Question 1

A charged particle of mass 0.0040 kg is subjected to a 4.0 T magnetic field that acts at a right angle to its motion.If the particle moves in a circle of radius 0.10 m at a speed of 2.0 m/s,what is the magnitude of the charge on the particle?
A)0.020 C
B)50 C
C)0.00040 C
D)2500 C

Accepted Answer

The correct answer is A because the magnitude of the charge on the particle is given by the equation q = mvr/B, where m is the mass of the particle, v is its speed, r is the radius of the circle in which it moves, and B is the magnetic field strength. Plugging in the given values, we get q = (0.0040 kg)(2.0 m/s)(0.10 m)/(4.0 T) = 0.020 C.

Question 2

-A solenoid is wound with 470 turns on a form 4 cm in diameter and 50 cm long.The windings carry a current in the sense that is shown.The current produces a magnetic field,of magnitude 4.1 mT,at the center of the solenoid.In the figure,the current in the solenoid windings is closest to:
A)3.5 A
B)3.0 A
C)2.6 A
D)4.3 A
E)3.9 A

Accepted Answer

The magnetic field at the center of a solenoid is given by:B = μ₀nIwhere:μ₀ is the permeability of free space (4π × 10^-7 T·m/A)n is the number of turns per unit length (turns/m)I is the current (A)In this case, we have:n = 470 turns / 0.5 m = 940 turns/mB = 4.1 mT = 4.1 × 10^-3 TSubstituting these values into the equation above, we get:4.1 × 10^-3 T = 4π × 10^-7 T·m/A * 940 turns/m * ISolving for I, we get:I = 3.5 ATherefore, the current in the solenoid windings is closest to 3.5 A.

Question 3

A rigid rectangular loop,which measures 0.30 m by 0.40 m,carries a current of 9.9 A,as shown.A uniform external magnetic field of magnitude 1.8 T in the negative x-direction is present.Segment CD is in the x-z plane and forms a 19° angle with the z-axis,as shown.The y-component of the magnetic force on segment AB is closest to:
A)+5.1 N
B)-5.1 N
C)+1.7 N
D)-1.7 N
E)zero

Accepted Answer

A

Question 4

Alpha particles (charge = +2e,mass = 6.68 × 10^-27 kg)are accelerated in a cyclotron to a final orbit radius of 0.30 m.The magnetic field in the cyclotron is 0.80 T.The period of circular motion of the alpha particles is closest to: A)0.16 μs B)0.25 μs C)0.33 μs D)0.40 μs E)0.49 μs

Accepted Answer

The period of circular motion of a charged particle in a magnetic field is given by:T = 2πr/vwhere r is the radius of the orbit, v is the speed of the particle, and q is the charge of the particle.We can find the speed of the alpha particle using the centripetal force equation:mv²/r = qvBwhere m is the mass of the alpha particle, v is the speed of the alpha particle, r is the radius of the orbit, q is the charge of the alpha particle, and B is the magnetic field strength.Solving for v, we get:v = qBr/mSubstituting this expression for v into the equation for the period, we get:T = 2πr/(qBr/m)Simplifying, we get:T = 2πm/(qB)Plugging in the given values, we get:T = 2π(6.68 × 10^-27 kg)/(2e)(0.80 T)T ≈ 0.16 μsTherefore, the period of circular motion of the alpha particles is closest to 0.16 μs.

Question 5

-A solenoid of length 18 cm consists of a wire wrapped tightly around a wooden core.The magnetic field strength is 2.0 T inside the solenoid.If the solenoid is stretched to 29 cm by applying a force to it,what does the magnetic field become?
A)1.2 T
B)3.2 T
C)15 T
D)2.0 T

Accepted Answer

The magnetic field strength inside a solenoid is inversely proportional to its length when the number of turns and current are constant. Stretching the solenoid increases its length, thereby decreasing the magnetic field strength. The new field strength is calculated by the ratio of the original and new lengths: (18 cm / 29 cm) * 2.0 T = 1.2 T.

Question 6

A wire carrying a current is shaped in the form of a circular loop of radius 4.0 mm.If the magnetic field strength at its center is 1.1 mT with no external magnetic fields contributing to it,what is the magnitude of the current that flows through the wire?
A)7.0 A
B)28 A
C)14 A
D)17 A

Accepted Answer

The magnetic field strength at the center of a circular loop of wire is given by:B = (μ0 * I) / (2 * π * r)where:B is the magnetic field strengthμ0 is the permeability of free space (4π × 10^-7 T·m/A)I is the current flowing through the wirer is the radius of the loopSolving for I, we get:I = (B * 2 * π * r) / μ0Plugging in the given values, we get:I = (1.1 mT * 2 * π * 4.0 mm) / (4π × 10^-7 T·m/A)I ≈ 7.0 ATherefore, the magnitude of the current that flows through the wire is approximately 7.0 A.

Question 7

An electron in a magnetic field has a cyclotron frequency of 8.0 × 10¹² Hz.What is the magnetic field strength? (The mass of an electron is 9.1 × 10^-31kg,and the charge of an electron is 1.6 × 10^-19C. ) A)290 T B)4.5 T C)0.22 T D)0.0035 T

Accepted Answer

The cyclotron frequency is given by:$$f_c = \frac{qB}{2\pi m}$$where:* $f_c$ is the cyclotron frequency* $q$ is the charge of the electron* $B$ is the magnetic field strength* $m$ is the mass of the electronSolving for $B$, we get:$$B = \frac{2\pi m f_c}{q}$$Plugging in the given values, we get:$$B = \frac{2\pi (9.1 \times 10^{-31} \text{ kg})(8.0 \times 10^{12} \text{ Hz})}{1.6 \times 10^{-19} \text{ C}} = 0.22 \text{ T}$$Therefore, the correct answer is C.

Question 8

A very short solenoid contains N = 20 circular loops of wire.The radius of the loop is R = 0.5 meters,and the length of the solenoid is d = 0.02 meters.Approximately what is the magnetic field in the center of the solenoid? A)µ_oN I/d B)µ_oN I/(2R) C)µ_oN I/(2dR) D)µ_oN I/(2pR)

Accepted Answer

The magnetic field in the center of a solenoid can be approximated by the equation B = µN*I/(2R), where N is the number of loops, I is the current, R is the radius of the loops, and µ is the permeability of free space. Plugging in the given values, we get B = 4π*10^-7 * 20*I/(2*0.5) = 2.5*10^-5 I T. Thus, the magnetic field in the center of the solenoid is proportional to the current, and the correct choice is B, which matches this formula.

Question 9

A circular coil of wire of 200 turns and diameter 6 cm carries a current of 7 A.It is placed in a magnetic field of 0.90 T with the plane of the coil making an angle of 30° with the magnetic field.What is the torque on the coil?
A)3.1 N ∙ m
B)1.8 N ∙ m
C)5.9 N ∙ m
D)8.2 N ∙ m
E)1.5 N ∙ m

Accepted Answer

The torque on a coil is given by τ = n * I * A * B * sin(θ), where n is the number of turns, I is the current, A is the area of the coil, B is the magnetic field, and θ is the angle between the normal to the coil and the magnetic field. Here, n = 200, I = 7 A, B = 0.90 T, θ = 30°, and the area A = π * (0.03 m)^2. Calculating this gives τ = 3.1 N ∙ m.

Question 10

Two parallel,straight wires are 7.0 cm apart and each carries a 18.0 A current in the same direction.One wire is securely anchored,and the other is attached in the center to a movable cart.If the force needed to move the wire when it is not attached to the cart is negligible,with what force does the wire pull on the cart? Both wires are 50.0 m long.
A)4600 dynes
B)370 dynes
C)660 dynes
D)9300 dynes

Accepted Answer

The force between two parallel wires carrying current in the same direction is given by:F = (μ0 * I1 * I2 * L) / (2π * d)where:μ0 is the permeability of free space (4π x 10^-7 T·m/A)I1 and I2 are the currents in the wires (18.0 A)L is the length of the wires (50.0 m)d is the distance between the wires (7.0 cm = 0.07 m)Plugging in the values, we get:F = (4π x 10^-7 T·m/A * 18.0 A * 18.0 A * 50.0 m) / (2π * 0.07 m)F = 4600 dynesTherefore, the force with which the wire pulls on the cart is 4600 dynes.

Question 11

How far from the center of a wire carrying a 2.0 A current does the magnetic field that results from the current have a magnitude of 4.0 × 10^-5 T? A)1.0 mm B)0.5 mm C)0.16 mm D)13 mm

Accepted Answer

The magnetic field B around a long straight wire is given by B = (μ₀I)/(2πr), where μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), I is the current, and r is the distance from the wire. Solving for r with B = 4.0 × 10⁻⁵ T and I = 2.0 A gives r = 1.0 mm.

Question 12

Alpha particles (charge = +2e,mass = 6.68 × 10^-27 kg)are accelerated in a cyclotron to a final orbit radius of 0.50 m.The magnetic field in the cyclotron is 0.50 T.The kinetic energy of an alpha particle in the final orbit is closest to: A)3.0 MeV B)2.6 MeV C)3.4 MeV D)3.9 MeV E)4.3 MeV

Accepted Answer

The kinetic energy (KE) of an alpha particle in a cyclotron can be calculated using the formula KE = (q^2 * B^2 * r^2) / (2m), where q is the charge, B is the magnetic field, r is the radius, and m is the mass. Substituting the given values (q = 2e = 2 * 1.6 × 10^-19 C, B = 0.50 T, r = 0.50 m, m = 6.68 × 10^-27 kg) into the formula gives a kinetic energy closest to 3.0 MeV.

Question 13

A rigid rectangular loop,which measures 0.30 m by 0.40 m,carries a current of 5.5 A,as shown.A uniform external magnetic field of magnitude 2.9 T in the negative x-direction is present.Segment CD is in the x-z plane and forms a 35° angle with the z-axis,as shown.An external torque applied to the loop keeps it in static equilibrium.The magnitude of the external torque is closest to:
A)1.1 N ∙ m
B)0.73 N ∙ m
C)1.3 N ∙ m
D)1.4 N ∙ m
E)1.6 N ∙ m

Accepted Answer

The answer of  A rigid rectangular loop,which measures 0.30...

Question 14

A very long wire generates a magnetic field of 0.0050 × 10^-4T at a distance of 3.0 mm.What is the magnitude of the current? A)7.5 mA B)1.7 mA C)3300 mA D)24,000 mA

Accepted Answer

The magnetic field of a long wire is given by:$$B = \frac{\mu_0 I}{2\pi r}$$where:* B is the magnetic field in teslas (T)* μ0 is the permeability of free space (4π × 10^-7 T·m/A)* I is the current in amperes (A)* r is the distance from the wire in meters (m)We can rearrange this equation to solve for the current:$$I = \frac{2\pi rB}{\mu_0}$$Plugging in the given values, we get:$$I = \frac{2\pi (3.0 	imes 10^{-3} 	ext{ m})(0.0050 	imes 10^{-4} 	ext{ T})}{4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m}/	ext{A}} = 7.5 	imes 10^{-3} 	ext{ A} = 7.5 	ext{ mA}$$Therefore, the magnitude of the current is 7.5 mA.

Question 15

The magnitude of a magnetic field a distance 3.0 µm from a wire is 20.0 × 10^-4 T.How much current is flowing through the wire? Assume the wire is the only contributor to the magnetic field. A)30 mA B)188 mA C)19 mA D)377 mA

Accepted Answer

The magnetic field around a long, straight wire is given by:$$B = \frac{\mu_0 I}{2\pi r}$$where:* B is the magnetic field strength in teslas (T)* μ0 is the permeability of free space (4π × 10^-7 T·m/A)* I is the current in the wire in amperes (A)* r is the distance from the wire in meters (m)We are given that:* B = 20.0 × 10^-4 T* r = 3.0 µm = 3.0 × 10^-6 mSubstituting these values into the equation above, we get:$$20.0 × 10^-4 T = \frac{(4π × 10^-7 T·m/A) I}{2π (3.0 × 10^-6 m)}$$Solving for I, we get:$$I = 30 mA$$Therefore, the correct answer is A.

Question 16

A uniform magnetic field of magnitude 0.80 T in the negative z-direction is present in a region of space.A uniform electric field is also present.In the figure,the electric field is set at 20,800 V/m in the positive y-direction.An electron is projected with an initial velocity vo = 2.6 × 10⁴ m/s.in the positive x-direction.The y-component of the initial force on the electron is closest to: A)- 7 × 10^-15 N B)+ 7 × 10^-15 N C)- 3 × 10^-15 N D)+ 3 × 10^-15N E)zero

Accepted Answer

The answer of  A uniform magnetic field of magnitude...

Question 17

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? (The mass of an electron is 9.1 × 10^-31 kg,and the charge of an electron is 1.6 × 10^-19C. ) A)2.1 × 10¹⁰ Hz B)4.8 × 10^-7 Hz C)2.1 × 10¹⁴Hz D)4.8 × 10^-11Hz

Accepted Answer

The frequency of circular motion is given by:

f = (qB)/(2πm)

where q is the charge of the particle, B is the magnetic field strength, and m is the mass of the particle.

Plugging in the values for the electron:

q = 1.6 × 10^-19 C
B = 0.75 T
m = 9.1 × 10^-31 kg

f = (1.6 × 10^-19 C × 0.75 T)/(2π × 9.1 × 10^-31 kg)

f = 2.1 × 10^10 Hz

Therefore, the electron traverses a circular path at a frequency of 2.1 × 10^10 Hz, which corresponds to answer choice A.

Question 18

A square wire 2.0 cm on each side is oriented so that a 5.00 mT magnetic field makes a 30.0º angle with the normal to the plane of the square.If a 1.0 mA current causes a 32.0 N · m torque on the wire,how many loops does the wire make? A)3.2 × 10¹⁰ B)3.2 × 10⁶ C)1.6 × 10¹⁰ D)4.5 × 10¹⁰

Accepted Answer

The torque on a current loop is given by:$$	au = NIAB\sin	heta$$where:* N is the number of loops* I is the current* A is the area of the loop* B is the magnetic field strength* θ is the angle between the normal to the loop and the magnetic fieldIn this problem, we are given:* I = 1.0 mA = 0.001 A* A = (2.0 cm)^2 = 4.0 cm^2 = 4.0 x 10^-4 m^2* B = 5.00 mT = 0.005 T* θ = 30.0º* τ = 32.0 N · mWe can solve for N by rearranging the equation:$$N = \frac{	au}{IAB\sin	heta}$$Plugging in the values we know, we get:$$N = \frac{32.0 	ext{ N} \cdot 	ext{m}}{(0.001 	ext{ A})(4.0 	imes 10^{-4} 	ext{ m}^2)(0.005 	ext{ T})(\sin 30.0º)}$$$$N = 3.2 	imes 10^{10}$$Therefore, the correct answer is A.

Question 19

A mass spectrograph is operated with deuterons, which have a charge of +e and a mass of 
3.34 × 10S1U1P1-27S1S1P0 kg. Deuterons emerge from the source, which is grounded with negligible velocity. The velocity of the deuterons as they pass through the accelerator grid is 8.0 × 10S1U1P15S1S1P0 m/s. A uniform magnetic field of magnitude B = 0.20 T, directed out of the plane, is present at the right of the grid.
-In the figure,the deuterons are in circular orbit in the magnetic field.The radius of the orbit and the initial sense of deflection are closest to:
A)60 mm,upward
B)70 mm,upward
C)60 mm,downward
D)70 mm,downward
E)80 mm,downward

Accepted Answer

The answer of 
A mass spectrograph is operated with deuterons,...

Question 20

A 5 m length of wire carrying a current of 5 A lies on a horizontal table with a rectangular top of dimensions 0.300 m × 0.400 m.The ends of the wire are attached to opposite ends of a diagonal of the rectangle.A vertical magnetic field of 0.30 T is present.What magnetic force acts on this segment of wire?
A)0.75 N
B)1.1 N
C)7.5 N
D)zero
E)The force cannot be determined without knowing the shape of the length of wire.

Accepted Answer

The magnetic force on a current-carrying wire in a magnetic field is given by F = BIL sin(θ). Here, B = 0.30 T, I = 5 A, L = 5 m, and θ = 90° (since the wire is horizontal and the magnetic field is vertical). Thus, F = 0.30 * 5 * 5 * sin(90°) = 0.75 N.

Quiz 24: Magnetic Fields and Forces

A charged particle of mass 0.0040 kg is subjected to a 4.0 T magnetic field that acts at a right angle to its motion.If the particle moves in a circle of radius 0.10 m at a speed of 2.0 m/s,what is the magnitude of the charge on the particle?

-A solenoid is wound with 470 turns on a form 4 cm in diameter and 50 cm long.The windings carry a current in the sense that is shown.The current produces a magnetic field,of magnitude 4.1 mT,at the center of the solenoid.In the figure,the current in the solenoid windings is closest to:

Alpha particles (charge = +2e,mass = 6.68 × 10^-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m.The magnetic field in the cyclotron is 0.80 T.The period of circular motion of the alpha particles is closest to:

-A solenoid of length 18 cm consists of a wire wrapped tightly around a wooden core.The magnetic field strength is 2.0 T inside the solenoid.If the solenoid is stretched to 29 cm by applying a force to it,what does the magnetic field become?

A wire carrying a current is shaped in the form of a circular loop of radius 4.0 mm.If the magnetic field strength at its center is 1.1 mT with no external magnetic fields contributing to it,what is the magnitude of the current that flows through the wire?

An electron in a magnetic field has a cyclotron frequency of 8.0 × 10¹² Hz.What is the magnetic field strength? (The mass of an electron is 9.1 × 10^-31kg,and the charge of an electron is 1.6 × 10^-19C. )

A very short solenoid contains N = 20 circular loops of wire.The radius of the loop is R = 0.5 meters,and the length of the solenoid is d = 0.02 meters.Approximately what is the magnetic field in the center of the solenoid?

A circular coil of wire of 200 turns and diameter 6 cm carries a current of 7 A.It is placed in a magnetic field of 0.90 T with the plane of the coil making an angle of 30° with the magnetic field.What is the torque on the coil?

How far from the center of a wire carrying a 2.0 A current does the magnetic field that results from the current have a magnitude of 4.0 × 10^-5 T?

Alpha particles (charge = +2e,mass = 6.68 × 10^-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.50 m.The magnetic field in the cyclotron is 0.50 T.The kinetic energy of an alpha particle in the final orbit is closest to:

A very long wire generates a magnetic field of 0.0050 × 10^-4T at a distance of 3.0 mm.What is the magnitude of the current?

The magnitude of a magnetic field a distance 3.0 µm from a wire is 20.0 × 10^-4 T.How much current is flowing through the wire? Assume the wire is the only contributor to the magnetic field.

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? (The mass of an electron is 9.1 × 10^-31 kg,and the charge of an electron is 1.6 × 10^-19C. )

A square wire 2.0 cm on each side is oriented so that a 5.00 mT magnetic field makes a 30.0º angle with the normal to the plane of the square.If a 1.0 mA current causes a 32.0 N · m torque on the wire,how many loops does the wire make?

Quiz 24: Magnetic Fields and Forces

A charged particle of mass 0.0040 kg is subjected to a 4.0 T magnetic field that acts at a right angle to its motion.If the particle moves in a circle of radius 0.10 m at a speed of 2.0 m/s,what is the magnitude of the charge on the particle?

-A solenoid is wound with 470 turns on a form 4 cm in diameter and 50 cm long.The windings carry a current in the sense that is shown.The current produces a magnetic field,of magnitude 4.1 mT,at the center of the solenoid.In the figure,the current in the solenoid windings is closest to:

Alpha particles (charge = +2e,mass = 6.68 × 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m.The magnetic field in the cyclotron is 0.80 T.The period of circular motion of the alpha particles is closest to:

-A solenoid of length 18 cm consists of a wire wrapped tightly around a wooden core.The magnetic field strength is 2.0 T inside the solenoid.If the solenoid is stretched to 29 cm by applying a force to it,what does the magnetic field become?

A wire carrying a current is shaped in the form of a circular loop of radius 4.0 mm.If the magnetic field strength at its center is 1.1 mT with no external magnetic fields contributing to it,what is the magnitude of the current that flows through the wire?

An electron in a magnetic field has a cyclotron frequency of 8.0 × 1012 Hz.What is the magnetic field strength? (The mass of an electron is 9.1 × 10-31 kg,and the charge of an electron is 1.6 × 10-19 C. )

A very short solenoid contains N = 20 circular loops of wire.The radius of the loop is R = 0.5 meters,and the length of the solenoid is d = 0.02 meters.Approximately what is the magnetic field in the center of the solenoid?

A circular coil of wire of 200 turns and diameter 6 cm carries a current of 7 A.It is placed in a magnetic field of 0.90 T with the plane of the coil making an angle of 30° with the magnetic field.What is the torque on the coil?

How far from the center of a wire carrying a 2.0 A current does the magnetic field that results from the current have a magnitude of 4.0 × 10-5 T?

Alpha particles (charge = +2e,mass = 6.68 × 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.50 m.The magnetic field in the cyclotron is 0.50 T.The kinetic energy of an alpha particle in the final orbit is closest to:

A very long wire generates a magnetic field of 0.0050 × 10-4 T at a distance of 3.0 mm.What is the magnitude of the current?

The magnitude of a magnetic field a distance 3.0 µm from a wire is 20.0 × 10-4 T.How much current is flowing through the wire? Assume the wire is the only contributor to the magnetic field.

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? (The mass of an electron is 9.1 × 10-31 kg,and the charge of an electron is 1.6 × 10-19 C. )