problem 27 - notendur.hi.isisa13/efni/einn/s.hl/edlisfraedi ii/heimadæmi 6.pdf · the figure shows...

Heimadæmi 6Due: 11:00pm on Thursday, February 25, 2016

You will receive no credit for items you complete after the assignment is due. Grading Policy

Problem 27.68

The rectangular loop shown in the figure is pivoted about the yaxis and carries a current of 15.0 in the direction indicated.

Part A

If the loop is in a uniform magnetic field with magnitude 0.48 in the xdirection, find the magnitude of the torquerequired to hold the loop in the position shown.

Express your answer using two significant figures.

ANSWER:

Correct

Part B

What is the direction of the torque required to hold the loop in the position shown.

ANSWER:

Correct

A

T +

= 3.0×10−2 τ N ⋅ m

+ i

− i

+ j

− j

+k

−k

Part C

Repeat part A for the case in which the field is in the zdirection.


ANSWER:

Correct

Part D

What is the direction of the torque required to hold the loop in the position shown.

ANSWER:

Correct

Part E

For the magnetic field in part A, what torque would be required if the loop were pivoted about an axis through its center,parallel to the yaxis?


ANSWER:

Correct

Part F

For the magnetic field in part C, what torque would be required if the loop were pivoted about an axis through its center,parallel to the yaxis?


ANSWER:

−

= 1.7×10−2 τ N ⋅ m

+ i

− i

+ j

− j

+k

−k

= 3.0×10−2 τ N ⋅ m

= 1.7×10−2 τ N ⋅ m

Correct

Problem 27.22

Part A

A thin copper rod that is 1.0 m long and has a mass of 0.050 kg is in a magnetic field of 0.10 T. What minimum currentin the rod is needed in order for the magnetic force to cancel the weight of the rod?

ANSWER:

Correct

Exercise 27.18

An alpha particle (a nucleus, containing two protons and two neutrons and having a mass of ) travelinghorizontally at 35.6 enters a uniform, vertical, 1.80 magnetic field.

Part A

What is the diameter of the path followed by this alpha particle?

Express your answer with the appropriate units.

ANSWER:

Correct

Part B

What effect does the magnetic field have on the speed of the particle?

ANSWER:

4.9 A

7.6 A

2.5 A

9.8 A

1.2 A

He 6.64 × kg10−27

km/s T

= 0.821 d mm

The speed remains constant.

The speed increases with time.

The speed decreases with time.

Correct

Part C

What is the magnitude of the acceleration of the alpha particle while it is in the magnetic field?


ANSWER:

Correct

Part D

What is the direction of this acceleration?

ANSWER:

Correct

Part E

Explain why the speed of the particle does not change even though an unbalanced external force acts on it.

ANSWER:

Correct

= 3.09×1012 a ms2

The acceleration is perpendicular to and and so is horizontal, out of the center of curvature of theparticle’s path.

The acceleration is perpendicular to and and so is horizontal, toward the center of curvature of theparticle’s path.

The acceleration is parallel to and perpendicular to and so is vertical, along the particle’s path.

The acceleration is perpendicular to and parallel to and so is horizontal, along the particle’s path.

v B

v B

v B

v B

The unbalanced force is parallel to so it changes the direction of but not its magnitude, which is the

speed.

The unbalanced force is perpendicular to so it changes the direction of but not its magnitude, which

is the speed.

The unbalanced force is parallel to so it changes the magnitude of but not its direction, which is the

speed.

The unbalanced force is perpendicular to so it changes the magnitude of but not its direction, which

is the speed.

FB→

v v

FB→

v v

FB→

v v

FB→

v v

Electromagnetic Velocity Filter

When a particle with charge moves across a magnetic field of magnitude , it experiences a force to the side. If the

proper electric field is simultaneously applied, the electric force on the charge will be in such a direction as to cancel themagnetic force with the result that the particle will travel in a straight line. The balancing condition provides a relationshipinvolving the velocity of the particle. In this problem you will figure out how to arrange the fields to create this balance andthen determine this relationship.

Part A

Consider the arrangement of ion source and electric field plates shown in the figure. The ion source sends particles withvelocity along the positive x axis. They encounter electricfield plates spaced a distance apart that generate a uniformelectric field of magnitude in the +y direction. To cancelthe resulting electric force with a magnetic force, a magneticfield (not shown) must be added in which direction? Using therighthand rule, you can see that the positive z axis isdirected out of the screen.

Choose the direction of .

Hint 1. Method for determining direction

Assume a sign for the charge. Since both the electric force and magnetic force depend on , in particular, theyalso depend on its sign. So the sign doesn't matter here. Apply the righthand rule to the equation for themagnetic force, .

Hint 2. Righthand rule

Curl the fingers of your right hand from the first vector to the second in the product. Your outstretched thumbthen points in the direction of the crossproduct vector.

ANSWER:

Correct

Part B

q B

E

v

v d

E

B

q

= q ×F M v B

i

− i

j

− j

k

−k

Now find the magnitude of the magnetic field that will cause the charge to travel in a straight line under the combinedaction of electric and magnetic fields.

Express the magnetic field that will just balance theapplied electric field in terms of some or all of thevariables , , and .

Hint 1. Find the magnetic force

What is , the magnitude of the force due to a magnetic field (with a magnitude of ) interacting with acharge moving at velocity (a speed of )?

Express in terms of some or all of the variables , , and .

ANSWER:

Hint 2. Find the force due to the electric field

What is , the magnitude of a force on a charge due to an applied electric field ?

Express in terms of one or both of the variables and .

ANSWER:

ANSWER:

Correct

Part C

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (Forexample, would the velocity of a neutral particle be selected by passage through this device?) The explanation of thisis that the mass and the charge control the resolution of the deviceparticles with the wrong velocity will be acceleratedaway from the straight line and will not pass through the exit slit. If the acceleration depends strongly on the velocity,then particles with just slightly wrong velocities will feel a substantial transverse acceleration and will not exit theselector. Because the acceleration depends on the mass and charge, these influence the sharpness (resolution) of thetransmitted particles.

Bbal

q v E

FM B Bq v v

FM q B v

= FM qBv

FE q E

FE q E

= FE qE

= BbalEv

Assume that you want a velocity selector that will allow particles of velocity to pass straight through withoutdeflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to selectthe velocity . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmittedparticles) you would want to use particles with __________.

Assume that the selector is short enough so that particles that move away from the axis do not have time tocome back to it.

Hint 1. Use Newton's law

If the velocity is "wrong" the forces won't balance and the resulting transverse force will cause a transverseacceleration. Use to determine how this acceleration will depend on and . You want particles withthe incorrect velocity to have the maximum possible deviation in the y direction so that they will not go througha slit placed at the right end. This means that the acceleration should be maximum.

ANSWER:

CorrectYou want particles with the incorrect velocity to have the maximum possible deviation in the y direction so thatthey will not go through a slit placed at the right end. The deviation will be maximum when the acceleration ismaximum. The acceleration is directly proportional to and inversely proportional to :

.So for maximum deviation, should be large and small.

Exercise 27.49

The figure shows a portion of a silver ribbon with = 11.0 and = 0.20 , carrying a current of 140 in the direction. The ribbon lies in a uniform magnetic field, in the direction, with magnitude 0.91 . Apply the simplified model ofthe Hall effect.

Part A

If there are 5.85×1028 free electrons per cubic meter, find the magnitude of the drift velocity of the electrons in the direction.

v

v

a = F/m q m

both and large

large and small

small and large

both and small

q m

q m

q m

q m

q m

a = = ( + × )+F E F

Mm

qm E v B

q m

z1 mmy1 mm A +x

yT

x


ANSWER:

Correct

Part B

Find the magnitude of the electric field in the direction due to the Hall effect.


ANSWER:

Correct

Part C

Find the direction of the electric field in the direction due to the Hall effect.

ANSWER:

Correct

Part D

Find the Hall emf.


ANSWER:

Correct

Exercise 26.35

The resistance of a galvanometer coil is 30.0 , and the current required for fullscale deflection is 500 .

Part A

Show in a diagram how to convert the galvanometer to an ammeter reading 25.0 full scale.

= 6.8×10−3 vd m/s

z

= 6.2×10−3 E V/m

z

−z

+z

= 6.8×10−5 EHall V

Ω μA

mA

ANSWER:

Correct

Part B

Compute the shunt resistance.

ANSWER:

Correct

Part C

Show how to convert the galvanometer to a voltmeter reading 500 full scale.

ANSWER:

Correct

Part D

Compute the series resistance.

ANSWER:

Correct

Problem 26.21

Part A

add a shunt resistor in parallel with the galvanometer coil

add a resistor in series with the galvanometer coil

= 0.612 Rs Ω

mV

add a shunt resistor in parallel with the galvanometer coil

add a resistor in series with the galvanometer coil

= 970 Rs Ω

(a) For the circuit shown in the figure, determine the current in the 7.0 resistor.

Express your answer using three significant figures.

ANSWER:

Correct

Part B

(b) For the circuit shown in the figure, determine the current in the 8.0 resistor.


ANSWER:

Correct

Part C

(c) For the circuit shown in the figure, determine the current in the 4.0 resistor.


ANSWER:

Correct

PSS 27.1: Magnetic Forces

Learning Goal:

To practice ProblemSolving Strategy 27.1: Magnetic Forces.

A particle with mass 1.81×10−3 and a charge of 1.22×10−8 has, at a given instant, a velocity . What are the magnitude and direction of the particle’s acceleration produced by a uniform

magnetic field ?

ProblemSolving Strategy 27.1: Magnetic Forces

Ω

1.55 A

Ω

1.27 A

Ω

0.284 A

kg C= (3.00 × m/s)v 104 j

= (1.63 T) + (0.980 T)B i j

IDENTIFY the relevant concepts: The righthand rule allows you to determine the magnetic force on a moving charged particle.

SET UP the problem using the following steps:

1. Draw the velocity vector and magnetic field with their tails together so that you can visualize the plane inwhich these two vectors lie.

2. Identify the angle between the two vectors.3. Identify the target variables. This may be the magnitude and direction of the force, the velocity, or themagnetic field.

EXECUTE the solution as follows:

1. Express the magnetic force using the equation . The magnitude of the force is given by .

2. Remember that is perpendicular to the plane of the vectors and . The direction of is determinedby the righthand rule. If is negative, the force is opposite to .

EVALUATE your answer: Whenever you can, solve the problem in two ways. Verify that the results agree.

IDENTIFY the relevant concepts

The problem asks for the acceleration of a moving charged particle. Since acceleration is related to force, you will need todetermine the magnetic force acting on the particle.

SET UP the problem using the following steps

Part A

Draw the velocity and magnetic field vectors. Since they have different units, their relative magnitudes aren'trelevant. Be certain they have the correct orientations relative to the given coordinate system. The dot in the center ofthe image represents the particle.

Recall that , , and are the unit vectors in the x, y, and z directions, respectively.

ANSWER:

v B

ϕ

= q ×F v B F = qvB sin ϕ

F v B ×v B

q ×v B

v B

i j k

Correct

The strategy points out that there are two ways to solve problems with magnetic forces. In this problem, youalready have the components of the vectors, so the cross product method will be much easier. This means thatyou do not need to find the value of , the angle between and .

Also, note that the coordinate system in the vector drawing applet is twodimensional. To make it threedimensional, add a positive z axis oriented out of the screen.

EXECUTE the solution as follows

Part B

Find the acceleration vector for the charge.

Enter the x, y, and z components of the acceleration in meters per second squared separated by commas.

Hint 1. How to find cross products

Recall that the cross product distributes like a regular scalar product:

You will also need to use the following relations for products of unit vectors:

Finally, remember that the cross product of any vector with itself is zero. For example, .

Hint 2. Find

Calculate , in terms of its components.

ϕ v B

× ( + ) = × + ×A B C A B A C

× = × = −i j k j i k

× = × = −j k i k j i

× = × = −k i j i k j

× = 0j j

×v B

×v B

Enter the x, y, and z components of in tesla meters per second separated by commas.

Hint 1. How to find cross products

Recall that the cross product distributes like a regular scalar product:

You will also need to use the following relations for products of unit vectors:

Finally, remember that the cross product of any vector with itself is zero. For example, .

ANSWER:

Correct

ANSWER:

Correct

EVALUATE your answer

Part C

You can check your result by comparing its magnitude to the magnitude the acceleration would have if the particle'svelocity had the same magnitude but it was perpendicular to the magnetic field.Find the value of the expression (the magnitude of when is perpendicular to ), where is the magnitudeof the charge, is the magnitude of the velocity, is the magnitude of the magnetic field, and is the mass of theparticle.

Express your answer in meters per second squared.

Hint 1. Find the magnitude of the velocity

What is the value of ? Remember that the magnitude of a vector is given by

Express your answer in meters per second squared.

ANSWER:

×v B

× ( + ) = × + ×A B C A B A C

× = × = −i j k j i k

× = × = −j k i k j i

× = × = −k i j i k j

× = 0j j

= 0,0,−4.89×104 ×v B T ⋅ m/s

= 0,0,0.330 a m/s2

qvB/m a v B qv B m

v + +vx i vy j vzk

+ +v2x v2

y v2z

− −−−−−−−−−√

= 3.00×104 v m/s2

Hint 2. Find the magnitude of the magnetic field

What is the value of ? Remember that the magnitude of a vector is given by

Express your answer in teslas.

ANSWER:

ANSWER:

Correct

This quantity is of similar size to the magnitude of your answer from Part B. If you wanted to check precisely, youcould find the value of and multiply the value you calculated above by . You would find that you had thesame magnitude as the magnitude of the acceleration vector you found in Part B. Note that the magnitude of themagnetic force, and therefore the magnitude of the particle's acceleration, is at its maximum when isperpendicular to , so it is not surprising that your answer to Part C is somewhat larger than the magnitude of theacceleration calculated in Part B.

To check the direction of your answer from Part B, use the righthand rule. Point the fingers of your right handparallel to in your answer to Part A and then turn your wrist so that you can curl those fingers down toward .You will find that your thumb points into the screen, which is the negative z direction. Thus, your answer from PartB has the proper direction as well as the proper magnitude.

± Determining the Velocity of a Charged Particle

A particle with a charge of 5.10 is moving in a uniform magnetic field of 1.25 ) . The magnetic force on

the particle is measured to be 3.80×10−7 7.60×10−7 .

Part A

Are there components of the velocity that cannot be determined by measuring the force?

Hint 1. Magnetic force on a moving charged particle

Recall the following formula:.

If you know , does uniquely define ?

ANSWER:

Correct

B + +Bx i By j Bzk

+ +B2x B2

y B2z

− −−−−−−−−−−√

= 1.90 B T

= 0.385 qvB/m m/s2

ϕ sin ϕ

v

B

v B

− nC = −(B T k

= −(F N ) + (i N ) j

= q ×F v B

B ×v B v

yes

no

Part B

Calculate the x component of the velocity of the particle.

Express your answer in meters per second to three significant figures.

Hint 1. Relation between and

Which component of the force depends on the x component of the velocity?

ANSWER:

ANSWER:

Correct

Part C

Calculate the y component of the velocity of the particle.

Express your answer in meters per second to three significant figures.

Hint 1. Relation between and

Which component of the force depends on the y component of the velocity?

ANSWER:

ANSWER:

Correct

Part D

Calculate the scalar product . Work the problem out symbolically first, then plug in numbers after you've simplifiedthe symbolic expression.

Express your answer in watts to three significant figures.

v F

x

y

= 119 vx m/s

v F

x

y

= 59.6 vy m/s

⋅v F

Hint 1. Formula for dot product

The dot product of two vectors and is given by

.

ANSWER:

Correct

Part E

What is the angle between and ?

Express your answer in degrees to three significant figures.

Hint 1. Another dot product formula

Recall that

,

where is the angle between and .

ANSWER:

Correct

Notice that the dot product of the velocity and the force is zero. This will always be the case. Since , must be perpendicular to both and . This result is important because it implies that magnetic fields can only

change the direction of a charged particle's velocity, not its speed.

Exercise 27.38

A straight, vertical wire carries a current of 2.30 downward in a region between the poles of a large superconductingelectromagnet, where the magnetic field has a magnitude of = 0.594 and is horizontal.

Part A

What is the magnitude of the magnetic force on a 1.00 section of the wire that is in this uniform magnetic field, ifthe magnetic field direction is east?


ANSWER:

A B

⋅ = + +A B AxBx AyBy AzBz

0 W

v F

⋅ = cos θA B ∣∣A

∣∣∣∣B

∣∣

θ A B

90 ∘

= q ×F v B

F v B

AB T

cm

= 1.37×10−2 F N

Correct

Part B

What is the direction of this magnetic force?

ANSWER:

Correct

Part C

What is the magnitude of the magnetic force on a 1.00 section of the wire that is in this uniform magnetic field, ifthe magnetic field direction is south?


ANSWER:

Correct

Part D


ANSWER:

Correct

Part E

What is the magnitude of the magnetic force on a 1.00 section of the wire that is in this uniform magnetic field, ifthe magnetic field direction is 30.0 south of west?


ANSWER:

west

south

north

east

cm

= 1.37×10−2 F N

north

east

west

south

cm∘

Correct

Part F


ANSWER:

Correct

Score Summary:Your score on this assignment is 98.8%.You received 10.87 out of a possible total of 11 points.

= 1.37×10−2 F N

= 60.0 north of west ϕ ∘

problem 27 - notendur.hi.isisa13/efni/einn/s.hl/edlisfraedi ii/heimadæmi 6.pdf · the figure shows...

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