Question 1

An electron moving with a velocity = 5.0 × 10⁷ m/s Enters a region of space where perpendicular electric and a magnetic fields are present.The electric field is = )What magnetic field will allow the electron to go through the region without being deflected? A) = +2.0 × 10^-4 T B) = -2.0 × 10^-4 T C) = +2.0 × 10^-4 T D) = -2.0 × 10^-4 T E) = +5.0 × 10^-4 T

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Question 2

A wire segment 1.2 m long carries a current I = 3.5 A and is oriented as shown in the figure.A uniform magnetic field of magnitude 0.50 T pointing toward the -x direction is present as shown.The +z-axis points directly into the page.What is the magnetic force vector on the wire segment? 
A) +1.6 N 
B) -1.6 N 
C) +1.6 N 
D) (+1.3  - 1.6
 ) N
E) (-1.3  + 1.6
 N

Accepted Answer

The magnetic force on a wire segment is given by:

F = IlBsinθ

where I is the current, L is the length of the wire segment, B is the magnetic field, and θ is the angle between the direction of the current and the magnetic field.

In this case, the wire segment is oriented at an angle of 90 degrees to the magnetic field (sin 90 = 1). Therefore, the magnetic force is:

F = (3.5 A)(1.2 m)(0.50 T) = 2.1 N

Since the magnetic field points in the -x direction, the force vector points in the +y direction (using the right-hand rule). Therefore, the correct answer is C: +1.6 N. The other answer choices are either incorrect or involve calculations that are not necessary.

Question 3

In a mass spectrometer,a singly-charged particle (charge e)has a speed of 1.0 × 10⁶ m/s and enters a uniform magnetic field of 0.20 T. The radius of the circular orbit of the particle is 0.020 m. What is the mass of this particle? (e = 1.60 × 10^-19C) A) 3.2 × 10^-28 kg B) 6.4 × 10^-28 kg C) 1.7 × 10^-27 kg D) 4.5 × 10^-27 kg

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Question 4

An L-shaped metal machine part is made of two equal-length segments that are perpendicular to each other and carry a 4.50-A current as shown in the figure.This part has a total mass of 3.80 kg and a total length of 3.00 m,and it is in an external 1.20-T magnetic field that is oriented perpendicular to the plane of the part,as shown.What is the magnitude of the NET magnetic force that the field exerts on the part? 
A) 8.10 N
B) 11.5 N
C) 16.2 N
D) 22.9 N
E) 32.4 N

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Question 5

A beam of electrons is accelerated through a potential difference of 10 kV before entering a region having uniform electric and magnetic fields that are perpendicular to each other and perpendicular to the direction in which the electron is moving. If the magnetic field in this region has a value of 0.010 T,what magnitude of the electric field is required if the particles are to be undeflected as they pass through the region? A) 2.3 × 10³ V/m B) 7.9 × 10³ V/m C) 5.9 × 10⁵ V/m D) 6.0 × 10⁵ V/m E) 7.2 × 10⁶ V/m

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Question 6

A proton is first accelerated from rest through a potential difference V and then enters a uniform 0.750-T magnetic field oriented perpendicular to its path.In this field,the proton follows a circular arc having a radius of curvature of 1.84 cm.What was the potential difference V? (m_proton = 1.67 × 10^-27 kg, e = 1.60 × 10^-19C)

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Question 7

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 current in the rod is needed in order for the magnetic force to cancel the weight of the rod?
A) 1.2 A
B) 2.5 A
C) 4.9 A
D) 7.6 A
E) 9.8 A

Accepted Answer

The weight of the rod can be calculated using the formula:

w = mg

where w is the weight, m is the mass, and g is the acceleration due to gravity (9.8 m/s^2).

w = 0.050 kg × 9.8 m/s^2 = 0.49 N

The magnetic force can be calculated using the formula:

F = BIL

where F is the magnetic force, B is the magnetic field, I is the current, and L is the length of the rod.

For the magnetic force to cancel the weight, we need:

F = w

BIL = mg

Solving for I:

I = mg/BL

Plugging in the values:

I = 0.050 kg × 9.8 m/s^2 / (0.10 T × 1.0 m)

I = 4.9 A

Therefore, the minimum current needed is 4.9 A, which corresponds to choice C.

Question 8

A doubly charged ion (charge 2e)with velocity 6.9 × 10⁶ m/s moves in a circular path of diameter 60)0 cm in a magnetic field of 0.80 T in a mass spectrometer. What is the mass of this ion? (e = 1.60 × 10^-19C) A) 11 × 10^-27 kg B) 6.7 × 10^-27 kg C) 4.5 × 10^-27 kg D) 3.3 × 10^-27 kg

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Question 9

A straight wire that is 0.60 m long is carrying a current of 2.0 A.It is placed in a uniform magnetic field of strength 0.30 T.If the wire experiences a force of 0.18 N,what angle does the wire make with respect to the magnetic field?
A) 25°
B) 30°
C) 35°
D) 60°
E) 90°

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Question 10

A charged particle of mass 0.0020 kg is subjected to a 6.0 T magnetic field which acts at a right angle to its motion.If the particle moves in a circle of radius 0.20 m at a speed of 5.0 m/s,what is the magnitude of the charge on the particle?
A) 0.0083 C
B) 120 C
C) 0.00040 C
D) 2500 C

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Question 11

A wire in the shape of an "M" lies in the plane of the paper.It carries a current of 2.0 A,flowing from points A to E,as shown in the figure.It is placed in a uniform magnetic field of 0.75 T in the same plane,directed as shown on the right side of the figure.The figure indicates the dimensions of the wire.What are the magnitude and direction of the force acting on  
(a)section AB of this wire?
(b)section BC of this wire?
(c)section CD of this wire?
(d)section DE of this wire?
(e)the entire wire?

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Question 12

The figure shows a velocity selector that can be used to measure the speed of a charged particle.A beam of particles is directed along the axis of the instrument.A parallel plate capacitor sets up an electric field E,which is oriented perpendicular to a uniform magnetic field B.If the plates are separated by 2)0 mm and the value of the magnetic field is 0.60 T,what voltage between the plates will allow particles of speed 5.0 × 10⁵ m/s to pass straight through without deflection? A) 600 V B) 1900 V C) 3800 V D) 190 V E) 94 V

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Question 13

A straight 15.0-g wire that is 2.00 m long carries a current of 8.00 A.This wire is aligned horizontally along the west-east direction with the current going from west to east.You want to support the wire against gravity using the weakest possible uniform external magnetic field.
(a)Which way should the magnetic field point?
(b)What is the magnitude of the weakest possible magnetic field you could use?

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Question 14

Alpha particles (charge = +2e,mass = 6.68 × 10^-²⁷ kg,e = 1.60 × 10^-19C)0.87 μs. A) 0.44 μs. B) 0.67 μs. C) are accelerated in a cyclotron to a final orbit radius of 0.90 m. The magnetic field in the cyclotron is 0.30 T. The period of the circular motion of the alpha particles is closest to D) 1.1 μs.

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Question 15

An electron moving perpendicular to a uniform magnetic field of 3.2 × 10^-2 T moves in a circle of radius 0.40 cm.How fast is this electron moving? (m_el = 9.11 × 10^-31 kg.e = 1.60 × 10^-19C) A) 2.2 × 10⁷ m/s B) 1.9 × 10^-2 m/s C) 1.9 × 10^-30 m/s D) 3.0 × 10⁶ m/s

Accepted Answer

The correct answer is A.The magnetic force on the electron is given by:$$F = qvB$$where:* q is the charge of the electron* v is the speed of the electron* B is the magnetic field strengthThe centripetal force on the electron is given by:$$F = mv^2/r$$where:* m is the mass of the electron* v is the speed of the electron* r is the radius of the circleSetting the magnetic force and the centripetal force equal to each other, we get:$$qvB = mv^2/r$$Solving for v, we get:$$v = qBr/m$$Plugging in the given values, we get:$$v = (1.60 × 10^{-19} C)(3.2 × 10^{-2} T)(0.40 × 10^{-2} m)/(9.11 × 10^{-31} kg)$$$$v = 2.2 × 10^7 m/s$$

Question 16

An alpha particle is moving at a speed of 5.0 × 10⁵ m/s in a direction perpendicular to a uniform magnetic field of strength 0.040 T.The charge on an alpha particle is 3.2 × 10^-19 C and its mass is 6.6 × 10^-27 kg. (a)What is the radius of the path of the alpha particle? (b)How long does it take the alpha particle to make one complete revolution around its path?

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Question 17

As shown in the figure,a small particle of charge q = -7.0 × 10^-6 C and mass m = 3.1 × 10^-12 kg has velocity v₀ = 9.4 × 10³ m/s as it enters a region of uniform magnetic field.The particle is observed to travel in the semicircular path shown,with radius R = 5.0 cm.Calculate the magnitude and direction of the magnetic field in the region.

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Question 18

A wire carries a 4.0-A current along the +x-axis through a magnetic field  = (5.0 
+ 7)0 
)T) If the wire experiences a force of 30 N 
As a result,how long is the wire?
A) 1.1 m
B) 0.87 m
C) 1.5 m
D) 0.63 m

Accepted Answer

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Question 19

An electron enters a magnetic field of 0.75 T with a velocity perpendicular to the direction of the field.At what frequency does the electron traverse a circular path? (m_el = 9.11 × 10^-31 kg,e = 1.60 × 10^-19C) A) 2.1 × 10¹⁰ Hz B) 4.8 × 10^-7 Hz C) 2.1 × 10¹⁴ Hz D) 4.8 × 10^-11 Hz

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Question 20

A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (m_proton = 1.67 × 10^-27 kg,e = 1.60 × 10^-19C) A) 0.080 m B) 0.11 m C) 0.14 m D) 0.17 m

Accepted Answer

First, we can use the formula for the speed of a particle moving through a potential difference V:

v = sqrt(2eV/m)

where e is the charge of the particle, m is its mass, and V is the potential difference. Plugging in the values for a proton:

v = sqrt((2×1.60×10^-19 C × 1000 V) / 1.67×10^-27 kg) = 1.84×10^7 m/s

Next, we can use the formula for the radius of a charged particle moving through a magnetic field:

r = mv / (eB)

where B is the magnetic field strength, and we use the velocity we just found. Plugging in the values:

r = (1.67×10^-27 kg × 1.84×10^7 m/s) / (1.60×10^-19 C × 0.040 T) = 0.11 m

Therefore, the answer is B) 0.11 m, as this is the calculated radius of the proton's resulting orbit.

Quiz 1: A: Foundations

An electron moving with a velocity = 5.0 × 10⁷ m/s Enters a region of space where perpendicular electric and a magnetic fields are present.The electric field is = ) What magnetic field will allow the electron to go through the region without being deflected?

A wire segment 1.2 m long carries a current I = 3.5 A and is oriented as shown in the figure.A uniform magnetic field of magnitude 0.50 T pointing toward the -x direction is present as shown.The +z-axis points directly into the page.What is the magnetic force vector on the wire segment?

In a mass spectrometer,a singly-charged particle (charge e) has a speed of 1.0 × 10⁶ m/s and enters a uniform magnetic field of 0.20 T. The radius of the circular orbit of the particle is 0.020 m. What is the mass of this particle? (e = 1.60 × 10^-19C)

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 current in the rod is needed in order for the magnetic force to cancel the weight of the rod?

A doubly charged ion (charge 2e) with velocity 6.9 × 10⁶ m/s moves in a circular path of diameter 60) 0 cm in a magnetic field of 0.80 T in a mass spectrometer. What is the mass of this ion? (e = 1.60 × 10^-19C)

A straight wire that is 0.60 m long is carrying a current of 2.0 A.It is placed in a uniform magnetic field of strength 0.30 T.If the wire experiences a force of 0.18 N,what angle does the wire make with respect to the magnetic field?

A charged particle of mass 0.0020 kg is subjected to a 6.0 T magnetic field which acts at a right angle to its motion.If the particle moves in a circle of radius 0.20 m at a speed of 5.0 m/s,what is the magnitude of the charge on the particle?

Alpha particles (charge = +2e,mass = 6.68 × 10^-²⁷ kg,e = 1.60 × 10^-19C) 0.87 μs.

An electron moving perpendicular to a uniform magnetic field of 3.2 × 10^-2 T moves in a circle of radius 0.40 cm.How fast is this electron moving? (m_el = 9.11 × 10^-31 kg.e = 1.60 × 10^-19C)

A wire carries a 4.0-A current along the +x-axis through a magnetic field = (5.0 + 7) 0 ) T) If the wire experiences a force of 30 N As a result,how long is the wire?

An electron enters a magnetic field of 0.75 T with a velocity perpendicular to the direction of the field.At what frequency does the electron traverse a circular path? (m_el = 9.11 × 10^-31 kg,e = 1.60 × 10^-19C)

A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (m_proton = 1.67 × 10^-27 kg,e = 1.60 × 10^-19C)

Quiz 1: A: Foundations

An electron moving with a velocity = 5.0 × 107 m/s Enters a region of space where perpendicular electric and a magnetic fields are present.The electric field is = ) What magnetic field will allow the electron to go through the region without being deflected?

A wire segment 1.2 m long carries a current I = 3.5 A and is oriented as shown in the figure.A uniform magnetic field of magnitude 0.50 T pointing toward the -x direction is present as shown.The +z-axis points directly into the page.What is the magnetic force vector on the wire segment?

In a mass spectrometer,a singly-charged particle (charge e) has a speed of 1.0 × 106 m/s and enters a uniform magnetic field of 0.20 T. The radius of the circular orbit of the particle is 0.020 m. What is the mass of this particle? (e = 1.60 × 10-19C)

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 current in the rod is needed in order for the magnetic force to cancel the weight of the rod?

A doubly charged ion (charge 2e) with velocity 6.9 × 106 m/s moves in a circular path of diameter 60) 0 cm in a magnetic field of 0.80 T in a mass spectrometer. What is the mass of this ion? (e = 1.60 × 10-19C)