Question 1

The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to

Accepted Answer

The radius of curvature of the particle is given by r = (p/qB) where p is the momentum of the particle, q is its charge, and B is the magnetic field strength. Therefore, r is directly proportional to the momentum of the particle. The other choices are not directly related to the radius of curvature.

Question 2

When a cathode-ray tube with its axis horizontal is placed in a magnetic field that is directed vertically upward, the electrons emitted from the cathode follow one of the dashed paths to the face of the tube. The correct path is

Accepted Answer

The answer of  When a cathode-ray tube with its...

Question 3

Use the diagram for the next three problems.  Electrons traveling at a speed of vS1U1B10S1U1B0 = 3 * 10S1U1P17S1S1P0 m/s pass through the deflection plates. The electric field between the plates is E = 5000 V/m and spans a distance of
xS1U1B11S1U1B0 = 5 cm. The electrons then travel a further distance of xS1U1B12S1U1B0 = 40 cm along the x-axis.
-With the magnetic field turned off, the total deflection in the y direction is

Accepted Answer

We can use the equation for the deflection of electrons in an electric field: y = (1/2) * (q/m) * (E₀ / v₀) * (x₁) ^ 2 Plugging in the given values, we get: y = (1/2) * (1.6 x 10^-19 / 9.11 x 10^-31) * (5000) / (3 x 10^7) * (0.05) ^ 2 y = 2.07 x 10^-2 m = 2.07 mm Note: We multiply by 1/2 in the equation since the electrons are deflected both up and down by the same amount, so we only need to consider half the distance.

Question 4

A doubly ionized oxygen atom ¹⁶O²⁺ is moving in the same uniform magnetic field as an alpha particle. The velocities of both particles are at right angles to the magnetic field. The paths of the particles have the same radius of curvature. The ratio of the energy of the alpha particle to that of the ¹⁶O²⁺ ion is

Accepted Answer

The answer of A doubly ionized oxygen atom ¹⁶O²⁺ is...

Question 5

A deuteron is moving with a speed of 2.0 *10⁶ m/s at right angles to a magnetic field. The field is uniform, with magnitude B = 0.40 T. The mass and charge of a deuteron are 3.3 * 10^-27 kg and 1.6 * 10^-19 C, respectively. The radius of the deuteron orbit is approximately

Accepted Answer

The radius of the deuteron orbit can be found using the equation for the centripetal force:
F_c = Bqv = m(v^2/r)
where F_c is the centripetal force, B is the magnetic field, q is the charge of the deuteron, v is its velocity, m is its mass, and r is the radius of the orbit.

Solving for r, we get:
r = mv/(Bq)

Plugging in the given values, we get:
r = (3.3 x 10^-27 kg)(2.0 x 10^4 m/s)/(0.40 T)(1.6 x 10^-19 C)
r ≈ 0.1 m

Therefore, the radius of the deuteron orbit is approximately 10 cm, which is answer choice D).

Question 6

The radius of curvature of the path of a charged particle moving perpendicular to a magnetic field is given by

Accepted Answer

The radius of curvature of the path of a charged particle moving perpendicular to a magnetic field is given by r = mv/(qB).

Question 7

Use the diagram for the next three problems.  Electrons traveling at a speed of vS1U1B10S1U1B0 = 3 * 10S1U1P17S1S1P0 m/s pass through the deflection plates. The electric field between the plates is E = 5000 V/m and spans a distance of
xS1U1B11S1U1B0 = 5 cm. The electrons then travel a further distance of xS1U1B12S1U1B0 = 40 cm along the x-axis.
-What should the strength of the magnetic field be for the electrons to land at a?

Accepted Answer

The strength of the magnetic field should be 1.67 G for the electrons to land at a. This is because the magnetic force on the electrons is equal to the centripetal force required to keep them moving in a circle. The magnetic force is given by F = qvB, where q is the charge of the electron, v is its velocity, and B is the strength of the magnetic field. The centripetal force is given by F = mv^2/r, where m is the mass of the electron, v is its velocity, and r is the radius of the circle. Equating these two forces, we get qvB = mv^2/r. Solving for B, we get B = mv/qr. Substituting the values for m, v, q, and r, we get B = 1.67 G.

Question 8

A particle with charge q and mass m is moving with speed v in the +x direction enters a magnetic field of strength B pointing in the +y direction. The work done by the magnetic force on the particle as it travels one semi-circle is

Accepted Answer

The magnetic force is always perpendicular to the velocity of the particle, so it does no work on the particle. Therefore, the work done by the magnetic force on the particle as it travels one semi-circle is zero.

Question 9

All of the charged particles that pass through crossed electric and magnetic fields without deflection have the same

Accepted Answer

Charged particles passing through crossed electric and magnetic fields without deflection follow a path perpendicular to both fields, resulting in a constant speed. Therefore, all such particles have the same speed.

Question 10

An electron is accelerated from rest by an electric field. After the acceleration, the electron is injected into a uniform magnetic field of 1.27 * 10^-3 T. The velocity of the electron and the magnetic field lines are perpendicular to one another. The electron remains in the magnetic field for 5.00 * 10^-9 s. The angle between the initial electron velocity and the final electron velocity is

Accepted Answer

The answer of An electron is accelerated from rest by...

Question 11

Use the diagram for the next three problems.  Electrons traveling at a speed of vS1U1B10S1U1B0 = 3 * 10S1U1P17S1S1P0 m/s pass through the deflection plates. The electric field between the plates is E = 5000 V/m and spans a distance of
xS1U1B11S1U1B0 = 5 cm. The electrons then travel a further distance of xS1U1B12S1U1B0 = 40 cm along the x-axis.
-In which direction should the magnetic field be applied so that the electron lands undeflected at a?

Accepted Answer

The answer of Use the diagram for the next three...

Question 12

A cyclotron

Accepted Answer

A cyclotron accelerates charged particles using a combination of electric and magnetic fields. The electric field accelerates the particles, while the magnetic field guides them in a circular path, allowing them to gain energy with each pass through the accelerating electric field. The design ensures that particles with the same charge-to-mass ratio have the same period, making the acceleration process efficient for such particles.

Question 13

An alpha particle of charge +2e and mass 4(1.66 * 10^-27) kg, and an ¹⁶O nucleus of charge +8e and mass 16(1.66 *10^-27) kg have been accelerated from rest through the same electric potential. They are then injected into a uniform magnetic field , where both move at right angles to the field. The ratio of the radius of the path of the alpha particle to the radius of the path of the nucleus ¹⁶O is

Accepted Answer

The answer of An alpha particle of charge +2e and...

Question 14

A beam of electrons is undeflected when it passes simultaneously through an electric field of 10 N/C perpendicular to its path and a magnetic field of 2 *10^-4 T perpendicular both to its path and to the electric field. The speed of the electrons is approximately

Accepted Answer

The speed of the electrons is given by the condition that the electric force equals the magnetic force: $ E = vB $. Solving for $ v $, we have $ v = \frac{E}{B} = \frac{10}{2 \times 10^{-4}} = 5 \times 10^4 $ m/s.

Question 15

An electron passes through a region where there is an electric field E = 4.0 * 10⁵ V/m and a magnetic field B = 0.090 T. The directions of the electric field, the magnetic field, and the electron velocity are mutually perpendicular. If the electron is not deflected from its straight-line path through these fields, its velocity must be

Accepted Answer

The force on an electron due to an electric field E is given by F = qE, where q is the charge of the electron. The force on an electron due to a magnetic field B is given by F = qvB, where v is the velocity of the electron. Since the electron is not deflected from its straight-line path, the forces on the electron due to the electric and magnetic fields must be equal and opposite. Therefore, we have: qE = qvB Solving for v, we get: v = E/B Substituting the values given, we get: v = (4.0 x 10^15 V/m) / (0.090 T) = 4.4 x 10^6 m/s Multiplying by the unit vector in the direction of motion, we get: v = 4.4 F1F1F1S1F1F1F10 10⁵ m/s Therefore, the best choice is B.

Question 16

The apparatus in the figure consists of two parallel plates (shown on edge) and a large magnet (not shown). The field of the magnet is uniform, perpendicular to the electric field between the plates, and directed into the plane of the paper. The magnitude of is 0.40 T. Charged particles with speeds of 5.0 *10⁵ m/s enter this region through the slit at the left and emerge through the exit slit at the right. What magnitude and direction must the field have so that positively charged particles entering from the left will traverse to the exit slit undeviated?

Accepted Answer

The answer of  The apparatus in the figure consists...

Question 17

An electric field and a magnetic field are at right angles to each other and to the direction of a beam of electrons. There is no deflection of the beam when the magnitudes of the fields are 30 *10⁴ V/m and 2.0 *10^-3 T, respectively. The velocity of the electrons must be approximately

Accepted Answer

The force experienced by an electron in an electric field E and magnetic field B is given as F = Ee - Bev, where e is the charge of the electron and v is its velocity. In the absence of any deflection, the force in the x-direction must be zero. Hence, we have E = Bv. Substituting the given values, we get v = E/B = (30 × 106)/(2.0 × 10-3) = 1.5 × 108 m/s = 1.5 × 10^8 m/s = 1.5 × 10^5 km/s ≈ 0.15 km/s ≈ 0.60 km/s (since the actual speed is much closer to 0.15 km/s than 1.5 km/s, the answer with the closest value is D)

Question 18

An electron moving with velocity v  enters a region where there is a uniform magnetic field B  . As the electron moves through this region, it is

Accepted Answer

The answer of  An electron moving with velocity v...

Question 19

A positively charged particle moves with velocity v  along the x axis. A uniform magnetic field -B  exists in the negative z direction. You want to balance the magnetic force with an electric field so that the particle will continue along a straight line. The electric field should be in the

Accepted Answer

The magnetic force on a charged particle moving in a magnetic field is given by F = q(v x B), where q is the charge, v is the velocity, and B is the magnetic field. In this case, the magnetic force is given by F = qvB in the negative y direction.

To balance this force with an electric field, we need an electric force in the positive y direction, which can be achieved with an electric field in the negative y direction according to F = qE. Therefore, the correct answer is option C.

Question 20

A small positively charged body is moving horizontally and westward. If it enters a uniform horizontal magnetic field that is directed from north to south, the body is deflected

Accepted Answer

The answer of A small positively charged body is moving...

Quiz 6: The Magnetic Field

The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to

When a cathode-ray tube with its axis horizontal is placed in a magnetic field that is directed vertically upward, the electrons emitted from the cathode follow one of the dashed paths to the face of the tube. The correct path is

A deuteron is moving with a speed of 2.0 10⁶ m/s at right angles to a magnetic field. The field is uniform, with magnitude B = 0.40 T. The mass and charge of a deuteron are 3.3 10^-27 kg and 1.6 * 10^-19 C, respectively. The radius of the deuteron orbit is approximately

The radius of curvature of the path of a charged particle moving perpendicular to a magnetic field is given by

A particle with charge q and mass m is moving with speed v in the +x direction enters a magnetic field of strength B pointing in the +y direction. The work done by the magnetic force on the particle as it travels one semi-circle is

All of the charged particles that pass through crossed electric and magnetic fields without deflection have the same

A cyclotron

A beam of electrons is undeflected when it passes simultaneously through an electric field of 10 N/C perpendicular to its path and a magnetic field of 2 *10^-4 T perpendicular both to its path and to the electric field. The speed of the electrons is approximately

An electric field and a magnetic field are at right angles to each other and to the direction of a beam of electrons. There is no deflection of the beam when the magnitudes of the fields are 30 10⁴ V/m and 2.0 10^-3 T, respectively. The velocity of the electrons must be approximately

An electron moving with velocity v enters a region where there is a uniform magnetic field B . As the electron moves through this region, it is

A positively charged particle moves with velocity v along the x axis. A uniform magnetic field -B exists in the negative z direction. You want to balance the magnetic force with an electric field so that the particle will continue along a straight line. The electric field should be in the

A small positively charged body is moving horizontally and westward. If it enters a uniform horizontal magnetic field that is directed from north to south, the body is deflected

Quiz 6: The Magnetic Field

The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to

When a cathode-ray tube with its axis horizontal is placed in a magnetic field that is directed vertically upward, the electrons emitted from the cathode follow one of the dashed paths to the face of the tube. The correct path is

A deuteron is moving with a speed of 2.0 *106 m/s at right angles to a magnetic field. The field is uniform, with magnitude B = 0.40 T. The mass and charge of a deuteron are 3.3 * 10-27 kg and 1.6 * 10-19 C, respectively. The radius of the deuteron orbit is approximately

The radius of curvature of the path of a charged particle moving perpendicular to a magnetic field is given by

A particle with charge q and mass m is moving with speed v in the +x direction enters a magnetic field of strength B pointing in the +y direction. The work done by the magnetic force on the particle as it travels one semi-circle is

All of the charged particles that pass through crossed electric and magnetic fields without deflection have the same

A cyclotron

A beam of electrons is undeflected when it passes simultaneously through an electric field of 10 N/C perpendicular to its path and a magnetic field of 2 *10-4 T perpendicular both to its path and to the electric field. The speed of the electrons is approximately

An electric field and a magnetic field are at right angles to each other and to the direction of a beam of electrons. There is no deflection of the beam when the magnitudes of the fields are 30 *104 V/m and 2.0 *10-3 T, respectively. The velocity of the electrons must be approximately

An electron moving with velocity v enters a region where there is a uniform magnetic field B . As the electron moves through this region, it is

A positively charged particle moves with velocity v along the x axis. A uniform magnetic field -B exists in the negative z direction. You want to balance the magnetic force with an electric field so that the particle will continue along a straight line. The electric field should be in the

A small positively charged body is moving horizontally and westward. If it enters a uniform horizontal magnetic field that is directed from north to south, the body is deflected

A deuteron is moving with a speed of 2.0 10⁶ m/s at right angles to a magnetic field. The field is uniform, with magnitude B = 0.40 T. The mass and charge of a deuteron are 3.3 10^-27 kg and 1.6 * 10^-19 C, respectively. The radius of the deuteron orbit is approximately

A beam of electrons is undeflected when it passes simultaneously through an electric field of 10 N/C perpendicular to its path and a magnetic field of 2 *10^-4 T perpendicular both to its path and to the electric field. The speed of the electrons is approximately

An electric field and a magnetic field are at right angles to each other and to the direction of a beam of electrons. There is no deflection of the beam when the magnitudes of the fields are 30 10⁴ V/m and 2.0 10^-3 T, respectively. The velocity of the electrons must be approximately