Solutions and Explanations for HW Chapter 28 BB Prepared by Prof. Omar Chmaissem and TA Abhilash Sajja Question 1: The magnetic field at a distance of 2 cm from a current carrying wire is 4 μT. What is the magnetic field at a distance of 4 cm from the wire? A) 4 μT B) 6 μT C) 2 μT D) 8 μT E) 10 μT Solution and Explanation: To calculate the magnetic field near a long wire, I would need to know the current magnitude going through it

and the distance to the point I am interested in. The magnetic field equation is r2π

IμB o

In this problem, the current is not given but I was given two situations that I can use to solve my problem;

1

o

1r2π

IμB and

2

o

2r2π

IμB . Since the current is the same in both cases, taking the ratio of the two fields is

straight forward: 1

2

2

1

r

r

B

B

Substituting the parameters B1 = 4 μT, r1 = 2 cm, r2 = 4 cm, we get B2 = 2 μT.

Question 2: At what distance from a long straight wire carrying a current of 5.0 A is the magnitude of the magnetic field due to the wire equal to the strength of the Earth's magnetic field of about 5.0 × 10-5 T? A) 4.0 cm B) 3.0 cm C) 5.0 cm D) 1.0 cm E) 2.0 cm Solution and Explanation: Straightforward. The same equation we use above. Magnetic Field of infinite length current carrying wire at

distance r is given by r

IB o

2

Substitute the parameters μo = 4 x 10-7 T.m/A, I = 5 A, B = 5.0 x 10-5 T in the above equation, we get r = 2.0 cm Question 3: A straight wire carries a 12 A current eastward and a second straight wire carries a 14 A current westward. The wires are separated by a distance of 42 cm. The force per unit length on the wire carrying 12 A is A) 5.1 × 10-4 N/m B) 5.1 × 10-7 N/m C) 8.0 × 10-7 N/m. D) 8.0 × 10-5 N/m E) 5.1 × 10-6 N/m

Solution and Explanation: If I1, I2 are in the same direction, the force is attractive If I1, I2 are in the opposite direction, the force is repulsive. Magnetic force per unit length in terms of the current is given by

r 2π

II μF 21o

Substituting the parameters μo = 4 x 10-7 T.m/A, I1 = 12 A, I2 = 14A, r = 0.42 m, we get

Force per unit length

F = 8.0 × 10-5 N/m

Question 4: Extra Credit: A solenoid with n1 = 1800 turns/m and a current I1 = 3 A is filled with a paramagnetic material at a temperature T1 = 250 K. A second solenoid with n2 = 1500 turns/m and a current I2= 1.1 A is filled with the same paramagnetic material at a different temperature T2. The magnetizations are the same in both cases. What is the value of T2? A) 76 K B) 150 K C) 137 K D) 100 K E) 200 K Solution and Explanation:

For a solenoid, magnetic field is given by nIB o

Also for paramagnetic materials, magnetization is given by

T

BCM (Curie’s law)

In the above question magnetization is said to be the same. Hence M1 = M2

2

2

1

1

T

B

T

B

2

22

1

11

T

In

T

In oo

Substituting the values n1 = 1800 turns/m, n2 = 1500 turns/m, I1 = 3A, I2 = 1.1A, T1 = 250K, we get T2 = 76 K

Question 5: Two coaxial circular coils of radius R = 6 cm are positioned a distance d = 2 cm apart, as shown in Fig. 28-11. Calculate the magnetic field half-way between the coils. Each coil carries a current I = 4 A.

FIGURE 28-11

A) 15.2 x 10-5 T B) 2.1 x 10-5 T C) 4.0 x 10-5 T D) 8.0 x 10-5 T E) 6.2 x 10-5 T

Solution and Explanation: Before anything, we need to figure out the directions of the magnetic field produced at midpoint by the two rings. Using the right hand rule (thumb in the direction of the current and wrapping our fingers around the loop) we find that the magnetic field is directed from left to right through each loop. At midpoint, both loops produce magnetic field contributions pointing in the same direction to the right side of this paper. Let’s calculate the magnitudes and add them directly up. The magnetic field produced by a circular coil is given by (check the slides and the corresponding solved example in the textbook).

3/222

2

o

Rx 4π

R 2π IμB

where R is the radius of the loop and x is the distance from the loop center to the point of

interest on the loop axis, here x = d/2 Substituting the parameters μo = 4 x 10-7 T.m/A, R = 0.06 m, x = d/2 = 0.01 m (I = 4 A, we get B = 4 x 10-5 T Since the magnetic fields of both the circular loops add up. Magnetic field half way between them is 2B = 2 (4 x 10-5) = 8 x 10-5 T.

Question 6: Two long parallel wires carry currents of 5.0 A and 8.0 A in the same direction. The wires are separated by 0.30 m. Find the magnetic force per unit length between the two wires. A) 2.7 x 10-5 N/m repulsive B) 7.2 x 10-5 N/m repulsive C) 7.2 x 10-5 N/m attractive D) 3.4 x 10-5 N/m repulsive E) 2.7 x 10-5 N/m attractive Solution and Explanation: Same as Question 3 but here the force is attractive (currents in the same direction). Magnetic force per unit

length between two wires is given by r 2π

II μF 21o

Again:

If I1, I2 are in the same direction, the force is attractive If I1, I2 are in the opposite direction, the force is repulsive. Substituting the parameters μo = 4 x 10-7 T.m/A, I1 = 5 A, I2 = 8A, r = 0.30 m, we get

Force per unit length

F = 2.7 × 10-5 N/m attractive.

Question 7: What is the value of the magnetic field at the center of a square wire of length a on the side, if the current in the wire is I ?

A) a

IB

B) B = 0

C) a

IB

2

D) aπ

Iμ22B

E) a

IB

2

Solution and Explanation: The square can be thought of as an assembly of 4 finite length current carrying wires. The magnetic field of a finite length wire can be obtained by adjusting the boundary conditions of the long wire problem we did in class. Draw the square together with faint lines from the square corners to its center. Let’s look at the left vertical side of the square. The starting boundary angle is 45 degrees and the final boundary angle is 135o.

414.1)707()707.0(cosR4π

Iμ

R4π

Iμ

R4π

Iμdθsinθ

R4π

IμB oo135

45

o135

45

o

1

P P

45o

135o

R=a/2

This is the magnetic field produced by the left side of the square Using the right hand rule, the magnetic fields produced by all the sides of the square add up to form 4xB1

14 BB

414.14

R4π

IμB o

with R = a/2 so the total field is:

aπ

Iμ2

a4π

IμB oo 2

22/

4

Question 8: Fig. 28-6 shows two long, parallel current-carrying wires. The wires carry equal currents I1 = I2 = 20 A in the directions indicated and are located a distance d = 0.5 m apart. Calculate the magnitude and direction of the magnetic field at a point located at an equal distance d from each wire. FIGURE 28-6

A) 4μT to the right B) 8μT upward C) 8μT downward D) 4μT downward E) 4μT upward

Solution and Explanation:

Magnetic Field of a current carrying wire at a distance d is given by d

IB o

2

Each of the two wires produces a circular magnetic field around it. At the point of interest (the bottom vertex of the equilateral triangle) we draw magnetic field vectors that are tangent to the circles. These are the magnetic field contributions from the two wires at this particular point. As you can see, the horizontal components will cancel out while the vertical components (upward) will add up.

To find the vertical components, we need the angle between the vectors and the red vertical axis. This can be determined as follows: 1- The triangle labeled with sides = d is an equilateral triangle with each of its angles being 60o. The red line

cut through it and divide it by two equal halves. So the angle between the red line and the left dashed side of the triangle (or the right one) equal 30o.

2- Next, we know that the tangent to the circle (i.e., the vectors plotted on the figure) must be perpendicular to the radii of the circles. For example, the angle between the left vector and the right dashed side of the triangle (this is the radius of the right circle) is 90 degrees.

3- Finally, subtract the right half of the vertex angle (30o) from 90o, we end up with 60o being the angle between the vector direction and the red vertical axis.

Having determined the angle, the vertical component of the magnetic vector is then B cos(60o) or

Td

IB oo 7104060cos

2

Parameters substituted here were μo = 4 x 10-7 T.m/A, I = 20 A, d = 0.5 m. We have two vertical components (one from each wire) that add up, so multiplying the previous result by 2, we get B = 8 μT pointing upward.

Question 9: Extra credit: Aluminum has a magnetic susceptibility = 1.7 × 10-5 at T = 300 K. What will be the magnetization of a small sample of aluminum placed in a 1.5 T magnetic field at T = 150 K? A) 81.1 A/m B) 10.2 A/m C) 1.5 A/m D) 20.3 A/m E) 40.6 A/m

Solution and Explanation:

Magnetic susceptibility H

MThis can be written as HM eq(9.1)

Also magnetization is given by

T

BCM (Curie’s law)

But HB

Hence, T

HCM

eq(9.2)

Comparing eq(9.1) and eq(9.2)

Curie’s constant o

TC

eq(9.3)

Substituting the parameters 1.7 x 10-5, μo = 4 x 10-7 T.m/A, T = 300 K in eq(9.3), we get C = 4.06 x 103 A.K/m.T

From Curie’s law T

BCM

Substituting the parameters B = 1.5T, T = 150 K, C= 4.06 x 103 A.K/m.T in eq(9.4), we get M = 40.6 A/m.

Question 10: How many turns should a 10 cm long solenoid have if it is to generate a 1.5 × 10-3 T magnetic field on 1.0 A of current? A) 15 B) 12 C) 1200 D) 3200 E) 120 Solution and Explanation:

Magnetic field anywhere inside the solenoid is given by nIB o where n is the number of turns per unit

length. Hence, the equation can be written as

NIB o

Substituting the parameters μo = 4 x 10-7 T.m/A, I = 1A, l = 0.1m, B = 1.5 x 10-3 T into the above equation, we

get N = 120 turns (approximately).

Question 11: A circular loop of wire of radius 10 cm carries a current of 6 A. The magnitude of the magnetic field at the center of the loop is A) 1.2 × 10-5 T B) 1.2 × 10-7 T C) 3.8 × 10-8 T D) 3.8 × 10-7 T E) 3.8 × 10-5 T Solution and Explanation:

Magnetic field produced by a circular coil at the center of the loop is given by R

IB o

2

Substituting the parameters μo = 4 x 10-7 T.m/A, I = 6A, R = 0.1m in the above equation, we get B = 3.8 x 10-5 T

Question 12: Fig. 28-9 shows the cross-section of a hollow cylinder of inner radius a = 5 cm and outer radius b = 7 cm. A uniform current density j = 1 A/cm2 flows through the cylinder. Calculate the magnitude of the magnetic field at a distance d = 10 cm from the axis of the cylinder. FIGURE 28

A) 2.5 × 10-4 T B) 0 C) 4.5 × 10-4 T D) 1.5 × 10-4 T E) 0.5 × 10-4 T Solution and Explanation: For this problem, we need to apply Ampere’s law (at least the first time to understand how it works and what equation to use when you encounter such a wire with a uniform current density). The magnetic field around the hollow cylindrical wire is circular (just like with solid wires). The magnitude of the magnetic field around the circle is constant because the distance to the center of the wire is constant. Assume the current is pointing into the page so the magnetic field direction it generates is clockwise. Now, let’s choose a circular path that mimics the magnetic field shape with exactly the same radius. The circular path encloses the entire current. To integrate around the closed path, let’s walk around the path also in a clockwise direction. With this choice, the vectors B and dl are always pointing in the same direction with the

angle between the two vectors equal 0 over the entire closed path.

)2(0cos dBdBBddB o

The above equation was the left side of Ampere’s law.

Now to the right side of the equation (o Ienclosed) which requires knowing the current enclosed. In this problem, we were not given the current but we were given a uniform current density J. We know that J = I/A or I = JA so we just need to multiply the current density by the cross section area A of the hollow cylinder.

A = (b2-a2)

The current Ienclosed is then )(* 22 abJI

)57(*1 22 I

3.75I A Equating the left side of the equation with its right side we get: enclIdB )2(

Hence, the magnetic field is given by r

IB enclo

2 which is the same as the equation for a regular straight wire.

Substituting the parameters μo = 4 x 10-7 T.m/A, I =75.3A, r = 0.1 cm (convert to meters) in the above equation, we get B = 1.5 x 10-4 T Question 13: A wire of thickness d = 6 mm is tightly wound 150 times around a cylindrical core to form a solenoid. A current I = 0.2 A is sent through the wire. What is the magnetic field on the axis of the solenoid? A) 1.2 × 10-5 T B) 4.2 × 10-5 T C) 9.8 × 10-5 T D) 2.1 × 10-5 T E) 5.7 × 10-5 T

Solution and Explanation:

Magnetic field on the axis or anywhere inside the solenoid is given by nIB o where n is the number of

turns per unit length. Hence, the equation can be written as

NIB o eq(13.1)

Length l of the solenoid is obtained as l = N(number of turns) x d(thickness of wire)

l = 150 x 0.006 = 0.9 m

Substituting parameters μo = 4 x 10-7 T.m/A, N = 150, I = 0.2A, l = 0.9m in eq(13.1), we get

B = 4.2 x 10-5 T Question 14: A semicircular wire of radius R connects two straight wire segments. If a current I flows along the wire, the magnitude of the magnetic field at the center of the semicircular wire due to the current in the semicircular wire is A) μoI/2R B) μoI/4πR C) μoI/8πR D) μoI/4R E) μoI/8R Solution and Explanation: We’ve done this one in class using a quarter circle. Follow the same steps and change the boundary limits. Brief, the magnetic field at the center of a semicircular wire will be given by

dR

IB o

24

RR

IB o

24 (integral around a semicircle is R )

R

IB o

4

Try thinking about other circular segments and see how you need to change the boundary limits accordingly. Question 15: A solenoid with 300 turns has a radius of 0.040 m and is 40 cm long. If this solenoid carries a current of 12 A, what is the magnitude of the magnetic field at the center of the solenoid? A) 4.9 mT B) 11 mT C) 6.0 mT D) 16 mT E) 9.0 mT Solution and Explanation: Straightforward. Magnetic field on the axis of the solenoid is given by

nIB o where n is the number of turns per unit length

Hence, the equation can be written as

NIB o

Substituting the parameters μo = 4 x 10-7 T.m/A, N = 300, I = 12A, = 0.4 m in the above equation, we get B = 11 mT.

Question 16: Two long parallel wires carry currents of 20 A and 5.0 A in opposite directions. The wires are separated by 0.20 m. What is the magnetic field midway between the two wires? A) 5.0 × 10-5 T B) 1.0 × 10-5 T C) 3.0 × 10-5 T D) 2.0 × 10-5 T E) 4.0 × 10-5 T Solution and Explanation:

Magnetic Field of infinite length current carrying wire at a distance r is given by r

IB o

2

Magnetic Field of wire 1 r

IB o

2

1

1 eq(16.1)

Magnetic Field of wire 2 r

IB o

2

2

2 eq(16.2)

The direction of the magnetic fields produced by both the wires is the same at midpoint (check with the right hand rule) and they get added up.

Hence, the total magnetic field 21 BBB

r

I

r

IB oo

22

21

Substituting the parameters μo = 4 x 10-7 T.m/A, r = 0.1 m (this is the distance to the midpoint between the two wires), I1 = 20A, I2 = 5A in the above equation, we get B = 5.0 x 10-5 T Question 17: Four long parallel wires each carry 2.0 A in the same direction. They are parallel to the z-axis, and they pass through the corners of a square of side 4.0 cm positioned in the x-y plane. What magnetic field does one of the wires experience due to the other wires? A) 2.1 × 10-6 T B) 0 T C) 1.2 × 10-5 T D) 1.2 × 10-6 T E) 2.1 × 10-5 T Solution and Explanation: Draw circles, radii and tangents. The tangent vectors must be perpendicular to the radii at the wire location. You should obtain three magnetic field vectors (one from each wire). These can be qualitatively added up (vector addition) then calculated numerically as follows:

Magnetic Field of infinite length current carrying wire at distance r is given by r

IB o

2

Magnetic Field due to adjacent wire 04.02

21

oB

Magnetic Field due to another adjacent wire 04.02

22

oB

Magnetic Field due to diagonally placed wire 204.02

23

oB

The total magnetic field on a wire is then the sum of these three vectors. Notice that B1 and B2 are equal and magnitude and would add up in the diagonal direction. This is the same direction as B3. So the sum of B1 and B2 will be directly added to B3.

321 45cos45cos BBBB oo

B = 2.1 x 10-5 T

Question 18: A long wire carrying a 2.0-A current is placed along the y axis. What is the magnitude of the magnetic field at a point 0.60 m along the x-axis? A) 6.7 T B) 12 T C) 6.7 × 10-7 T D) 1.2 × 10-7 T E) 13 × 10-7 T Solution and Explanation:

Straightforward. Magnetic Field of infinite length current carrying wire at distance r is given by r

IB o

2

Substituting the parameters μo = 4 x 10-7 T.m/A, I = 2 A, r = 0.60 m in the above equation, we get B = 6.7 x 10-7 T

Question 19: A high power line carrying 1000 A generates what magnetic field at the ground, 10 m away? A) 2.0 × 10-5 T B) 3.2 × 10-6 T C) 5.6 × 10-5 T D) 4.7 × 10-6 T E) 6.4 × 10-6 T Solution and Explanation:

Magnetic Field of infinite length current carrying wire at distance r is given by r

IB o

2

Again, to show you how powerful is this equation and what type of questions may it may be useful for. Substituting the parameters μo = 4 x 10-7 T.m/A, I = 1000 A, r = 10 m in the above equation, we get B = 2.0 x 10-5 T

Question 20: Fig. 28-10 shows a segment of wire of length L = 1.25 m, carrying a current I = 12 A. Calculate the magnitude of the magnetic field produced by this wire at a point P located at a distance D = 1.25 m from one of the ends of the wire. FIGURE 28-10

A) 1.1 × 10-7 T B) 2.7 × 10-7 T C) 6.8 × 10-7 T D) 5.4 × 10-7 T E) 7.9 × 10-7 T Solution and Explanation: This is a finite length wire so we need to adjust the boundary limits like we did with the square problem above. Notice that the point of interest is shifted upward (opposite the top end of the wire). Fortunately, the distances L and D are the same so we pretty much can construct a square with L and being the two sides. Draw a line from the bottom end of the wire to point P. This is the distance r. The unit vector r (hat) points towards P. The angle between dl and r-hat is 45 degrees. This is the lower boundary limit. As dl slides all the way to the top, the distance r becomes the same as the dashed line labeled D. Vector dl points upward while r-hat

points to the right. The upper boundary angle is therefore 90 degrees. Let’s plug these into the equation:

Substituting the parameters μo = 4 x 10-7 T.m/A, I = 12A, D = 1.25 m in the equation above, we get B = 6.8 x 10-7 T

707.0)707()0(cosD4π

Iμ

D4π

Iμ

D4π

Iμdθsinθ

D4π

IμB oo90

45

o90

45

o1

Variations of the problem: Try making L and D different so you would end up with a rectangle instead of a square. Use trig functions to determine the lower angle boundary. You can also shift the point P to other places (perhaps opposite the center of the wire or something easy to deal with).

Question 21: Two long parallel wires carry currents of 10 A in opposite directions. They are separated by 40 cm. What is the magnetic field in the plane of the wires at a point that is 20 cm from one wire and 60 cm from the other? A) 3.3 × 10-6 T B) 3.3 × 10-5 T C) 1.5 × 10-6 T D) 6.7 × 10-6 T E) 6.7 × 10-5 T Solution and Explanation: This is neat. The point of interest is not between the wires. It’s to the side of one of the two wires. Draw the two parallel lines and label the distances. Now apply the right hand rule one wire at a time. You will see that at our point of interest the magnetic fields produced by the two wires are in OPPOSITE directions. Notice they will be in the same direction between the wires. Let me emphasize that this analysis is only valid with the currents pointing in opposite directions. If they were to be in the same directions, things are different (please check).

Magnetic Field of infinite length current carrying wire at distance r is given by r

IB o

2

Magnetic Field due to wire-1 2.02

101

oB

Magnetic Field due to wire-2 6.02

102

oB

Total magnetic field at that point 21 BBB

6.02

10

2.02

10

ooB

B = 6.7 x 10-6 T Question 22: Two parallel segments of wire, of length L = 0.8 m are positioned a distance d = 1 m apart as shown in Fig. 28-8. Calculate the magnetic field at point P located half-way between the two segments. (Each segment carries an electric current I = 10 A in the directions indicated.) FIGURE 28-8

A) 8 μT B) 3 μT C) 0 D) 0.5 μT E) 1 μT Solution and Explanation:

The truth is the test bank solved this problem the wrong way which would yield an answer of 8 T. The test bank authors used the long wire equation to determine the magnetic field at midpoint. The lines we have here are short and of a finite length. We should use the method used a few times above.

We need to determine the lower and upper boundary angles. The low angle can be obtained using tan() =

(d/2)/(L/2) = 0.5/0.4 = 1.25 or = 51.34o. The top boundary angle can be easily determined (from symmetry) to be 128.66o.

25.1)625.0()625.0(cosR4π

Iμ

R4π

Iμ

R4π

Iμdθsinθ

R4π

IμB oo128.66

51.34

o128.66

51.34

o1

Consider magnetic field due to wire 1 to be B1 and due to wire 2 to be B2

By symmetry B1 = B2 and pointing in the same direction. The total magnetic field

25.1221

R4π

IμoBBB

Substituting the parameters μo = 4 x 10-7 T.m/A, I = 10A, R = d/2 = 0.5m into the above equation, we get B = T5

Note: You will get credit for this problem if you selected T8 . There is no option available for the correct

answer so selecting any other answer does not make sense and it would simply mean that you were off anyway.