Apply Motion in EM Fields - IB Physics

1. A particle of charge q and mass m moves in a circular path in a uniform magnetic field B. The period of its revolution is T. Which of the following quantities is the period T independent of?
   Answer: The speed of the particle, v. The magnetic force provides the centripetal force: qvB = mv²/r. The speed is related to the period by v = 2πr/T. Substituting v into the force equation gives q(2πr/T)B = m(2πr/T)²/r. Simplifying this equation leads to T = 2πm / (qB). This expression shows that the period T depends on m, q, and B, but is independent of the particle's speed v and the radius of its path r.
2. 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?
   Answer: The top surface of the strip becomes negative and the bottom becomes positive. 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.
3. 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?
   Answer: 0 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.
4. 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?
   Answer: Work done is zero, exit kinetic energy is K. 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.
5. 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?
   Answer: Net force is zero, net torque is non-zero. 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.
6. An alpha particle ($q = +2e$, $m = 6.64 \times 10^{-27} \text{ kg}$) is accelerated from rest through a potential difference of $2500 \text{ V}$ and then enters a region with parallel electric and magnetic fields. The electric field is $E = 1.8 \times 10^4 \text{ N/C}$ and the magnetic field is $B = 0.30 \text{ T}$, both pointing in the same direction perpendicular to the particle's initial velocity. What is the magnitude of the particle's acceleration immediately upon entering this region?
   Answer: $1.1 \times 10^{12} \text{ m/s}^2$ First, find the velocity after acceleration: $qV = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times 2 \times 1.6 \times 10^{-19} \times 2500}{6.64 \times 10^{-27}}} = \sqrt{2.41 \times 10^{11}} = 4.91 \times 10^5 \text{ m/s}$. Upon entering the field region, both electric and magnetic forces act perpendicular to velocity. The electric force is $F_E = qE = 2 \times 1.6 \times 10^{-19} \times 1.8 \times 10^4 = 5.76 \times 10^{-15} \text{ N}$. The magnetic force is $F_B = qvB = 2 \times 1.6 \times 10^{-19} \times 4.91 \times 10^5 \times 0.30 = 4.72 \times 10^{-14} \text{ N}$. Since both forces are perpendicular to velocity and in the same direction, $F_{total} = F_E + F_B = 5.29 \times 10^{-14} \text{ N}$. Therefore, $a = \frac{F}{m} = \frac{5.29 \times 10^{-14}}{6.64 \times 10^{-27}} = 1.1 \times 10^{12} \text{ m/s}^2$. Choice A neglects the electric force. Choice C neglects the magnetic force. Choice D uses incorrect charge value.
7. 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?
   Answer: It is deflected in the direction of the magnetic force. 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.
8. A charged particle moves in a region where both electric and magnetic fields are present. The electric field is uniform with magnitude $E = 6.0 \times 10^4 \text{ N/C}$, and the magnetic field is uniform with magnitude $B = 0.25 \text{ T}$. Both fields are perpendicular to each other and to the particle's initial velocity. If the particle follows a cycloid path and returns to its starting height after traveling a horizontal distance of $8.4 \text{ cm}$, what is the particle's charge-to-mass ratio?
   Answer: $2.9 \times 10^7 \text{ C/kg}$ When you encounter a charged particle moving in crossed electric and magnetic fields, you're dealing with cycloid motion - a combination of circular motion (from the magnetic force) and linear drift (from the electric field). The key insight is that the horizontal distance traveled when returning to the starting height equals one complete cycloid wavelength. For cycloid motion, the wavelength formula is $\lambda = \frac{2\pi mE}{qB^2}$, where the particle drifts due to the electric field while executing circular arcs due to the magnetic field. Since the particle returns to its starting height after traveling 8.4 cm horizontally, this distance is one wavelength. Setting up the equation: $0.084 = \frac{2\pi mE}{qB^2}$ Solving for charge-to-mass ratio: $\frac{q}{m} = \frac{2\pi E}{0.084 \times B^2} = \frac{2\pi \times 6.0 \times 10^4}{0.084 \times (0.25)^2} = 2.9 \times 10^7 \text{ C/kg}$ This confirms answer C is correct. Answer A ($1.8 \times 10^7$) likely comes from incorrectly using half the wavelength or making an error in the cycloid formula. Answer B ($4.7 \times 10^7$) might result from confusing cycloid motion with simple circular motion equations. Answer D ($3.6 \times 10^7$) could come from arithmetic errors in the calculation or using incorrect field relationships. Study tip: Remember that cycloid motion problems always involve the interplay between $E$ and $B^2$ in the wavelength formula. Practice identifying when horizontal distance corresponds to complete wavelengths versus partial cycles.
9. 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?
   Answer: r is proportional to √V. 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.
10. 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?
    Answer: It follows a helical path. 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.
11. 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?
    Answer: BIL 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.
12. 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)
    Answer: 1.0 cm 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.
13. 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?
    Answer: Decreasing the initial horizontal velocity v. 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)/($mv^2$) 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.
14. A charged particle with charge-to-mass ratio $\frac{q}{m} = 2.0 \times 10^7 \text{ C/kg}$ enters a velocity selector with $E = 3.0 \times 10^4 \text{ N/C}$ and $B = 0.15 \text{ T}$. After exiting the selector, it enters a second region with only a magnetic field of $0.25 \text{ T}$ 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?
    Answer: $1.0$ In the velocity selector, particles travel undeflected when $qE = qvB_1$, giving $v = \frac{E}{B_1} = \frac{3.0 \times 10^4}{0.15} = 2.0 \times 10^5 \text{ m/s}$. 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 ($\vec{F} \cdot \vec{v} = 0$), kinetic energy is conserved. Therefore, the kinetic energy remains the same: $\frac{KE_2}{KE_1} = 1.0$. Choice B incorrectly uses the ratio of magnetic fields ($0.25/0.15$). Choice C uses the ratio of field strengths squared. Choice D uses the inverse ratio of magnetic fields.
15. A proton enters a uniform magnetic field of magnitude $0.50 \text{ T}$ with velocity $2.0 \times 10^6 \text{ m/s}$ at an angle of $30°$ to the field lines. After completing one full helical orbit, what is the net displacement of the proton parallel to the magnetic field direction?
    Answer: $4.1 \times 10^{-2} \text{ m}$ The proton follows a helical path with constant velocity component parallel to B. The parallel component is $v_{||} = v\cos(30°) = 2.0 \times 10^6 \times \frac{\sqrt{3}}{2} = 1.73 \times 10^6 \text{ m/s}$. The period for one orbit is $T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.50} = 1.31 \times 10^{-7} \text{ s}$. The displacement parallel to B is $d = v_{||}T = 1.73 \times 10^6 \times 1.31 \times 10^{-7} = 4.1 \times 10^{-2} \text{ m}$. Choice B uses $\sin(30°)$ instead of $\cos(30°)$. Choice C uses the total velocity instead of the parallel component. Choice D assumes the displacement is half the correct value.
16. 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?
    Answer: The velocity component parallel to the field changes at a constant rate; the component perpendicular to the field is constant. 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.
17. A rectangular conducting loop with dimensions $2.0 \text{ cm} \times 3.0 \text{ cm}$ moves with constant velocity $v = 0.50 \text{ m/s}$ perpendicular to a uniform magnetic field boundary. The magnetic field has magnitude $B = 0.80 \text{ T}$ and exists only in a region $5.0 \text{ cm}$ wide. The loop enters the field region with its $2.0 \text{ cm}$ edge leading. What is the total impulse delivered to the loop during its complete passage through the magnetic field region?
    Answer: Zero As the loop enters the field, the induced EMF is $\varepsilon = Blv = 0.80 \times 0.02 \times 0.50 = 8.0 \times 10^{-3} \text{ V}$, creating a current that produces a magnetic force opposing motion (Lenz's law). When the loop is completely inside the uniform field, there's no flux change, so no current and no force. As the loop exits, the EMF has the same magnitude but opposite direction, creating an equal and opposite impulse. The total impulse is zero because the momentum change during entry exactly cancels the momentum change during exit. Choice A calculates impulse during entry only. Choice B uses half the correct entry calculation. Choice C doubles the entry impulse incorrectly.
18. 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?
    Answer: 0.10 N 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.
19. A cyclotron is designed to accelerate protons to a final kinetic energy of $25 \text{ MeV}$. The magnetic field strength is $B = 1.5 \text{ T}$. At the moment when a proton has kinetic energy of $16 \text{ MeV}$, what is the ratio of its orbital radius to its final orbital radius?
    Answer: $0.80$ When you encounter cyclotron problems, focus on the relationship between particle energy, velocity, and orbital radius. In a cyclotron, charged particles follow circular paths whose radii depend on their momentum. The key insight is that orbital radius $r = \frac{mv}{qB}$, where $m$ is mass, $v$ is velocity, $q$ is charge, and $B$ is magnetic field strength. Since kinetic energy $KE = \frac{1}{2}mv^2$, we can express velocity as $v = \sqrt{\frac{2KE}{m}}$. Substituting this into the radius formula gives us $r = \frac{m\sqrt{\frac{2KE}{m}}}{qB} = \frac{\sqrt{2mKE}}{qB}$. This means radius is proportional to $\sqrt{KE}$. Therefore, the ratio of radii equals the square root of the ratio of kinetic energies: $\frac{r_{16}}{r_{25}} = \sqrt{\frac{16 \text{ MeV}}{25 \text{ MeV}}} = \sqrt{0.64} = 0.80$ This confirms answer C) $0.80$ is correct. The wrong answers represent common mistakes: A) $0.89$ might come from taking $\sqrt{0.8}$ instead of $\sqrt{0.64}$. B) $0.64$ results from incorrectly assuming radius is directly proportional to energy rather than its square root. D) $0.75$ could arise from arithmetic errors or confusion with other energy relationships. Remember: in cyclotron problems, radius scales with $\sqrt{KE}$, not linearly with energy. This square root relationship appears frequently in IB Physics when dealing with energy-dependent motion parameters.
20. A proton and an electron enter a region of uniform magnetic field with the same initial velocity, which is perpendicular to the field. Which statement correctly compares their paths?
    Answer: They curve in opposite directions, with the proton's path having a larger radius. The direction of the magnetic force depends on the sign of the charge, so the proton (+e) and electron (-e) will be deflected in opposite directions. The radius of the circular path is given by r = mv/|q|B. Since v, |q|, and B are the same for both particles, the radius is directly proportional to the mass (r ∝ m). As a proton is much more massive than an electron ($m_p$ >> $m_e$), the radius of the proton's path will be much larger than the radius of the electron's path.

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