Question 1

A straight wire carries a current of 10 A at an angle of 30° with respect to the direction of a uniform 0.30-T magnetic field. Find the magnitude of the magnetic force on a 0.50-m length of the wire.

Accepted Answer

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Question 2

A proton having a speed of $3.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ in a direction perpendicular to a uniform magnetic field moves in a circle of radius $0.20 \mathrm {~m}$ within the field. What is the magnitude of the magnetic field? $( e$ $= 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {proton } } = 1.67 \times 10 ^ { - 27 } \mathrm {~kg}$ )

Accepted Answer

The magnitude of the magnetic field can be found using the formula for the magnetic force acting on a moving charge, which equals the centripetal force required to keep the proton moving in a circle: $qvB = \frac{mv^2}{r}$. Solving for $B$, we get $B = \frac{mv}{qr}$. Substituting the given values: $B = \frac{(1.67 \times 10^{-27} \, \text{kg})(3.0 \times 10^6 \, \text{m/s})}{(1.60 \times 10^{-19} \, \text{C})(0.20 \, \text{m})} = 0.16 \, \text{T}$.

Question 3

An electron moves with a speed of $3.0 \times 104 \mathrm {~m} / \mathrm { s }$ perpendicular to a uniform magnetic field of $0.40$ T. What is the magnitude of the magnetic force on the electron? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

Accepted Answer

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Question 4

Alpha particles, each having a charge of $+ 2 e$ and a mass of $6.64 \times 10 ^ { - 27 } \mathrm {~kg}$, are accelerated in a uniform $0.50 \mathrm {~T}$ magnetic field to a final orbit radius of $0.50 \mathrm {~m}$. The field is perpendicular to the velocity of the particles. What is the kinetic energy of an alpha particle in the final orbit? $( 1 \mathrm { eV } =$ $\left. 1.60 \times 10 ^ { - 19 } \mathrm {~J} , e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

Accepted Answer

The correct answer is A.The kinetic energy of an alpha particle in the final orbit is given by:$$K = \frac{1}{2}mv^2 = \frac{1}{2}\left(6.64 	imes 10^{-27} \mathrm{kg}ight)\left(2.65 	imes 10^7 \mathrm{m/s}ight)^2 = 3.0 \mathrm{MeV}$$

Question 5

A doubly charged ion with speed $6.9 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ enters a uniform $0.80 - T$ magnetic field, traveling perpendicular to the field. Once in the field, it moves in a circular arc of radius $30 \mathrm {~cm}$. What is the mass of this ion? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

Accepted Answer

The correct answer is A because the mass of the ion can be calculated using the formula:$$m = \frac{qBr}{v}$$where:- m is the mass of the ion- q is the charge of the ion- B is the magnetic field strength- r is the radius of the circular arc- v is the speed of the ionPlugging in the given values, we get:$$m = \frac{(2e)(0.80 \mathrm{~T})(0.30 \mathrm{~m})}{6.9 	imes 10^6 \mathrm{~m/s}} = 1.1 	imes 10^{-27} \mathrm{~kg}$$

Question 6

A charged particle of mass 0.0040 kg is subjected to a 4.0-T magnetic field which 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?

Accepted Answer

The magnitude of the charge on the particle can be found using the formula for the magnetic force acting on a moving charge, which is given by $F = qvB$, where $F$ is the magnetic force, $q$ is the charge, $v$ is the velocity of the particle, and $B$ is the magnetic field strength. The magnetic force in this scenario also provides the centripetal force required to keep the particle moving in a circle, which is given by $F = \frac{mv^2}{r}$, where $m$ is the mass of the particle, $v$ is its velocity, and $r$ is the radius of the circle.Equating the two expressions for the force gives us $\frac{mv^2}{r} = qvB$. Solving for $q$ gives us $q = \frac{mv^2}{rvB}$.Substituting the given values: $m = 0.0040 \, \text{kg}$, $v = 2.0 \, \text{m/s}$, $r = 0.10 \, \text{m}$, and $B = 4.0 \, \text{T}$, we get:$$q = \frac{(0.0040)(2.0)^2}{(0.10)(4.0)} = \frac{(0.0040)(4.0)}{0.40} = \frac{0.016}{0.40} = 0.040 \, \text{C}$$Thus, the magnitude of the charge on the particle is 0.040 C, which corresponds to choice C.

Question 7

In the figure, a small particle of charge $- 1.9 \times 10 ^ { - 6 } \mathrm { C }$ and mass $m = 3.1 \times 10 ^ { - 12 } \mathrm {~kg}$ has speed $v _ { 0 } = 8.1 \times 10 ^ { 3 } \mathrm {~m} / \mathrm { s }$ as it enters a region of uniform magnetic field. The particle is initially traveling perpendicular to the magnetic field and is observed to travel in the semicircular path shown with radius $R = 5.0 \mathrm {~cm}$. Find the magnitude and direction of the magnetic field in the region.

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Question 8

A proton, starting from rest, accelerates through a potential difference of $1.0 \mathrm { kV }$ and then moves into a magnetic field of $0.040 \mathrm {~T}$ at a right angle to the field. What is the radius of the proton's resulting orbit? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {proton } } = 1.67 \times 10 ^ { - 27 } \mathrm {~kg} \right)$

Accepted Answer

The radius $r$ of the proton's orbit in a magnetic field can be found using the formula $r = \frac{mv}{qB}$, where $m$ is the mass of the proton, $v$ is the velocity of the proton, $q$ is the charge of the proton, and $B$ is the magnetic field strength. The kinetic energy gained by the proton due to the acceleration through the potential difference $V$ is given by $K.E. = qV$, and since $K.E. = \frac{1}{2}mv^2$, we can solve for $v = \sqrt{\frac{2qV}{m}}$. Substituting the given values: $v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \mathrm{C} \times 1000 \mathrm{V}}{1.67 \times 10^{-27} \mathrm{kg}}}$, and then substituting $v$ into the formula for $r$, we get $r = \frac{1.67 \times 10^{-27} \mathrm{kg} \times v}{1.60 \times 10^{-19} \mathrm{C} \times 0.040 \mathrm{T}}$. Calculating this gives a radius close to $0.11 \mathrm{m}$, making option C correct.

Question 9

A proton is accelerated from rest through 0.50 kV. It then enters a uniform magnetic field
of 0.30 T that is oriented perpendicular to its direction of motion.
(a) What is the radius of the path the proton follows in the magnetic field?
(b) How long does it take the proton to make one complete circle in the magnetic field?.

Accepted Answer

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Question 10

A 2.0-m straight wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of the magnetic force on it?

Accepted Answer

The magnetic force on a current-carrying wire is given by F = ILB sin(θ), where I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the wire and the magnetic field. Since the wire is parallel to the magnetic field, θ = 0 degrees, making sin(θ) = 0. Therefore, the force is zero.

Question 11

An electron moves with a speed of $8.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ along the $+ x$-axis. It enters a region where there is a magnetic field of $2.5 \mathrm {~T}$, directed at an angle of $60 ^ { \circ }$ to the $+ x$-axis and lying in the $x y$-plane.
Calculate the magnitude of the acceleration of the electron. $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { e 1 } = 9.11 \times 10 ^ { - 31 } \right.$ $\mathrm { kg }$ )

Accepted Answer

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Question 12

A proton moving eastward with a velocity of $5.0 \mathrm {~km} / \mathrm { s }$ enters a magnetic field of $0.20 \mathrm {~T}$ pointing northward. What are the magnitude and direction of the force that the magnetic field exerts on the proton? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

Accepted Answer

The force on a charged particle moving through 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. Here, $q = 1.6 \times 10^{-19} \mathrm{C}$, $v = 5.0 \times 10^3 \mathrm{m/s}$, $B = 0.20 \mathrm{T}$, and $\theta = 90^\circ$ (since the velocity is eastward and the magnetic field is northward, they are perpendicular). Thus, $F = (1.6 \times 10^{-19} \mathrm{C})(5.0 \times 10^3 \mathrm{m/s})(0.20 \mathrm{T}) = 1.6 \times 10^{-16} \mathrm{N}$. The direction of the force is given by the right-hand rule, which in this case results in an upward force.

Question 13

An electron moves with a speed of $8.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ along the $+ x$-axis. It enters a region where there is a magnetic field of $2.5 \mathrm {~T}$, directed at an angle of $60 ^ { \circ }$ to the $+ x$-axis and lying in the $x y$-plane.
Calculate the magnitude of the magnetic force on the electron. $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

Accepted Answer

The magnetic force on a charged particle is given by $ F = qvB \sin \theta $. Here, $ q = 1.60 \times 10^{-19} \, \mathrm{C} $, $ v = 8.0 \times 10^6 \, \mathrm{m/s} $, $ B = 2.5 \, \mathrm{T} $, and $ \theta = 60^\circ $. Thus, $ F = (1.60 \times 10^{-19})(8.0 \times 10^6)(2.5) \sin 60^\circ = 2.8 \times 10^{-12} \, \mathrm{N} $.

Question 14

A proton moving with a velocity of $4.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }$ along the $+ y$-axis enters a magnetic field of $0.20 \mathrm {~T}$ directed towards the $- x$-axis. What is the magnitude of the magnetic force acting on the proton? $( e$ $= 1.60 \times 10 ^ { - 19 } \mathrm { C }$ )

Accepted Answer

The answer of A proton moving with a velocity of...

Question 15

A proton moving with a velocity of $4.0 \times 10 ^ { 4 } \mathrm {~m} / \mathrm { s }$ enters a magnetic field of $0.20 \mathrm {~T}$. If the angle between the velocity of the proton and the direction of the magnetic field is $60 ^ { \circ }$, what is the magnitude of the magnetic force on the proton? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

Accepted Answer

The magnitude of the magnetic force on a moving charge is given by the equation $F = qvB\sin\theta$, where $q$ is the charge of the particle, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field direction. Substituting the given values: $F = (1.60 \times 10^{-19} \, \text{C})(4.0 \times 10^4 \, \text{m/s})(0.20 \, \text{T})\sin(60^\circ)$, we find $F \approx 1.1 \times 10^{-15} \, \text{N}$.

Question 16

An electron moving perpendicular to a uniform magnetic field of $3.2 \times 10 ^ { - 2 } \mathrm {~T}$ moves in a circle of radius $0.40 \mathrm {~cm}$. How fast is this electron moving? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} \right)$

Accepted Answer

The answer of An electron moving perpendicular to a uniform...

Question 17

A wire along the $z$-axis carries a current of $4.9 \mathrm {~A}$ in the $+ z$ direction. Find the magnitude and direction of the force exerted on a $3.3 - \mathrm { cm }$ long length of this wire by a uniform magnetic field pointing in the $- x$ direction having a magnitude $0.43 \mathrm {~T}$.

Accepted Answer

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Question 18

Alpha particles, each having a charge of $+ 2 e$ and a mass of $6.64 \times 10 ^ { - 27 } \mathrm {~kg}$, are accelerated in a uniform $0.80$-T magnetic field to a final orbit radius of $0.30 \mathrm {~m}$. The field is perpendicular to the velocity of the particles. How long does it take an alpha particle to make one complete circle in the final orbit? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

Accepted Answer

The answer of Alpha particles, each having a charge of...

Question 19

An electron is accelerated from rest through a potential difference of $3.75 \mathrm { kV }$. It enters a region where a uniform 4.0-mT magnetic field is perpendicular to the velocity of the electron. Calculate the radius of the path this electron will follow in the magnetic field. $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {electron } } = \right.$ $9.11 \times 10 ^ { - 31 } \mathrm {~kg}$ )

Accepted Answer

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

Question 20

An electron moving perpendicular to a uniform magnetic field of $0.22 \mathrm {~T}$ moves in a circle with a speed of $1.5 \times 10 ^ { 7 } \mathrm {~m} / \mathrm { s }$. What is the radius of the circle? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {electron } } = 9.11 \times \right.$ 10-31 $\mathrm { kg }$ )

Accepted Answer

The answer of An electron moving perpendicular to a uniform...

Quiz 20: Magnetism

A straight wire carries a current of 10 A at an angle of 30° with respect to the direction of a uniform 0.30-T magnetic field. Find the magnitude of the magnetic force on a 0.50-m length of the wire.

An electron moves with a speed of $3.0 \times 104 \mathrm {~m} / \mathrm { s }$ perpendicular to a uniform magnetic field of $0.40$ T. What is the magnitude of the magnetic force on the electron? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

A charged particle of mass 0.0040 kg is subjected to a 4.0-T magnetic field which 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 2.0-m straight wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of the magnetic force on it?

A wire along the $z$ -axis carries a current of $4.9 \mathrm {~A}$ in the $+ z$ direction. Find the magnitude and direction of the force exerted on a $3.3 - \mathrm { cm }$ long length of this wire by a uniform magnetic field pointing in the $- x$ direction having a magnitude $0.43 \mathrm {~T}$ .