dx

′)

Bhor(r) = µ0I

4π

(∫ x2

x1

−sy0dx′

(y20 + x′2)3/2

)z

= −µ0Isy0

4π

(x′

y20√y2

0 + x′2

∣∣∣∣∣x2

x1

)z

= −µ0Is

4πy0

(x2√y2

0 + x22− x1√

y20 + x2

1

)z

We can now find the contribution to the magnetic field for each wire. We label them by their length: a,b, c,..., as shown in the picture.

Wire a: x1 = −a, x2 = 0, y0 = −(R1 + b), s = +1

Ba = µ0I

4π(R1 + b)

(a√

(R1 + b)2 + a2

)z

Wire b: This wire creates not magnetic field at the origin because, in this case, dl × r = 0 since dl and rare parallel.

Wire c: x1 = 0, x2 = c, y0 = −(R1 − (f − d− b)), s = +1

Bc = µ0I

4π(R1 − f + d+ b))

(c√

(R1 − f + d+ b)2 + c2

)z

Wire d: y1 = −(R1 − f + d+ b)), y2 = −(R1 − f + b), x0 = c, s = +1

Bd = µ0I

4πc

(−(R1 − f + b)√c2 + (R1 − f + b)2

+ R1 − f + d+ b√c2 + (R1 − f + d+ b)2

)z

10

Wire e: x1 = −a, x2 = c, y0 = −(R1 − f + b), s = −1

Be = µ0I

4π(f −R1 − b)

(c√

(R1 − f + b)2 + c2+ a√

(R1 − f + b)2 + a2

)z

Wire f: y1 = −(R1 + b)), y2 = −(R1 − f + b), x0 = −a, s = −1

Bf = µ0I

4πa

(−(R1 − f + b)√a2 + (R1 − f + b)2

+ R1 + b√a2 + (R1 + b)2

)z

The total magnetic field is then given by:

Btot = B1 +B2 +Ba +Bb +Bc +Bd +Be +Bf

We now solve the problem of figure 5. First of all, the horizontal wires will have no contribution because,for these parts, dl× r = 0 since dl and r are parallel. As was calculated above, the magnetic field at thecenter of a semi-circle of radius R carrying a current I is given by:

Bsemi-circle = −µ0I

4R z

The total magnetic field will then be given by:

B = −µ0I

4

∞∑n=0

(−1)n

R+ naz

= −µ0I

4a Φ(−1, 1, Ra

)

where Φ(z, s, a) is the well-known Lerch zeta-function!

2The solutions for question 2) are provided by Keshav.

11