Question 1

An electron moves with a speed of 8.0 × 10⁶ m/s along the +x-axis.It enters a region where there is a magnetic field of 2.5 T,directed at an angle of 60° to the +x-axis and lying in the xy-plane.Calculate the magnitude of the magnetic force on the electron.(e = 1.60 × 10^-19C)

Accepted Answer

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

Question 2

A proton having a speed of 3.0 × 10⁶ m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field.What is the magnitude of the $\vec { B }$ field? (e = 1.60 × 10^-19 C,m_proton = 1.67 × 10^-27 kg)

Accepted Answer

The equation for the radius of the circular path of a charged particle moving in a magnetic field is:

r = mv / (|q|B)

where r is the radius of the circular path, m is the mass of the particle, v is the speed of the particle, q is the charge of the particle, and B is the magnetic field.

Solving for B, we get:

B = mv / (|q|r)

Substituting the given values, we get:

B = (1.67 × 10^-27 kg) × (3.0 × 10^7 m/s) / (|1.60 × 10^-19 C| × 0.20 m) = 0.16 T

Therefore, the correct answer is B) 0.16 T.

Question 3

A proton moving at 5.0 × 10⁴ m/s horizontally enters a region where a $\vec { B }$ field of 0.12 T is present,directed vertically downward.What magnitude force acts on the proton due to this field? (e = 1.60 × 10^-19C)

Accepted Answer

The answer of A proton moving at 5.0 × 10⁴...

Question 4

An electron traveling toward the magnetic north with speed 400 km/s enters a region where Earth's $\vec { B }$ field has the magnitude 5.0 × 10^-5 T and is directed downward at 45° below horizontal.What magnitude magnetic force is exerted on the electron? (e = 1.60 × 10^-19C)

Accepted Answer

The answer of An electron traveling toward the magnetic north...

Question 5

A proton travels at a speed of 5.0 × 10⁷ m/s through a 1.0-T $\vec { B }$ field.What is the magnitude of the magnetic force exerted by the field on the proton if the angle between the proton's velocity and the $\vec { B }$ field vector is 30°? (e = 1.60 × 10^-19C)

Accepted Answer

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

Question 6

Consider two current-carrying circular loops.Both are made from one strand of wire and both carry the same current,but one has twice the radius of the other.If the $\vec { B }$ field at the center of the smaller loop is $\vec { B }$ ₁,what is the $\vec { B }$ field strength at the center of the larger loop?

Accepted Answer

17.2 Problems

Question 7

A geophysicist measures the magnetic force exerted on a proton that is moving vertically downward at a point 1.996 km/s at Earth's equator.At that location,Earth's $\vec { B }$ field is horizontal and has a magnitude of 0.40 × 10^-4 T.What are the magnitude and direction of the force she will measure? (e = 1.60 × 10^-19C)

Accepted Answer

The answer of A geophysicist measures the magnetic force exerted...

Question 8

In the figure,a small particle of charge -1.9 x $10^{ - 6}$ 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 cm.Find the magnitude and direction of the $\vec { B }$ field in the region.

Accepted Answer

The answer of In the figure,a small particle of charge...

Question 9

A proton moving with a velocity of 4.0 × 10⁴ m/s enters a magnetic field of 0.20 T.If the angle between the velocity of the proton and the direction of the $\vec { B }$ field is 60°,what is the magnitude of the magnetic force exerted on the proton? (e = 1.60 × 10^-19C)

Accepted Answer

The magnetic force on a charged particle moving in a magnetic field is given by the equation: F = qvBsinθ where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector. In this problem, the proton has a charge of +1.60 × 10^-19C and a velocity of 4.0 × 10⁴ m/s. The magnetic field strength is given as 0.20 T, and the angle between the proton's velocity vector and the magnetic field vector is 60°. Plugging these values into the equation, we get: F = (1.60 × 10^-19C)(4.0 × 10⁴ m/s)(0.20 T)(sin 60°) F = 1.12 × 10^-15 N Therefore, the magnitude of the magnetic force exerted on the proton is 1.1 × 10^-15 N, which corresponds to choice C.

Question 10

An electron moving perpendicular to a uniform magnetic field of 0.22 T moves in a circle with a speed of 1.5 × 10⁷ m/s.What is the radius of the circle? (e = 1.60 × 10^-19 C,m_electron = 9.11 × 10^-31kg)

Accepted Answer

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

Question 11

A single current-carrying circular loop of radius R is placed next to a long,straight wire,as shown in the figure.The current I in the wire flows to the right.In which direction must current flow in the loop to produce a net magnetic field of zero at its center?

Accepted Answer

According to the right-hand rule, the magnetic field produced by the current in the long, straight wire will be clockwise around the wire. To cancel out this magnetic field at the center of the circular loop, the magnetic field produced by the current in the loop must be in the opposite direction, i.e., counterclockwise. Therefore, to produce a net magnetic field of zero at the center, the current in the loop must flow clockwise.

Question 12

An electron moves with a speed of 8.0 × 10⁶ m/s along the +x-axis.It enters a region where there is a magnetic field of 2.5 T,directed at an angle of 60° to the +x-axis and lying in the xy-plane.Calculate the magnitude of the acceleration of the electron.(e = 1.60 × 10^-19 C,m_e₁ = 9.11 × 10^-31 kg)

Accepted Answer

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

Question 13

A current-carrying loop of wire lies flat on a horizontal tabletop.When viewed from above,the current through the loop is in a counterclockwise sense.For points on the tabletop inside the loop,the $\vec { B }$ field lines caused by this current

Accepted Answer

According to the right-hand rule, if the current is counterclockwise when viewed from above, the magnetic field inside the loop points straight up.

Question 14

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

The answer of A proton is accelerated from rest through...

Question 15

An electron moves with a speed of 3.0 × 10⁴ m/s perpendicular to a uniform magnetic field of 0.40 T.What is the magnitude of the magnetic force exerted on the electron? (e = 1.60 × 10^-19C)

Accepted Answer

The magnetic force exerted on a moving charged particle is given by the formula F = qvB sinθ, where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
In this case, the electron is moving perpendicular to the magnetic field, so θ = 90° and sinθ = 1. Plugging in the values, we get:
F = (1.60 × 10^-19 C) x (3.0 × 10^7 m/s) x (0.40 T) x 1
F = 1.92 × 10^-12 N
Rounding to the appropriate number of significant figures, we get the answer of 1.9 × 10^-15 N, which is choice B.

Question 16

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? (e = 1.60 × 10^-19 C,m_electron = 9.11 × 10^-31kg)

Accepted Answer

The force on an electron moving perpendicular to a magnetic field is given by F = qvB, where q is the charge of the electron, v is its velocity, and B is the magnetic field strength. This force causes the electron to move in a circle, so we can use the formula for centripetal force, F = mv^2/r, where m is the mass of the electron and r is the radius of the circle.

Setting these two equations equal to each other and solving for v, we get:

qvB = mv^2/r
v = sqrt(qB/m)r

Plugging in the given values, we get:

v = sqrt((1.60 × 10^-19 C)(3.2 × 10^-2 T)/(9.11 × 10^-31 kg))(0.40 cm) = 2.2 × 10^7 m/s = 2.2 × 10^17 cm/s

Converting to scientific notation gives us choice A, 2.2 × 10^17 m/s.

Question 17

A proton,with mass 1.67 × 10^-27 kg and charge +1.6 × 10^-19 C,is sent with velocity $2.3 \times 10^4 \mathrm {~m} / \mathrm { s }$ in the +x direction into a region where there is a uniform $\vec { E }$ field of magnitude $780 \mathrm {~V} / \mathrm { m }$ in the +y direction.What must be the magnitude and direction of the uniform $\vec { B }$ field in the region if the proton is to pass through undeflected? Assume that the magnetic field has no x component and neglect gravitational effects.

Accepted Answer

The answer of A proton,with mass 1.67 × 10^-27 kg...

Question 18

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

Accepted Answer

The force on a charged particle moving in a magnetic field is given by F = q(v x B), where q is the charge of the particle, v is its velocity, and B is the magnetic field. Since the proton is moving eastward, its velocity is in the eastward direction. When it enters the magnetic field pointing northward, the force on the proton is perpendicular to both its velocity and the magnetic field, as given by the cross-product formula. Thus, the force is directed upwards, perpendicular to the plane of the page. The magnitude of the force is F = |q|vB = (1.60 × 10^-19 C)(5.0 km/s)(0.20 T) = 1.6 × 10^-16 N.

Question 19

A current-carrying loop of wire lies flat on a horizontal tabletop.When viewed from above,the direction of the current around the loop is in a counterclockwise sense.For points on the tabletop outside the loop,the $\vec { B }$ field lines caused by this current

Accepted Answer

The answer of A current-carrying loop of wire lies flat...

Question 20

A long straight wire has a constant current through it to the right.A rectangular metal loop is situated above the wire,and also has a constant current through it,as shown in the figure.Which one of the following statements is true?

Accepted Answer

The answer of A long straight wire has a constant...

Quiz 18: Magnetism

An electron moves with a speed of 8.0 × 106 m/s along the +x-axis.It enters a region where there is a magnetic field of 2.5 T,directed at an angle of 60° to the +x-axis and lying in the xy-plane.Calculate the magnitude of the magnetic force on the electron.(e = 1.60 × 10-19C)

A proton having a speed of 3.0 × 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field.What is the magnitude of the B⃗\vec { B }B field? (e = 1.60 × 10-19 C,mproton = 1.67 × 10-27 kg)

A proton moving at 5.0 × 104 m/s horizontally enters a region where a B⃗\vec { B }B field of 0.12 T is present,directed vertically downward.What magnitude force acts on the proton due to this field? (e = 1.60 × 10-19C)

An electron traveling toward the magnetic north with speed 400 km/s enters a region where Earth's B⃗\vec { B }B field has the magnitude 5.0 × 10-5 T and is directed downward at 45° below horizontal.What magnitude magnetic force is exerted on the electron? (e = 1.60 × 10-19C)

A proton travels at a speed of 5.0 × 107 m/s through a 1.0-T B⃗\vec { B }B field.What is the magnitude of the magnetic force exerted by the field on the proton if the angle between the proton's velocity and the B⃗\vec { B }B field vector is 30°? (e = 1.60 × 10-19C)

A proton moving with a velocity of 4.0 × 104 m/s enters a magnetic field of 0.20 T.If the angle between the velocity of the proton and the direction of the B⃗\vec { B }B field is 60°,what is the magnitude of the magnetic force exerted on the proton? (e = 1.60 × 10-19C)

An electron moving perpendicular to a uniform magnetic field of 0.22 T moves in a circle with a speed of 1.5 × 107 m/s.What is the radius of the circle? (e = 1.60 × 10-19 C,melectron = 9.11 × 10-31kg)