Question 1

The magnetic force on a charged particle moving at a constant speed in a uniform magnetic field will be halved when the angle between the velocity of the particle and the direction of the magnetic field changes from 25.0° to

Accepted Answer

D

Question 2

A proton moving with a speed of 6.0 × 10⁵ m/s perpendicular to a uniform magnetic field of 0.20 T will follow which of the paths described below? (q_p = 1.6 × 10−¹⁹ C and m_p = 1.67 × 10−²⁷ kg)

Accepted Answer

B

Question 3

A proton is released such that it has an initial speed of 5.0 × 10⁵ m/s from left to right across the page.A magnetic field of 1.2 T is present at an angle of 30° to the horizontal direction (or positive x axis).What is the magnitude of the force experienced by the proton? (q_p = 1.6 × 10−¹⁹ C)

Accepted Answer

C

Question 4

A proton moves across the Earth's equator in a northwesterly direction.At this point the Earth's magnetic field has a direction due north and is parallel to the surface.What is the direction of the force acting on the proton at this instant?

Accepted Answer

The answer of A proton moves across the Earth's equator...

Question 5

A proton moving at a speed of 3.8 × 10⁶ m/s cuts across the lines of a magnetic field at an angle of 40°.The strength of the field is 0.25 × 10−⁴ T.What is the magnitude of the force acting on the proton? (q_p = 1.6 × 10−¹⁹ C)

Accepted Answer

The answer of A proton moving at a speed of...

Question 6

If an electron is released at the equator and falls toward the Earth under the influence of gravity,the magnetic force on the electron will be toward the:

Accepted Answer

The answer of If an electron is released at the...

Question 7

A proton,mass 1.67 × 10−²⁷ kg and charge +1.6 × 10−¹⁹ C,moves in a circular orbit perpendicular to a uniform magnetic field of 0.50 T.Find the time for the proton to make one complete circular orbit.

Accepted Answer

The answer of A proton,mass 1.67 × 10−²⁷ kg and...

Question 8

A proton and an alpha particle are moving perpendicular to a uniform magnetic field with the same speed.If the velocity of the proton is at an angle of 20° with respect to the magnetic field direction,at what angle to the magnetic field direction would the alpha particle be moving if it experiences the same magnitude acceleration as the proton? The alpha particle has 4 times the mass of the proton and twice its charge.

Accepted Answer

We know that the force on a charged particle moving in a magnetic field is given by $F_B=qvB\sin	heta$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field, and $	heta$ is the angle between the velocity and the magnetic field. Since both particles experience the same magnitude acceleration, the force on each particle must be the same. The force is proportional to the charge of the particle, so we have:

$q_p v_p \sin 20° = q_{\alpha} v_{\alpha} \sin	heta$

Since the particles have the same speed, $v_p = v_{\alpha}$. We can also use the fact that the alpha particle has twice the charge of the proton:

$\dfrac{\sin 20°}{\sin	heta} = \dfrac{q_{\alpha}}{q_p} \cdot \dfrac{1}{2} = \dfrac{2}{1}$

Solving for $	heta$ gives $	heta \approx 42.86°$, which is closest to choice A.

Question 9

A proton and an alpha particle are moving perpendicular to a uniform magnetic field with the same speeds.If the velocity of the proton is at an angle of 20° with respect to the magnetic field direction,at what angle to the magnetic field direction would the alpha particle be moving if it experiences the same magnitude magnetic force as the proton? The alpha particle has 4 times the mass of the proton and twice its charge.

Accepted Answer

The magnetic force experienced by a charged particle moving in a magnetic field is given by $F = qvB\sin(\theta)$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field direction. For the proton and alpha particle to experience the same magnitude of force, and given that the alpha particle has twice the charge of the proton, the sine of the angle for the alpha particle must be half that of the proton to compensate for the double charge. Since the sine function is not linear, we cannot directly halve the angle, but knowing the relationship between charge, force, and angle, we can deduce that the angle for the alpha particle must be smaller and adjust for its greater charge. The correct answer must account for this non-linear relationship, leading to the selection of 9.8°, which is significantly smaller than the proton's angle, reflecting the necessary adjustment for the alpha particle's greater charge and mass, while ensuring the same force magnitude.

Question 10

Three identical bar magnets,each 2 cm long,are placed in a straight line on a frictionless surface with their ends equally spaced with 2 cm between ends.They are constrained so they cannot rotate,but they can slide along the straight line.The left magnet has its north pole nearest the north pole of the middle magnet,and the right magnet has its south pole toward the middle magnet's south pole.Once they are in place and released from rest,which of the following best describes the motion of the magnets?

Accepted Answer

The north pole of the left magnet and the south pole of the right magnet will attract the opposite pole of the middle magnet, causing it to move towards the left magnet. As a result, the right magnet will also move towards the left magnet. This will cause a decrease in the distance between the left and middle magnet and an increase in the distance between the middle and right magnet. Therefore, the end magnets will move away from the center magnet.

Question 11

At the Fermilab accelerator in Weston,Illinois,singly-charged ions with momentum 3.2 × 10−¹⁶ kg⋅m/s are held in a circular orbit of radius 1 km by an upward magnetic field.What B-field must be used to maintain the ions in this orbit? (qion = 1.6 × 10−¹⁹ C)

Accepted Answer

The answer of At the Fermilab accelerator in Weston,Illinois,singly-charged ions...

Question 12

A proton with initial kinetic energy E is moving in circular motion in a uniform magnetic field.When it has completed one fourth of a revolution,what is its kinetic energy?

Accepted Answer

The kinetic energy of a charged particle moving in a uniform magnetic field remains constant because the magnetic force does no work on the particle. Thus, after one fourth of a revolution, the kinetic energy of the proton is still E.

Question 13

A bar magnet is attached solidly to a frictionless surface and its length is aligned with the x axis.To the right of the first magnet a short distance away is a second bar magnet with its center placed on the x axis and its length perpendicular to the x axis.The second magnet is free to move.Once placed in position at rest,which best describes the initial motion of the second magnet?

Accepted Answer

The second magnet will start to rotate due to the magnetic field produced by the first magnet. This is because the poles of the second magnet will experience forces in opposite directions due to the magnetic field, causing it to rotate to align with the field lines of the first magnet.

Question 14

An electron which moves with a speed of 7.5 × 10³ m/s parallel to a uniform magnetic field of 0.40 T experiences a force of what magnitude? (e = 1.6 × 10−¹⁹ C)

Accepted Answer

The answer of An electron which moves with a speed...

Question 15

A deuteron,with the same charge but twice the mass of a proton,moves with a speed of 6.0 × 10⁵ m/s perpendicular to a uniform magnetic field of 0.20 T.Which of the paths described below would it follow? (q_p = 1.6 × 10−¹⁹ C and m_d = 3.34 × 10−²⁷ kg)

Accepted Answer

The magnetic force on a charged particle moving perpendicular to a magnetic field is given by F = qvB, where q is the charge of the particle, v is its speed, and B is the magnetic field strength. The force is perpendicular to both v and B, so it causes the particle to move in a circle. The radius of the circle is given by r = mv/qB, where m is the mass of the particle. In this case, the particle is a deuteron, which has a charge of +e and a mass of 2mp. So, the radius of the circle is:r = mv/qB = (2mp)(6.0 × 10^5 m/s)/(1.6 × 10^-19 C)(0.20 T) = 6.3 cmTherefore, the correct answer is A: a circular path of 6.3-cm radius.

Question 16

A proton,which moves perpendicular to a magnetic field of 0.60 T in a circular path of radius 0.080 m,has what speed? (q_p = 1.6 × 10−¹⁹ C and m_p = 1.67 × 10−²⁷ kg)

Accepted Answer

The answer of A proton,which moves perpendicular to a magnetic...

Question 17

An electron is released such that its initial velocity is from right to left across this page.The electron's path,however,is deflected in a direction toward the bottom edge of the page due to the presence of a uniform magnetic field.What is the direction of this field?

Accepted Answer

According to the right-hand rule, the direction of the magnetic field is perpendicular to both the direction of the electron's velocity and the direction of the deflection. Since the electron is deflected towards the bottom edge of the page and its initial velocity is from right to left across the page, the direction of the magnetic field must be out of the page. Thus, the correct answer is choice D, out of the page.

Question 18

An electron is moving at a speed of 12 × 10⁶ m/s at an angle of 30° with respect to a uniform magnetic field of 8.0 × 10^-4 T.What is the radius of the resulting helical path? (m_e = 9.11 × 10^-31 kg,q_e = 1.6 × 10^-19 C)

Accepted Answer

The answer of An electron is moving at a speed...

Question 19

The magnetic force on a charged particle moving at a constant speed in a uniform magnetic field will be doubled when the angle between the velocity of the particle and the direction of the magnetic field changes from 25.0° to

Accepted Answer

The magnetic force on a charged particle moving in a magnetic field is given by $F = qvB\sin(\theta)$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field direction. To double the force, the sine of the angle must double, since $q$, $v$, and $B$ remain constant. Initially, $\sin(25.0°)$ is involved, and to double this value, you need to find an angle whose sine is twice that of $\sin(25.0°)$. The correct answer is the angle that achieves this doubling effect, which is $57.7°$.

Question 20

Assume that a uniform magnetic field is directed into this page.If a proton is released with an initial velocity directed from the bottom edge to the top edge of the page,which of the following describes the direction of the resultant force acting on the proton?

Accepted Answer

The direction of the magnetic force on a charged particle moving through a magnetic field is given by the right hand rule. Point the thumb in the direction of the particle's velocity (top to bottom in this case) and the fingers in the direction of the magnetic field (into the page). The resulting force will be perpendicular to both, in this case to the left. Therefore, the correct choice is B, to the left.

Quiz 20: Induced Voltages and Inductance: Part A

The magnetic force on a charged particle moving at a constant speed in a uniform magnetic field will be halved when the angle between the velocity of the particle and the direction of the magnetic field changes from 25.0° to

A proton moving with a speed of 6.0 × 105 m/s perpendicular to a uniform magnetic field of 0.20 T will follow which of the paths described below? (qp = 1.6 × 10−19 C and mp = 1.67 × 10−27 kg)

A proton moves across the Earth's equator in a northwesterly direction.At this point the Earth's magnetic field has a direction due north and is parallel to the surface.What is the direction of the force acting on the proton at this instant?

A proton moving at a speed of 3.8 × 106 m/s cuts across the lines of a magnetic field at an angle of 40°.The strength of the field is 0.25 × 10−4 T.What is the magnitude of the force acting on the proton? (qp = 1.6 × 10−19 C)

If an electron is released at the equator and falls toward the Earth under the influence of gravity,the magnetic force on the electron will be toward the:

A proton,mass 1.67 × 10−27 kg and charge +1.6 × 10−19 C,moves in a circular orbit perpendicular to a uniform magnetic field of 0.50 T.Find the time for the proton to make one complete circular orbit.

At the Fermilab accelerator in Weston,Illinois,singly-charged ions with momentum 3.2 × 10−16 kg⋅m/s are held in a circular orbit of radius 1 km by an upward magnetic field.What B-field must be used to maintain the ions in this orbit? (qion = 1.6 × 10−19 C)