Question 1

You stand near the Earth's equator.A positively charged particle that starts moving parallel to the surface of the Earth in a straight line directed east is initially deflected upwards.If you know there are no electric fields in the vicinity, a possible reason why the particle does not initially acquire a downward component of velocity is because near the equator the magnetic field lines of the Earth are directed:

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Question 2

A particle (m = 3.0 $\mu$g, q = 5.0 $\mu$C) moves in a uniform magnetic field given by () mT.At t = 0 the velocity of the particle is equal to () m/s.The subsequent path of the particle is:

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Question 3

A current loop is oriented in three different positions relative to a uniform magnetic field.In position 1 the plane of the loop is perpendicular to the field lines.In positions 2 and 3 the plane of the loop is parallel to the field as shown.The torque on the loop is maximum in:

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Question 4

A charged particle (m = 5.0 g, q = -70 $\mu$C) moves horizontally at a constant speed of 30 km/s in a region where the free fall gravitational acceleration is 9.8 m/s2 downward, the electric field is 700 N/C upward, and the magnetic field is perpendicular to the velocity of the particle.What is the magnitude of the magnetic field in this region?

Accepted Answer

The magnetic force must balance the gravitational and electric forces for the particle to move at a constant speed. Using the equation $ qvB = mg + qE $, solve for $ B $ to find the magnitude of the magnetic field.

Question 5

The boundary shown is that of a uniform magnetic field directed in the positive z direction.An electron enters the magnetic field with a velocity pointing along the x axis and exits 0.63 $\mu$s later at point A.What is the magnitude of the magnetic field?

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Question 6

A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field.What is the ratio of the radii of their orbits?

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Question 7

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80$\mu$T.The electron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm.What is the speed of this electron as it moves in this region?

Accepted Answer

The equation for the helical path followed by a charged particle in a uniform magnetic field is given by:

R = mv/qB

where R is the radius of the helical path, m is the mass of the particle, v is the velocity of the particle, q is the charge of the particle, and B is the magnetic field.

In this case, the radius R = 2.0 mm and the pitch p = 9.0 mm. We can find the angle that the path makes with the magnetic field by noting that one complete turn of the helix corresponds to a distance of the pitch, so:

2πR = pcosθ

where θ is the angle between the magnetic field and the velocity of the electron. Solving for θ, we find:

θ = cos^-1(2πR/p) = cos^-1(2π(2.0x10^-3 m)/(9.0x10^-3 m)) ≈ 86.4°

The velocity of the electron can be found from the centripetal force acting on it due to the magnetic field:

mv^2/R = qvB

Solving for v, we find:

v = qBR/m = (1.6x10^-19 C)(80 T)(2.0x10^-3 m) / (9.11x10^-31 kg) ≈ 3.54x10^7 m/s = 35 km/s

Therefore, the correct answer is D.

Question 8

A proton is accelerated from rest through a potential difference of 150 V.It then enters a region of uniform magnetic field and moves in a circular path (radius = 12 cm).What is the magnitude of the magnetic field in this region?

Accepted Answer

The velocity of the proton can be found using conservation of energy:
(1/2)mv^2=qV
where m is the mass of the proton, q is its charge, and V is the potential difference.
Solving for v gives:
v=sqrt(2qV/m)

Since the proton moves in a circular path, the magnetic force must provide the centripetal force:
Fm=qvB=r(mv^2/r)=mv^2/r
where B is the magnitude of the magnetic field, and r is the radius of the circular path.

Combining these two equations and solving for B gives:
B=mv/(qr)

Plugging in the given values and constants gives:
B=(1.67x10^-27 kg)(sqrt(2(1.6x10^-19 C)(150 V))/((1.6x10^-19 C)(0.12 m)))
B≈15 mT

Question 9

A velocity selector uses a fixed electric field of magnitude E and the magnetic field is varied to select particles of various energies.If a magnetic field of magnitude B is used to select a particle of a certain energy and mass, what magnitude of magnetic field is needed to select a particle of equal mass but twice the energy?

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Question 10

One reason why we know that magnetic fields are not the same as electric fields is because the force exerted on a charge +q:

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The force exerted on a charge in an electric field is parallel to the direction of the field, while the force due to a magnetic field is perpendicular to both the direction of the magnetic field and the velocity of the moving charge, as described by the right-hand rule. This distinction is a fundamental principle of electromagnetism.

Question 11

A magnetic field is directed out of the page.Two charged particles enter from the top and take the paths shown in the figure below.Which statement is correct?

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Question 12

What is the kinetic energy of an electron that passes undeviated through perpendicular electric and magnetic fields if E = 4.0 kV/m and B = 8.0 mT?

Accepted Answer

For an electron to pass undeviated through perpendicular electric (E) and magnetic (B) fields, the electric force (F_e = Ee) must equal the magnetic force (F_m = evB), where e is the charge of an electron (approximately 1.6 x 10^-19 C), and v is the velocity of the electron. Solving for v gives v = E/B. The kinetic energy (KE) of the electron can be calculated using KE = 0.5mv^2, where m is the mass of an electron (approximately 9.11 x 10^-31 kg). Substituting v = E/B into the kinetic energy formula and converting the units appropriately gives the correct kinetic energy value in electronvolts (eV), leading to the correct answer.

Question 13

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T.What is the radius of the resulting circular path? (deuteron: m = 3.3 *10-27 kg, q = 1.6*10-19 C)

Accepted Answer

The kinetic energy gained by the deuteron is equal to the work done by the electric field, which is qV. This energy is converted into kinetic energy, (1/2)mv^2. Solving for v gives v = sqrt(2qV/m). The radius of the circular path in a magnetic field is given by r = mv/(qB). Substituting the expression for v, we get r = sqrt(2mV/q)/B. Plugging in the given values, we find r ≈ 13 mm.

Question 14

A charged particle moves in a region of uniform magnetic field along a helical path (radius = 5.0 cm, pitch = 12 cm, period = 5.0 ms).What is the speed of this particle as it moves along this path?

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Question 15

Three particles of equal charge, X, Y, and Z, enter a uniform magnetic field B.X has velocity of magnitude v parallel to the field.Y has velocity of magnitude v perpendicular to the field.Z has equal velocity components v parallel and perpendicular to the field.Rank the radii of their orbits from least to greatest.

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Question 16

A coaxial cable has an inner cylindrical conductor surrounded by cylindrical insulation and an outer cylindrical conducting shell.The outer shell carries the same current but in the opposite direction from that in the inner conductor as shown.If the coaxial cable sits in a uniform magnetic field directed upwards with respect to the cable, the effect of the field on the cable is:

Accepted Answer

The magnetic field will exert a force on the current carrying inner conductor and the outer shell in opposite directions due to their opposite current directions. However, since the current in the outer shell is distributed over a larger area, the force per unit area is smaller than that in the inner conductor. Therefore, the inner conductor will experience a larger force to the left and the outer shell will experience a smaller force to the right. This will result in a slight shift of the inner conductor to the left and the outer conductor to the right, but no net force on the cable as a whole.

Question 17

The diagram below shows the position of a long straight wire perpendicular to the page and a set of directions labelled A through H.  When the current in the wire is directed up out of the page, the direction of the magnetic field at point P is:

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Question 18

An electron moves in a region where the magnetic field is uniform, has a magnitude of 60 $\mu$T, and points in the positive x direction.At t = 0 the electron has a velocity that has an x component of 30 km/s, a y component of 40 km/s, and a z component of zero.What is the radius of the resulting helical path?

Accepted Answer

The force experienced by an electron moving with velocity v in a magnetic field B is given by the equation F = qv × B, where q is the charge on the electron. Since the magnetic field is uniform and points in the positive x direction, we can write the force on the electron as F = -qvBj, where j is the unit vector in the y direction.
The acceleration experienced by the electron is given by a = F/m, where m is the mass of the electron. Substituting the values, we get a = (-1.6 x 10^-19 C)(40 x 10^3 m/s)(60 x 10^-6 T)/9.11 x 10^-31 kg = -4.95 x 10^16 m/s^2 in the z direction.
The radius of the helical path is given by R = mv⊥/|qB|, where v⊥ is the component of the velocity perpendicular to the magnetic field. In this case, the magnetic force produces circular motion in the x-z plane, so we can project the initial velocity onto that plane to find v⊥. Using Pythagoras, we have v⊥ = √((30 x 10^3 m/s)^2 + (0)^2) = 30 x 10^3 m/s.
Substituting the values, we get R = (9.11 x 10^-31 kg)(30 x 10^3 m/s)/(1.6 x 10^-19 C)(60 x 10^-6 T) = 3.8 x 10^-3 m = 3.8 mm. Therefore, the correct answer is C.

Question 19

Two single charged ions moving perpendicularly to a uniform magnetic field (B = 0.40 T) with speeds of 5000 km/s follow circular paths that differ in diameter by 5.0 cm.What is the difference in the mass of these two ions?

Accepted Answer

The radius of circular path of an ion moving perpendicular to a uniform magnetic field is given by r = mv/qB, where m is the mass of the ion, v is its velocity, q is its charge, and B is the magnetic field. Since the ions have the same charge and move with the same speed, their radius of circular path will differ only due to their mass. Let the masses of the two ions be m1 and m2, and let their radii be r1 and r2, respectively. Then we have:

r1 - r2 = 5.0 cm = 0.050 m
m1v/qB - m2v/qB = 0.050 m
(m1 - m2)v/qB = 0.050 m
(m1 - m2) = (0.050 m)(qB/v)
(m1 - m2) = (0.050 m)(1.6 x 10^-19 C)(0.40 T)/(5000 x 10^3 m/s)
(m1 - m2) = 3.2 x 10^-28 kg

Therefore, the difference in mass of the two ions is 3.2 x 10^-28 kg, which is closest to choice C.

Question 20

An electron follows a circular path (radius = 15 cm) in a uniform magnetic field (magnitude = 3.0 G).What is the period of this motion?

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Quiz 8: Magnetic Forces

A current loop is oriented in three different positions relative to a uniform magnetic field.In position 1 the plane of the loop is perpendicular to the field lines.In positions 2 and 3 the plane of the loop is parallel to the field as shown.The torque on the loop is maximum in:

A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field.What is the ratio of the radii of their orbits?

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 $\mu$ T.The electron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm.What is the speed of this electron as it moves in this region?

A proton is accelerated from rest through a potential difference of 150 V.It then enters a region of uniform magnetic field and moves in a circular path (radius = 12 cm) .What is the magnitude of the magnetic field in this region?

One reason why we know that magnetic fields are not the same as electric fields is because the force exerted on a charge +q:

A magnetic field is directed out of the page.Two charged particles enter from the top and take the paths shown in the figure below.Which statement is correct?

What is the kinetic energy of an electron that passes undeviated through perpendicular electric and magnetic fields if E = 4.0 kV/m and B = 8.0 mT?

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T.What is the radius of the resulting circular path? (deuteron: m = 3.3 10-27 kg, q = 1.610-19 C)

A charged particle moves in a region of uniform magnetic field along a helical path (radius = 5.0 cm, pitch = 12 cm, period = 5.0 ms) .What is the speed of this particle as it moves along this path?

The diagram below shows the position of a long straight wire perpendicular to the page and a set of directions labelled A through H. When the current in the wire is directed up out of the page, the direction of the magnetic field at point P is:

Two single charged ions moving perpendicularly to a uniform magnetic field (B = 0.40 T) with speeds of 5000 km/s follow circular paths that differ in diameter by 5.0 cm.What is the difference in the mass of these two ions?

An electron follows a circular path (radius = 15 cm) in a uniform magnetic field (magnitude = 3.0 G) .What is the period of this motion?

Quiz 8: Magnetic Forces

A particle (m = 3.0 μ\muμ g, q = 5.0 μ\muμ C) moves in a uniform magnetic field given by () mT.At t = 0 the velocity of the particle is equal to () m/s.The subsequent path of the particle is:

A current loop is oriented in three different positions relative to a uniform magnetic field.In position 1 the plane of the loop is perpendicular to the field lines.In positions 2 and 3 the plane of the loop is parallel to the field as shown.The torque on the loop is maximum in:

The boundary shown is that of a uniform magnetic field directed in the positive z direction.An electron enters the magnetic field with a velocity pointing along the x axis and exits 0.63 μ\muμ s later at point A.What is the magnitude of the magnetic field?

A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field.What is the ratio of the radii of their orbits?

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μ\muμ T.The electron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm.What is the speed of this electron as it moves in this region?

A proton is accelerated from rest through a potential difference of 150 V.It then enters a region of uniform magnetic field and moves in a circular path (radius = 12 cm) .What is the magnitude of the magnetic field in this region?

One reason why we know that magnetic fields are not the same as electric fields is because the force exerted on a charge +q:

A magnetic field is directed out of the page.Two charged particles enter from the top and take the paths shown in the figure below.Which statement is correct?

What is the kinetic energy of an electron that passes undeviated through perpendicular electric and magnetic fields if E = 4.0 kV/m and B = 8.0 mT?

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T.What is the radius of the resulting circular path? (deuteron: m = 3.3 *10-27 kg, q = 1.6*10-19 C)

A charged particle moves in a region of uniform magnetic field along a helical path (radius = 5.0 cm, pitch = 12 cm, period = 5.0 ms) .What is the speed of this particle as it moves along this path?

The diagram below shows the position of a long straight wire perpendicular to the page and a set of directions labelled A through H. When the current in the wire is directed up out of the page, the direction of the magnetic field at point P is:

Two single charged ions moving perpendicularly to a uniform magnetic field (B = 0.40 T) with speeds of 5000 km/s follow circular paths that differ in diameter by 5.0 cm.What is the difference in the mass of these two ions?

An electron follows a circular path (radius = 15 cm) in a uniform magnetic field (magnitude = 3.0 G) .What is the period of this motion?

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 $\mu$ T.The electron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm.What is the speed of this electron as it moves in this region?

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T.What is the radius of the resulting circular path? (deuteron: m = 3.3 10-27 kg, q = 1.610-19 C)