IB Physics Quiz: Apply Motion In Em Fields
20 questions · exam conditions
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Apply Motion In Em FieldsQuestion 1 of 20

A positron enters a uniform magnetic field with a velocity vector at an angle of 30° to the direction of the magnetic field lines. Which statement best describes the positron's subsequent motion?

It follows a parabolic path.
It follows a circular path.
It follows a helical path.
It moves in a straight line at constant velocity.
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IB Physics Quiz

IB Physics Quiz: Apply Motion In Em Fields

Practice Apply 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.

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

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Question 1

A positron enters a uniform magnetic field with a velocity vector at an angle of 30° to the direction of the magnetic field lines. Which statement best describes the positron's subsequent motion?

  1. It follows a parabolic path.
  2. It follows a circular path.
  3. It follows a helical path. (correct answer)
  4. It moves in a straight line at constant velocity.
Explanation: The velocity can be resolved into two components. The component parallel to the magnetic field experiences no magnetic force and remains constant, causing the particle to drift along the field lines. The component perpendicular to the magnetic field causes a magnetic force that acts as a centripetal force, resulting in circular motion. The combination of linear motion along the field lines and circular motion perpendicular to them is a helical path.

Question 2

A charged particle with kinetic energy K enters a region of uniform magnetic field B, with its velocity perpendicular to the field. The particle completes a semi-circular path and exits the region. What is the work done by the magnetic field on the particle and the kinetic energy of the particle as it exits?

  1. Work done is zero, exit kinetic energy is K. (correct answer)
  2. Work done is negative, exit kinetic energy is less than K.
  3. Work done is positive, exit kinetic energy is greater than K.
  4. Work done is zero, exit kinetic energy is zero.
Explanation: The magnetic force on a moving charge is always perpendicular to its velocity (F = qvB sin θ). Since work done is W = Fd cos θ, and the angle between force and displacement is always 90° for circular motion, the work done by the magnetic field is zero. According to the work-energy theorem (W_net = ΔK), if the net work done is zero, the change in kinetic energy is zero. Therefore, the particle exits with the same kinetic energy K.

Question 3

Two long, parallel wires, X and Y, are separated by a distance r and carry currents I_X and I_Y. The force per unit length on wire Y is F. The current in wire X is then doubled, the current in wire Y is halved, and the separation is increased to 3r. What is the new force per unit length on wire Y?

  1. F/6
  2. F/3 (correct answer)
  3. 2F/3
  4. F
Explanation: The force per unit length between two parallel wires is given by F/L = (μ₀ * I_X * I_Y) / (2πr). Thus, F is proportional to (I_X * I_Y) / r. The new force, F_new, will be proportional to ((2*I_X) * (I_Y/2)) / (3r) = (I_X * I_Y) / (3r). This is 1/3 of the original proportionality. Therefore, the new force is F/3.

Question 4

An electron is projected horizontally with velocity v into a uniform electric field directed vertically downwards. Its path is parabolic. Which change would cause the electron to follow a parabolic path with a greater vertical displacement for the same horizontal distance traveled?

  1. Increasing the initial horizontal velocity v.
  2. Replacing the electron with a proton of the same initial velocity.
  3. Decreasing the strength of the electric field.
  4. Decreasing the initial horizontal velocity v. (correct answer)
Explanation: The vertical displacement (y) is given by y = (1/2)at^2, where a = qE/m. The horizontal distance (x) is x = vt, so the time spent in the field is t = x/v. Substituting t, we get y = (1/2)(qE/m)(x/v)^2. To increase y for a fixed x, the term (qE)/(mv2mv^2) must increase. Decreasing the initial velocity v will increase this term, leading to a greater vertical displacement. Increasing v would decrease y. A proton has a much larger mass m, so q/m is smaller, decreasing y. Decreasing E would decrease y.

Question 5

A horizontal wire Y is suspended directly below a fixed parallel horizontal wire X. Wire X carries a current of 20 A. Wire Y has a mass per unit length of 1.0 × 10⁻⁴ kg m⁻¹ and carries a current of 5.0 A in the opposite direction to wire X. What separation distance is required for the magnetic force to balance the gravitational force on wire Y? (Use g ≈ 10 m s⁻²)

  1. 1.0 cm (correct answer)
  2. 2.0 cm
  3. 1.0 mm
  4. 2.0 mm
Explanation: For equilibrium, the magnetic force per unit length must equal the gravitational force per unit length: F_B/L = F_g/L. This gives (μ₀I_XI_Y)/(2πr) = (m/L)g. Rearranging for r: r = (μ₀I_XI_Y)/(2π(m/L)g). Substituting values: r = (4π×10⁻⁷ × 20 × 5.0)/(2π × 1.0×10⁻⁴ × 10) = (2×10⁻⁵)/(2.0×10⁻³) = 1.0×10⁻² m = 1.0 cm.

Question 6

A wire is bent into a semi-circular arc 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 semi-circle. What is the magnitude of the net magnetic force on the wire?

  1. πBIR
  2. 2BIR (correct answer)
  3. BIR
  4. 0
Explanation: The magnetic force on a non-straight wire in a uniform magnetic field depends on the vector displacement from the starting point to the ending point of the wire. For a semi-circular arc of radius R, this displacement vector is a straight line of length 2R (the diameter). The force is given by F = BIL_eff, where L_eff = 2R. Therefore, the magnitude of the net force is F = B * I * (2R).

Question 7

An electron is initially at rest in a region of space containing a uniform electric field E and a uniform magnetic field B. The fields E and B are not parallel to each other. What is the nature of the initial force on the electron?

  1. The net force is zero.
  2. There is only a magnetic force acting on it.
  3. There is both an electric and a magnetic force acting on it.
  4. There is only an electric force acting on it. (correct answer)
Explanation: The electric force on a charge is given by F_E = qE. This force acts on the electron regardless of its velocity. The magnetic force is given by F_B = qvBsin(θ). Since the electron is initially at rest, its velocity v is zero. Therefore, the initial magnetic force is zero. The only initial force acting on the electron is the electric force.

Question 8

A charged particle moves in a vacuum through a region of uniform electric field. The particle's initial velocity is perpendicular to the direction of the electric field. Which statement correctly describes the components of the particle's velocity?

  1. The velocity component parallel to the field is constant; the component perpendicular to the field changes at a constant rate.
  2. The velocity component parallel to the field changes at a constant rate; the component perpendicular to the field is constant. (correct answer)
  3. Both components of the velocity change at a constant rate.
  4. Both components of the velocity remain constant.
Explanation: The electric field exerts a force F=qE only in the direction parallel to the field. This causes a constant acceleration (a=F/m) in that direction, so the velocity component parallel to the field changes at a constant rate (v_parallel = at). There is no force perpendicular to the field, so the acceleration in that direction is zero. Therefore, the velocity component perpendicular to the field remains constant, equal to its initial value.

Question 9

Two particles, X and Y, have the same magnitude of charge and move in the same uniform magnetic field in circular paths perpendicular to the field. The kinetic energy of X is four times that of Y (K_X = 4K_Y), and the mass of X is twice that of Y (m_X = 2m_Y). What is the ratio of the radius of path X to the radius of path Y, r_X / r_Y?

  1. 2
  2. 4
  3. 2\sqrt{2}
  4. 222\sqrt{2} (correct answer)
Explanation: The radius of a charged particle's path in a B-field is r = mv/(qB). Kinetic energy is K = (1/2)mv², so momentum p = mv = √(2mK). Substituting momentum into the radius equation gives r = p/(qB) = √(2mK)/(qB). Since q and B are the same for both particles, the radius r is proportional to √(mK). The ratio is r_X / r_Y = √(m_X * K_X) / √(m_Y * K_Y). Substituting the given relations: r_X / r_Y = √( (2m_Y) * (4K_Y) ) / √(m_Y * K_Y) = √(8 * m_Y * K_Y) / √(m_Y * K_Y) = √8 = 2√2.

Question 10

A proton enters a region with a uniform magnetic field B and a uniform electric field E. The fields are perpendicular to each other and to the proton's initial velocity v. The proton passes through undeviated. An alpha particle (charge +2e, mass ≈ 4mp) then enters the same region with an initial velocity of 2v along the same path. What is the subsequent motion of the alpha particle?

  1. It continues to move undeviated.
  2. It is deflected in the direction of the electric force.
  3. It is deflected in the direction of the magnetic force. (correct answer)
  4. It follows a complete circular path within the fields.
Explanation: For the proton to be undeviated, the electric force Fe = qE must balance the magnetic force Fb = qvB. This means v = E/B. The alpha particle has charge qa = 2q and velocity va = 2v. The electric force on it is F'e = (2q)E. The magnetic force on it is F'b = (2q)(2v)B = 4qvB. Since E = vB, we can write F'e = 2q(vB). Therefore, F'b = 2 * F'e. The magnetic force is stronger than the electric force, so the net force is in the direction of the magnetic force, causing deflection in that direction.

Question 11

A straight wire of length 20 cm carries a current of 5.0 A. The wire is oriented at a 30° angle to the direction of a uniform magnetic field of strength 0.40 T. The length of the wire that is actually inside the magnetic field is 10 cm. What is the magnitude of the magnetic force on the wire?

  1. 0.10 N (correct answer)
  2. 0.17 N
  3. 0.20 N
  4. 0.40 N
Explanation: The magnetic force on a current-carrying wire is given by F = BILsin(θ), where L is the length of the wire inside the field. Here, B = 0.40 T, I = 5.0 A, L = 10 cm = 0.10 m, and θ = 30°. So, F = (0.40 T)(5.0 A)(0.10 m)sin(30°) = (0.20 N)(0.5) = 0.10 N. The total length of the wire (20 cm) is irrelevant.

Question 12

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

  1. Net force is non-zero, net torque is zero.
  2. Net force is zero, net torque is zero.
  3. Net force is zero, net torque is non-zero. (correct answer)
  4. Net force is non-zero, net torque is non-zero.
Explanation: The forces on the sides of the loop that are parallel to the magnetic field are zero (since sin(0) or sin(180) is zero). The forces on the two sides perpendicular to the field are equal in magnitude (F=BIL) and opposite in direction. Thus, the net force on the loop is zero. However, these two forces act at different locations, forming a couple that creates a non-zero net torque, which will cause the loop to rotate.

Question 13

A flat, rectangular metal strip carries a current I to the right. The strip is placed in a uniform magnetic field B directed into the page. The charge carriers in the metal are electrons. What is the result of the magnetic force on the charge carriers?

  1. The top surface of the strip becomes negative and the bottom becomes positive. (correct answer)
  2. The top surface of the strip becomes positive and the bottom becomes negative.
  3. The left end of the strip becomes positive and the right end becomes negative.
  4. The charge carriers are unaffected as the net charge of the strip is zero.
Explanation: Conventional current (I) to the right means the charge carriers, electrons (negative charge), are moving to the left. The velocity (v) of electrons is to the left. The magnetic field (B) is into the page. Using the right-hand rule for a positive charge and reversing the direction for a negative charge (or using a left-hand rule), the force F = q(v x B) on the electrons is directed towards the top of the strip. This accumulation of electrons makes the top surface negative and leaves the bottom surface with a net positive charge.

Question 14

Three long, straight, parallel wires are arranged at the corners of an equilateral triangle. Wires 1 and 2 carry identical currents into the page. Wire 3 carries a current of the same magnitude out of the page. What is the direction of the net magnetic force on wire 3?

  1. Towards wire 1.
  2. Towards the midpoint of the line segment connecting wires 1 and 2.
  3. Away from the midpoint of the line segment connecting wires 1 and 2. (correct answer)
  4. The net force is zero.
Explanation: The force between wires with anti-parallel currents is repulsive. The force on wire 3 from wire 1 (F31F_31) is repulsive, directed along the line connecting them, away from wire 1. The force on wire 3 from wire 2 (F32F_32) is also repulsive and directed away from wire 2. Since the currents and distances are equal, the magnitudes of F_31 and F_32 are equal. The vector sum of these two forces points away from the side of the triangle formed by wires 1 and 2, specifically away from the midpoint of the line segment connecting them.

Question 15

A wire carrying current I consists of two straight segments of equal length L joined at a right angle. The wire lies in the xy-plane, with one segment along the positive x-axis (from origin to (L,0)) and the other along the positive y-axis (from origin to (0,L)). A uniform magnetic field B is directed along the positive x-axis. What is the magnitude of the net magnetic force on this L-shaped wire?

  1. 0
  2. BIL (correct answer)
  3. √2 BIL
  4. 2BIL
Explanation: The net force is the vector sum of the forces on the two segments. For the segment on the x-axis, the current is parallel to the magnetic field B, so the angle θ is 0, and the force F_x = BILsin(0) = 0. For the segment on the y-axis, the current is perpendicular to the magnetic field B, so θ = 90°, and the force F_y = BILsin(90) = BIL. The net force is the sum, which is 0 + BIL = BIL.

Question 16

Two parallel wires, P and Q, are separated by a distance d and carry equal currents I in the same direction. A third wire R, parallel to P and Q, is placed exactly midway between them. Wire R carries a current of 2I in the direction opposite to P and Q. What is the magnitude of the force per unit length on wire R?

  1. 0 (correct answer)
  2. 2μ0I2πd\frac{2\mu_0 I^2}{\pi d}
  3. 4μ0I2πd\frac{4\mu_0 I^2}{\pi d}
  4. μ0I2πd\frac{\mu_0 I^2}{\pi d}
Explanation: Wire R is midway, so its distance from both P and Q is d/2. The force between R and P is repulsive (opposite currents). Let's say this force pushes R towards Q. Its magnitude per unit length is F_RP/L = μ₀(2I)(I) / (2π(d/2)) = 2μ₀I² / (πd). The force between R and Q is also repulsive. This force pushes R towards P. Its magnitude per unit length is F_RQ/L = μ₀(2I)(I) / (2π(d/2)) = 2μ₀I² / (πd). The two forces are equal in magnitude and opposite in direction. Therefore, the net force on wire R is zero.

Question 17

A particle of charge q and mass m is accelerated from rest through a potential difference V. It then enters a uniform magnetic field B, perpendicular to its velocity, and moves in a circle of radius r. How does the radius r depend on the accelerating potential V?

  1. r is proportional to V.
  2. r is proportional to V².
  3. r is proportional to √V. (correct answer)
  4. r is proportional to 1/√V.
Explanation: First, find the velocity of the particle after acceleration. The work done by the electric field equals the kinetic energy gained: qV = (1/2)mv². So, v = √(2qV/m). Next, in the magnetic field, the magnetic force provides the centripetal force: qvB = mv²/r, which gives r = mv/(qB). Substituting the expression for v: r = (m/(qB)) * √(2qV/m) = (1/B) * √(2mV/q). Therefore, the radius r is proportional to the square root of the potential difference V.

Question 18

An electron moves with constant speed in a plane perpendicular to a magnetic field. The electron travels from a region of weak magnetic field into a region of strong magnetic field. How do the radius of its path and the period of its motion change as it enters the stronger field?

  1. Radius decreases and period decreases. (correct answer)
  2. Radius increases and period increases.
  3. Radius decreases and period increases.
  4. Radius increases and period decreases.
Explanation: The radius of the circular path is given by r = mv/(qB). Since m, v, and q are constant, the radius is inversely proportional to the magnetic field strength B. As B increases, r must decrease. The period of the motion is T = 2πm/(qB). Since m and q are constant, the period is also inversely proportional to B. As B increases, T must also decrease.

Question 19

An electron beam passes through crossed electric and magnetic fields. The electric field is E=4.0×104 N/CE = 4.0 \times 10^4 \text{ N/C} pointing upward, and the magnetic field is B=0.20 TB = 0.20 \text{ T} pointing into the page. If electrons travel undeflected in a straight horizontal line, what happens when the magnetic field strength is suddenly doubled while keeping the electric field constant?

  1. The electrons curve downward with radius r=5.7×10−3 mr = 5.7 \times 10^{-3} \text{ m} (correct answer)
  2. The electrons curve upward with radius r=5.7×10−3 mr = 5.7 \times 10^{-3} \text{ m}
  3. The electrons curve downward with radius r=1.1×10−2 mr = 1.1 \times 10^{-2} \text{ m}
  4. The electrons curve upward with radius r=1.1×10−2 mr = 1.1 \times 10^{-2} \text{ m}
Explanation: Initially, for undeflected motion: qE=qvBqE = qvB, so v=EB=4.0×1040.20=2.0×105 m/sv = \frac{E}{B} = \frac{4.0 \times 10^4}{0.20} = 2.0 \times 10^5 \text{ m/s}. When B doubles to 0.40 T0.40 \text{ T}, the magnetic force qvBqvB exceeds the electric force qEqE. Since electrons are negative, the magnetic force (using right-hand rule reversed) points downward, making the net force downward. The radius of curvature is r=mvq(Bnew)=9.1×10−31×2.0×1051.6×10−19×0.40=2.8×10−6/4.9×10−4=5.7×10−3 mr = \frac{mv}{q(B_{new})} = \frac{9.1 \times 10^{-31} \times 2.0 \times 10^5}{1.6 \times 10^{-19} \times 0.40} = 2.8 \times 10^{-6}/4.9 \times 10^{-4} = 5.7 \times 10^{-3} \text{ m}. Choice B has wrong direction. Choices C and D use the original magnetic field value in the radius calculation.

Question 20

A charged particle with charge-to-mass ratio qm=2.0×107 C/kg\frac{q}{m} = 2.0 \times 10^7 \text{ C/kg} enters a velocity selector with E=3.0×104 N/CE = 3.0 \times 10^4 \text{ N/C} and B=0.15 TB = 0.15 \text{ T}. After exiting the selector, it enters a second region with only a magnetic field of 0.25 T0.25 \text{ T} perpendicular to its velocity. What is the ratio of the particle's kinetic energy in the second region to its kinetic energy in the selector?

  1. 1.01.0 (correct answer)
  2. 1.71.7
  3. 2.82.8
  4. 0.600.60
Explanation: In the velocity selector, particles travel undeflected when qE=qvB1qE = qvB_1, giving v=EB1=3.0×1040.15=2.0×105 m/sv = \frac{E}{B_1} = \frac{3.0 \times 10^4}{0.15} = 2.0 \times 10^5 \text{ m/s}. When the particle enters the second magnetic field region, only the magnetic force acts on it, which is always perpendicular to velocity. Since magnetic forces do no work (F⃗⋅v⃗=0\vec{F} \cdot \vec{v} = 0), kinetic energy is conserved. Therefore, the kinetic energy remains the same: KE2KE1=1.0\frac{KE_2}{KE_1} = 1.0. Choice B incorrectly uses the ratio of magnetic fields (0.25/0.150.25/0.15). Choice C uses the ratio of field strengths squared. Choice D uses the inverse ratio of magnetic fields.