8.02pset8sol

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MASSCHUSETTS INSTITUTE OF TECHNOLOGY

ESG Physics

8.02 with Kai Spring 2003

Problem Set 8 Solution

Problem 1: 30-12

Determine the magnetic field (in terms of I , a and b) at the origin due to the current loopshown below.

Solution:

The magnetic field at point P due to a finite length is

( )01 2cos cos

4

I B

r

µ θ θ

π = − (1.1)

where 1 2andθ θ are the angles shown above.

So, to find the magnetic field at O, we can use the above formula. We let point P to be

the origin, and 2θ be the angle nearer to the x-axis.

The problem can be divided into 3 parts:

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Part I

Consider the left part of the wire. When the current is infinitely far away from the x-axis,

1 0θ → , so

1cos 1θ = (1.2)

and

22 2

cos d

d aθ =

+ (1.3)

Therefore

( )0 01 1 2

2 2cos cos 1

4 4

I I d B

a a d a

µ µ θ θ

π π

= − = −

+ (1.4)

And 1 B is out of page

Part II

Consider the part of the wire which is parallel to the x-axis. Denote 1θ to be the angle on

the left. We have r d = and

12 2

cos a

a d θ =

+ (1.5)

( )2 1 12 2

cos cos cos a

a d θ π θ θ = − = − = −

+ (1.6)

Therefore,

0 02

2 2 2 2 2 24 2

I Iaa a B

d a d a d d a d

µ µ

π π

= + =

+ + + (1.7)

2 B is into the page

Part III

Consider the part of the wire on the right. It is not difficult to observe that

3 1 B B= (1.8)

As a result, the total magnitude of the magnetic field is

1 2 3 1 22tot = + + = +B B B B B B

(1.9)

0 0

2 2 2 2

?12 2

I Iad a a d d a d

µ µ π π

= − − + +

k k (1.10)

( )2 20 0

2 2 2 2

?

2 2

I Iad a d

a a d d a d

µ µ

π π = + − −

+ +k k (1.11)

( )2 2 2 20

2 2

ˆ

2

I d a d d a

ad a d

µ

π = + − −

+k (1.12)

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Problem 2: 30-28

In the figure below, both currents are in the negative x direction.

(a)

Sketch the magnetic field pattern in the yz plane.

(b) At what distance d along the z axis is the magnetic field a maximum?

Solution

(a)

Currents are into the paper

(b)Combining the contribution of the magnetic field from each wire,

the z components cancel each other on the z axis. Therefore

ˆ B=B j

(2.1)

Denote1 B as the magnitude of the contribution of magnetic field

by one of the wire, then

( )0 0

1 2 2

22 sin 2

2 2

I Iz z B B

r r a z

µ µ θ

π π

= = =

+ (2.2)

where r is the distance from the point on the z axis to one of the

wire.

Therefore

( )0

2 2

Iz B

a z

µ

π =

+ (2.3)

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To find the maximum of B, we set 0dB

dz = , so we get

2

0

2 2 2 2

1 20

I dB z

dz a z a z

µ

π

= − =

+ + (2.4)

( )

2 20

22 2

0 I a z

a z

µ

π

− = +

(2.5)

which gives

z a= (2.6)

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Problem 3 30.34

A solenoid 2.50 cm in diameter and 30.0 cm long has 300 turns and carries 12.0 A.(a) Calculate the flux through the surface of a disk of radius 5.00 cm that is positioned

perpendicular to and centered on the axis of the solenoid.

(b) Figure P30.34b shows an enlarged end view of the same solenoid. Calculate the flux

through the blue area, which is defined by an annulus that has an inner radius of 0.400 cmand outer radius of 0.800 cm.

Solution:

(a)

B BAΦ = ⋅ =B A

(3.1)

where A is the cross sectional area of the solenoid.. Then

( )2 7.40 Wb1

o B

NI r

µ π µ

Φ = =

(3.2)

(b)

( )( )2 20

2 11

B NI BA r r µ π Φ = ⋅ = = − B A (3.3)

which gives

2.27 Wb B µ Φ = (3.4)

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Problem 4: 30-35

Consider the hemispherical closed surface in Figure P30.35. If the hemisphere is in a

uniform magnetic field that makes an angle θ with the vertical, calculate the magnetic

flux

(a)

through the flat surface 1S and(b) through the hemispherical surface 2S

Solution:

(a)

( )2 2

, cos 180 cos B flat B R B Rπ θ π θ Φ = ⋅ = ° − = −B A

(4.1)

(b) The net flux out of the closed surface is zero, therefore2

, , cos B curved B flat B Rπ θ Φ = −Φ = (4.2)

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Problem 5: 30.61

Solution:

(a)

40 2.74 10 T2

I Br

µ π

−= = × (5.1)

(b) At point C , conductor AB produces a field ( )( )412.74 10 T

2

−× −

, conductor DE

produces a field of ( )( )412.74 10 T

2

−× − j

, BD produces no field, and AE produces

negligible field. The totle field at C is ( )42.74 10 T−× −

(c)

31.15 10 N B I −= × = ×F L B (5.2)

(d)

( )-20.384msm

= =∑Fa i (5.3)

(e) The bar is already so far from AE that it moves through nearly constant magnetic

field. The force acting on the bar is constant, and therefore the bar’s acceleration

is constant

(f)

2 2 2 f iv v ax= + (5.4)

this gives

( )0.999m/s f v = i (5.5)

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Problem 6 : 30-67

A wire is bent into the shape shown below and the magnetic field is measures at 1 P when

the current in the wire is I .

The same wire is then formed into the shape shown below, and the magnetic field is

measured at 2 P when the current is again I .

If the total length of wire is the same in each case, what is the ratio of 1

2

B

B?

Solution:

For the first figure, we can divide the wire into four equal segments with length l . Then,

we can apply equation (2.1) with 90θ 1 = ° and 2 135θ = ° for the four segments, so the

total magnetic field at 1 P is

( )0 0 0

1 1 2

14 cos cos 04 2 2

I I I B l l l

µ µ µ θ θ π π π

= − = + = (7.1)

For the second figure, since the length of wire is the same, we have 4 R l π = , so

4l R

π = (7.2)

For a circular loop of wire, the magnitude of magnetic field at the center is

02

4

I B

R

µ θ

π = (7.3)

Since in this case, θ π = , thus

( )002 4 16

4

I I B l l

µ π µ π

π π

= =

(7.4)

Therefore

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0

1

202

8 221.15

16

I

B l

I B

l

µ

π

µ π π

= = ≈

(7.5)

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Problem 7: 31-20

Example 2

Consider the arrangement shown below.

Assume that 6.00 , 1.20 m R l = Ω = , and a uniform 2.50-T magnetic field is directed into

the page. At what speed should the bar be moved to produce a current of 0.500 A in theresistor?

Solution:Since we know that

Blvε = (9.1)

so we can express the current by

Blv I

R R

ε = = (9.2)

Substituting the given values, we can find that

1.00m/sv = (9.3)

8.02pset8sol

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