IB Physics Quiz: Understand Motion In Em Fields

20 questions · exam conditions

0:00

Understand Motion In Em FieldsQuestion 1 of 20

A proton moves in a circular path of radius $R$ in a uniform magnetic field $B$ with a speed $v$ . A second identical proton moves in a circular path of radius $R'$ in a uniform magnetic field of strength $B/3$ with a speed of $2v$ . What is the ratio $R'/R$ ?

2/3

3/2

1/6

← Back to quizzes

IB Physics Quiz

IB Physics Quiz: Understand Motion In Em Fields

Practice Understand Motion In Em Fields in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Understand Motion In Em Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

2/3
3/2
6 (correct answer)
1/6

Explanation: The magnetic force provides the centripetal force for the circular motion:

qvB = mv^2/r

. Solving for the radius

r

gives

r = mv/(qB)

. For the first proton,

R = mv/(qB)

. For the second proton, the new radius is

R' = m(2v)/(q(B/3)) = 6(mv/(qB))

. Since

R = mv/(qB)

, we have

R' = 6R

. Therefore, the ratio

R'/R

is 6.

Question 2

Two long, straight, parallel wires are separated by a distance $d$ and carry currents $I_1$ and $I_2$ . The magnitude of the force per unit length on each wire is $F$ . The current in the first wire is changed to $3I_1$ and the separation distance is changed to $2d$ . What is the new magnitude of the force per unit length on each wire?

$3F/2$ (correct answer)
$3F/4$
$6F$
$3F$

Explanation: The force per unit length between two parallel wires is given by

F/L = (\mu_0 I_1 I_2) / (2\pi d)

. Thus,

F

is directly proportional to the product of the currents and inversely proportional to the distance

d

. The new force per unit length,

F'

, will be proportional to

(3I_1)I_2 / (2d)

. This is

3/2

times the original proportionality

I_1 I_2 / d

. Therefore, the new force is

F' = (3/2)F

Question 3

A proton and an alpha particle enter a uniform magnetic field with the same velocity, which is perpendicular to the field lines. An alpha particle has a charge of $+2e$ and a mass of approximately $4m_p$ , where $e$ and $m_p$ are the charge and mass of a proton. What is the ratio of the magnitude of the acceleration of the alpha particle to that of the proton?

1/4
1/2 (correct answer)
1
2

Explanation: The magnetic force on a particle is

F = qvB

. According to Newton's second law, acceleration is

a = F/m = qvB/m

. The ratio of the accelerations is

a_{\alpha}/a_p = (q_{\alpha}vB/m_{\alpha}) / (q_p vB/m_p)

. Since

v

and

B

are the same, the ratio simplifies to

a_{\alpha}/a_p = (q_{\alpha}/m_{\alpha}) / (q_p/m_p) = (q_{\alpha}/q_p) \times (m_p/m_{\alpha})

. Substituting the given values:

(2e/e) \times (m_p/4m_p) = 2 \times (1/4) = 1/2

Question 4

A particle of charge $q$ and mass $m$ has an initial kinetic energy $E_k$ . It enters a region of uniform electric field $E$ with its initial velocity perpendicular to the field. What is the kinetic energy of the particle after it has been displaced by a distance $y$ in the direction of the electric field?

$E_k$
$qEy$
$E_k + qEy$ (correct answer)
$\sqrt{E_k^2 + (qEy)^2}$

Explanation: The electric field exerts a constant force

F = qE

on the particle in the direction of the field. As the particle is displaced by a distance

y

in this direction, the field does work

W = F y = qEy

on the particle. According to the work-energy theorem, the work done on an object equals the change in its kinetic energy (

W = \Delta E_k

). Therefore,

E_{k,final} - E_{k,initial} = qEy

. Rearranging gives

E_{k,final} = E_{k,initial} + qEy = E_k + qEy

Question 5

A horizontal copper wire is in a uniform magnetic field directed vertically downwards. The wire carries a current due to electrons flowing from east to west. What is the direction of the magnetic force on the wire?

North (correct answer)
South
East
West

Explanation: By convention, the direction of current is opposite to the direction of electron flow. If electrons flow from east to west, the conventional current

I

is directed from west to east. Using the right-hand rule (or Fleming's left-hand rule): point your fingers in the direction of the current (East), and curl them into the direction of the magnetic field (Down). Your thumb points in the direction of the force, which is North. Alternatively, for Fleming's rule: Field is down (first finger), Current is east (second finger), so the Force (thumb) is North.

Question 6

Three long, straight, parallel wires, X, Y, and Z, are equally spaced and lie in the same horizontal plane. Wire Y is between X and Z. The current in each wire is identical and flows in the same direction (into the page). What is the direction of the net magnetic force on wire Y?

Towards wire X
Towards wire Z
Vertically upwards, out of the plane
Zero (correct answer)

Explanation: Parallel currents in the same direction attract each other. Wire Y will be attracted to wire X, so there is a force on Y directed towards X. Wire Y will also be attracted to wire Z, so there is a force on Y directed towards Z. Since the currents are identical (

I_X = I_Y = I_Z

) and the wires are equally spaced (

d_{XY} = d_{YZ}

), the magnitudes of the two forces are equal according to the formula

F/L = (\mu_0 I_a I_b) / (2\pi d)

. As the two forces on wire Y are equal in magnitude and opposite in direction, the net force on wire Y is zero.

Question 7

A region of space contains a uniform magnetic field pointing north and a uniform electric field pointing east. An electron is placed at rest at a point in this region. What is the initial direction of the net force on the electron?

East
West (correct answer)
North
There is no net force.

Explanation: The magnetic force on a charge is given by

F_B = qvB\sin{\theta}

. Since the electron is initially at rest, its velocity

v=0

, and therefore the magnetic force on it is zero. The electric force is given by

F_E = qE

. The electric field

E

points east. Since the electron has a negative charge (

q = -e

), the electric force on it is in the direction opposite to the electric field. Thus, the electric force is directed west. The net force is the vector sum of the electric and magnetic forces, which in this case is just the electric force directed west.

Question 8

An alpha particle (charge $+2e$ , mass $4u$ ) and a proton (charge $+e$ , mass $1u$ ) are accelerated from rest through the same potential difference $V$ . They then enter a uniform magnetic field $B$ perpendicular to their velocities. What is the ratio of the radius of the alpha particle's path, $r_\alpha$ , to the radius of the proton's path, $r_p$ ?

1/2
$1/\sqrt{2}$
$\sqrt{2}$ (correct answer)
2

Explanation: First, find the speed after acceleration. The kinetic energy gained is

E_k = qV

, so

\frac{1}{2}mv^2 = qV

, which gives

v = \sqrt{2qV/m}

. The radius of the circular path is

r = mv/qB

. Substituting

v

, we get

r = m(\sqrt{2qV/m})/qB = (1/B) \sqrt{2mV/q}

. The ratio of the radii is

r_\alpha / r_p = \sqrt{(m_\alpha/q_\alpha)} / \sqrt{(m_p/q_p)} = \sqrt{(m_\alpha/m_p) \times (q_p/q_\alpha)}

. Substituting the values

m_\alpha = 4m_p

and

q_\alpha = 2q_p

, the ratio becomes

\sqrt{(4/1) \times (1/2)} = \sqrt{2}

Question 9

A proton is moving with an initial velocity $v$ in a region with a uniform electric field $E$ and a uniform magnetic field $B$ . All three vectors, $v$ , $E$ , and $B$ , are parallel and in the same direction. What is the motion of the proton?

It continues at a constant velocity.
It moves in a circular path at constant speed.
It moves in a helical path with constant pitch.
It accelerates in the direction of its initial velocity. (correct answer)

Explanation: The electric force is

F_E = qE

. Since the proton's charge

q

is positive and its velocity is parallel to

E

, this force acts in the direction of motion, causing the proton to accelerate. The magnetic force is

F_B = qvB\sin{\theta}

. Since the velocity

v

is parallel to the magnetic field

B

, the angle

\theta

is 0°, and

\sin(0°) = 0

. Therefore, the magnetic force is zero. The net force is just the electric force, which causes the proton to undergo linear acceleration.

Question 10

A proton moves in a circular path in a uniform magnetic field directed into the page. A uniform electric field is then applied, parallel to the magnetic field. How does the application of the electric field affect the proton's path?

The path remains a circle, but its radius increases.
The path becomes a helix with constant pitch.
The path becomes a straight line due to the electric force.
The path becomes a helix with an increasing pitch. (correct answer)

Explanation: The magnetic force provides the centripetal force for the circular motion in the plane perpendicular to the field. The electric field, being parallel to the magnetic field, exerts a force

F_E = qE

on the proton along the axis of the circular motion. This force causes the proton to accelerate in the axial direction. The combination of circular motion in the plane and accelerated linear motion along the axis results in a helical path. Since the axial velocity is increasing, the distance between successive turns of the helix (the pitch) also increases.

Question 11

A wire is bent into a semicircle of radius $R$ and carries a current $I$ . It is placed in a uniform magnetic field $B$ that is directed perpendicular to the plane of the semicircle. What is the magnitude of the magnetic force on the curved section of the wire?

0
$BI(2R)$ (correct answer)
$BI(\pi R)$
$2\pi RBI$

Explanation: The magnetic force on a current-carrying wire of any shape in a uniform magnetic field is equal to the force on a straight wire connecting the start and end points of the original wire, carrying the same current. For a semicircle of radius

R

, the start and end points are separated by the diameter, a distance

L_{eff} = 2R

. The force on this effective straight wire is given by

F = BIL_{eff}\sin{\theta}

. Since the field is perpendicular to the plane, it is perpendicular to this effective wire, so

\theta=90°

and

\sin{\theta}=1

. The force is

F = BI(2R)

Question 12

A positron with velocity $v$ enters a region of uniform magnetic field $B$ . The angle between the velocity vector $v$ and the magnetic field vector $B$ is 45°. Which statement best describes the path of the positron?

A circular path in a plane perpendicular to the magnetic field.
A straight line path along the direction of the magnetic field.
A helical path with its axis parallel to the magnetic field. (correct answer)
A parabolic path in a plane containing the magnetic field.

Explanation: The velocity vector can be resolved into two components: one parallel to the magnetic field (

v_\parallel = v \cos 45°

) and one perpendicular to it (

v_\perp = v \sin 45°

). The parallel component

v_\parallel

experiences no magnetic force, so the positron moves at a constant velocity in that direction. The perpendicular component

v_\perp

experiences a magnetic force that causes the positron to move in a circle. The combination of uniform linear motion along the field lines and circular motion perpendicular to them results in a helical (corkscrew) path.

Question 13

A singly-ionized carbon-12 atom ( $^{12}\text{C}^+$ ) passes undeflected through a velocity selector at a speed $v$ . The selector uses perpendicular electric and magnetic fields. A doubly-ionized carbon-12 atom ( $^{12}\text{C}^{2+}$ ) is then sent into the same velocity selector. At what speed must it travel to pass through undeflected?

$v/2$
$v$ (correct answer)
$v\sqrt{2}$
$2v$

Explanation: The condition for a charged particle to pass undeflected through a velocity selector is that the magnetic force must be equal and opposite to the electric force:

F_B = F_E

. This means

qvB = qE

. The charge

q

cancels from both sides of the equation, yielding

v = E/B

. This shows that the selection speed

v

depends only on the strengths of the electric and magnetic fields, not on the charge or mass of the particle. Therefore, the doubly-ionized atom must travel at the same speed

v

to pass through undeflected.

Question 14

A rectangular loop of wire carrying a steady current is placed in a uniform magnetic field. The plane of the loop is parallel to the direction of the magnetic field lines. What are the net force and net torque on the loop in this orientation?

The net force is zero and the net torque is zero.
The net force is non-zero and the net torque is zero.
The net force is zero and the net torque is non-zero. (correct answer)
The net force is non-zero and the net torque is non-zero.

Explanation: In a uniform magnetic field, the forces on opposite sides of a current loop are equal in magnitude and opposite in direction. This always results in a zero net force. The two sides of the loop parallel to the field experience zero force (

F = BIL\sin(0°)=0

). The two sides perpendicular to the field experience forces

F=BIL

in opposite directions. Because these forces do not act along the same line, they produce a couple, resulting in a non-zero net torque that will cause the loop to rotate. The torque is maximum in this orientation.

Question 15

Which statement correctly compares the trajectory of a charged particle entering a uniform electric field at a right angle to the field with that of a projectile launched horizontally in a uniform gravitational field, ignoring air resistance?

Both particles follow circular paths of constant radius.
The charged particle follows a parabolic path while the projectile follows a circular path.
The charged particle follows a circular path while the projectile follows a parabolic path.
Both particles follow parabolic paths. (correct answer)

Explanation: A charged particle in a uniform electric field experiences a constant force

F=qE

, and thus a constant acceleration

a=qE/m

in the direction of the field. A projectile in a uniform gravitational field experiences a constant force

F=mg

and thus a constant acceleration

g

downwards. In both cases, a particle with an initial velocity perpendicular to this constant acceleration will follow a parabolic trajectory. The motion is mathematically analogous.

Question 16

A straight conductor of length $L$ carrying current $I$ is placed in a uniform magnetic field $B$ at an angle of 30° to the field lines. The force on the conductor is $F$ . What is the force on a conductor of length $2L$ carrying current $2I$ placed at an angle of 90° to the same magnetic field?

$4F$
$8F$ (correct answer)
$F$
$2F$

Explanation: The magnetic force on a conductor is given by

F_{mag} = BIL\sin{\theta}

. Initially,

F = BIL\sin(30°)

which is

F = BIL(1/2)

. From this, we can state that

BIL = 2F

. For the new situation, the force

F'

F' = B(2I)(2L)\sin(90°)

. This simplifies to

F' = 4BIL(1) = 4BIL

. Substituting

BIL = 2F

into this new expression gives

F' = 4(2F) = 8F

Question 17

An electron moves at a constant speed and enters a region of uniform magnetic field. The electron's initial velocity is perpendicular to the direction of the magnetic field. What is the effect of the magnetic field on the kinetic energy and the momentum of the electron as it moves through the field?

Kinetic energy increases and the momentum vector changes.
Kinetic energy remains constant and the momentum vector remains constant.
Kinetic energy remains constant and the momentum vector changes. (correct answer)
Kinetic energy decreases and the momentum vector changes.

Explanation: The magnetic force on a charged particle is always perpendicular to its velocity (

\vec{F} = q\vec{v} \times \vec{B}

). Since the force is perpendicular to the displacement, the magnetic field does no work on the electron. By the work-energy theorem, if no work is done, the kinetic energy (

E_k = \frac{1}{2}mv^2

) remains constant. This also implies the speed of the electron is constant. However, the magnetic force continuously changes the direction of the electron's velocity. Since momentum (

\vec{p} = m\vec{v}

) is a vector quantity, a change in direction means the momentum vector changes, even if its magnitude (

mv

) is constant.

Question 18

An electron beam travels horizontally with speed $3.0 \times 10^{6}$ m/s through crossed electric and magnetic fields. The electric field has magnitude $1.5 \times 10^{3}$ N/C pointing vertically downward, and the magnetic field has magnitude 0.50 mT pointing into the page. If the electric field is suddenly turned off while maintaining the magnetic field, what happens to the electron's motion?

The electron follows a circular path with radius 34 mm moving clockwise when viewed from above
The electron follows a circular path with radius 34 mm moving counterclockwise when viewed from above (correct answer)
The electron follows a circular path with radius 17 mm moving counterclockwise when viewed from above
The electron continues in a straight line since the magnetic force was balanced by the electric force

Explanation: Initially, the electron travels straight because electric and magnetic forces balance. When the electric field is removed, only the magnetic force remains. The radius is r = mv/(|q|B) = (9.1×10⁻³¹)(3.0×10⁶)/[(1.6×10⁻¹⁹)(0.50×10⁻³)] = 0.034 m = 34 mm. Using the right-hand rule for the negative charge: velocity right, B into page, force upward initially, creating counterclockwise motion. Choice A has wrong direction. Choice C uses wrong mass value. Choice D incorrectly assumes motion stops when fields are unbalanced.

Question 19

An alpha particle (charge +2e, mass 4u where u = $1.66 \times 10^{-27}$ kg) is accelerated from rest through a potential difference of 1000 V and then enters a region with uniform magnetic field B = 0.20 T perpendicular to its velocity. What is the period of the alpha particle's circular motion in the magnetic field?

$6.5 \times 10^{-7}$ s
$1.3 \times 10^{-7}$ s
$3.3 \times 10^{-7}$ s
$2.6 \times 10^{-7}$ s (correct answer)

Explanation: The period of circular motion in a magnetic field is T = 2πm/(qB), which is independent of velocity. For an alpha particle: m = 4u = 4(1.66×10⁻²⁷) = 6.64×10⁻²⁷ kg, q = 2e = 2(1.6×10⁻¹⁹) = 3.2×10⁻¹⁹ C. Therefore: T = 2π(6.64×10⁻²⁷)/[(3.2×10⁻¹⁹)(0.20)] = 2.6×10⁻⁷ s. Choice A uses wrong charge value. Choice B uses wrong mass value. Choice C uses single charge instead of double charge.

Question 20

A mass spectrometer uses crossed electric and magnetic fields as a velocity selector, followed by a magnetic field region for mass analysis. In the velocity selector, E = $2.0 \times 10^{4}$ N/C and B₁ = 0.050 T. Selected ions then enter a magnetic field B₂ = 0.80 T and follow a semicircular path with radius 0.15 m. What is the mass-to-charge ratio of these ions?

$3.0 \times 10^{-7}$ kg/C (correct answer)
$1.5 \times 10^{-7}$ kg/C
$6.0 \times 10^{-7}$ kg/C
$4.8 \times 10^{-7}$ kg/C

Explanation: In the velocity selector: v = E/B₁ = (2.0×10⁴)/(0.050) = 4.0×10⁵ m/s. In the magnetic analyzer: r = mv/(qB₂), so m/q = rB₂/v = (0.15)(0.80)/(4.0×10⁵) = 3.0×10⁻⁷ kg/C. Choice B uses the wrong magnetic field value. Choice C doubles the correct answer. Choice D incorrectly combines both magnetic field values in the calculation.