Understand Motion in EM Fields - IB Physics

A rectangular current-carrying loop with dimensions 4.0 cm × 6.0 cm carries a current of 2.5 A and is placed in a uniform magnetic field of 0.80 T. The magnetic field makes an angle of 30° with the normal to the loop's plane. If the loop is free to rotate about an axis perpendicular to the magnetic field, what is the maximum kinetic energy the loop can achieve if released from rest at the initial position?
Answer: 8.3 × 10⁻⁴ J The magnetic dipole moment is μ = IA = (2.5)(0.04 × 0.06) = 6.0 × 10⁻³ A⋅m². The potential energy is U = -μB cos θ. Initially, θ = 30°, so U₁ = -(6.0 × 10⁻³)(0.80)cos(30°) = -4.16 × 10⁻³ J. At maximum kinetic energy, the loop is aligned with the field (θ = 0°), so U₂ = -(6.0 × 10⁻³)(0.80)cos(0°) = -4.80 × 10⁻³ J. By conservation of energy: KE_max = U₁ - U₂ = -4.16 × 10⁻³ - (-4.80 × 10⁻³) = 8.3 × 10⁻⁴ J. Choice A uses the total potential energy change incorrectly. Choice B uses wrong angle relationships. Choice D uses an incorrect reference point.
A charged particle with charge $+2.0 \times 10^{-6}$ C and mass $1.0 \times 10^{-3}$ kg is released from rest in a uniform electric field of magnitude $5.0 \times 10^{4}$ N/C pointing horizontally to the right. After traveling 0.20 m horizontally, the particle enters a region with a uniform magnetic field of magnitude 0.30 T pointing into the page, while the electric field remains present. What is the radius of curvature of the particle's path immediately upon entering the magnetic field region?
Answer: 0.67 m First, find the velocity when entering the magnetic field using kinematics. The electric force is F = qE = (2.0×10⁻⁶)(5.0×10⁴) = 0.10 N. The acceleration is a = F/m = 0.10/(1.0×10⁻³) = 100 m/s². Using v² = u² + 2as with u = 0, s = 0.20 m: v = √(2×100×0.20) = √40 = 6.32 m/s. In the magnetic field, the radius of circular motion is r = mv/(qB) = (1.0×10⁻³)(6.32)/[(2.0×10⁻⁶)(0.30)] = 0.67 m. Choice B uses the wrong velocity calculation. Choice C incorrectly combines electric and magnetic forces. Choice D doubles the correct answer.
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?
Answer: $3F/2$ 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 $F' = (3/2)F$ .
An ion is moving in a circular path in a uniform magnetic field. If the kinetic energy of the ion is doubled, what is the effect on the period of its circular motion?
Answer: It remains unchanged. The period of circular motion is given by $T = 2\pi r / v$ . The radius of the path is $r = mv/(qB)$ . Substituting the expression for $r$ into the equation for $T$ gives $T = 2\pi (mv/qB) / v = 2\pi m / (qB)$ . This shows that the period $T$ depends on the mass $m$ , charge $q$ , and magnetic field strength $B$ , but it is independent of the particle's speed $v$ and its kinetic energy. Therefore, doubling the kinetic energy will not change the period.
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?
Answer: The path becomes a helix with an increasing pitch. 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.
Two long, parallel, straight wires are suspended vertically. They attract each other with a magnetic force. What is the only valid deduction from this observation?
Answer: The current in both wires flows in the same direction. The rule for forces between parallel current-carrying conductors is that parallel currents (currents in the same direction) attract, and anti-parallel currents (currents in opposite directions) repel. The sign of the charge carriers (e.g., electrons or positive ions) does not affect this rule, as the direction of conventional current is what matters. The material of the wire is also not the primary reason for the force, which arises from the interaction of their magnetic fields.
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?
Answer: The electron follows a circular path with radius 34 mm moving counterclockwise when viewed from above 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.
Two parallel infinite wires carry currents $I_1 = 3.0$ A and $I_2 = 4.0$ A in opposite directions and are separated by distance d = 0.10 m. A charged particle with charge q = $+1.6 \times 10^{-19}$ C moves with velocity $\vec{v} = 2.0 \times 10^{5}$ m/s parallel to the wires at a point equidistant from both wires. What is the magnitude of the magnetic force on the particle?
Answer: $6.4 \times 10^{-18}$ N At the midpoint between antiparallel currents, the magnetic fields add. Each wire creates B = μ₀I/(2πr) at distance r = d/2 = 0.05 m. B₁ = (4π×10⁻⁷)(3.0)/(2π×0.05) = 1.2×10⁻⁵ T and B₂ = (4π×10⁻⁷)(4.0)/(2π×0.05) = 1.6×10⁻⁵ T. Since currents are opposite, fields point in the same direction, so B_total = 1.2×10⁻⁵ + 1.6×10⁻⁵ = 2.8×10⁻⁵ T. The force is F = qvB = (1.6×10⁻¹⁹)(2.0×10⁵)(2.8×10⁻⁵) = 6.4×10⁻¹⁸ N. Choice A uses only one wire. Choice C uses wrong field combination. Choice D incorrectly assumes cancellation for antiparallel currents.
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?
Answer: $2.6 \times 10^{-7}$ s 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.
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?
Answer: Kinetic energy remains constant and the momentum vector changes. 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.
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?
Answer: West 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.
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?
Answer: Zero 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.
A cyclotron is designed to accelerate protons using a magnetic field of 1.5 T. If the maximum radius of the proton's path is 0.25 m before extraction, what is the kinetic energy of the protons when they exit the cyclotron?
Answer: $6.8 \times 10^{-12}$ J In a cyclotron, r = mv/(qB), so v = qBr/m. For a proton: v = (1.6×10⁻¹⁹)(1.5)(0.25)/(1.67×10⁻²⁷) = 3.59×10⁷ m/s. The kinetic energy is KE = ½mv² = ½(1.67×10⁻²⁷)(3.59×10⁷)² = 6.8×10⁻¹² J. Choice B doubles the correct answer. Choice C uses the wrong mass value. Choice D uses an incorrect velocity calculation that squares the radius term.
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?
Answer: $BI(2R)$ 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 $F = BI(2R)$ .
A rectangular conducting loop with width w = 0.080 m moves with constant velocity v = 2.0 m/s through a non-uniform magnetic field. The magnetic field varies linearly with position according to B(x) = 0.10 + 0.50x, where x is measured in meters and B is in tesla. As the loop moves in the +x direction, what is the magnitude of the induced EMF when the leading edge of the loop is at position x = 0.20 m?
Answer: 0.080 V When a conducting loop moves through a non-uniform magnetic field, you need to consider how the magnetic flux through the loop changes with time. Since the field varies with position and the loop is moving, both the leading and trailing edges of the loop experience different magnetic field strengths simultaneously. The induced EMF equals the rate of change of magnetic flux. For a loop moving through a spatially varying field, this becomes: $\varepsilon = w \cdot v \cdot \Delta B$ , where $\Delta B$ is the difference in magnetic field between the leading and trailing edges of the loop. When the leading edge is at x = 0.20 m, the trailing edge is at x = 0.20 - 0.080 = 0.12 m (since the loop width is 0.080 m). Using B(x) = 0.10 + 0.50x: At leading edge: B(0.20) = 0.10 + 0.50(0.20) = 0.20 T At trailing edge: B(0.12) = 0.10 + 0.50(0.12) = 0.16 T Therefore: $\varepsilon = 0.080 \times 2.0 \times (0.20 - 0.16) = 0.160 \times 0.04 = 0.080$ V This confirms answer C is correct. Answer A (0.064 V) likely results from using an incorrect velocity or width value. Answer B (0.048 V) probably comes from miscalculating the field difference or using wrong positions. Answer D (0.032 V) suggests halving the correct result, possibly from considering only one edge of the loop. Remember: in non-uniform field problems, always consider what's happening at both edges of the loop simultaneously—the EMF depends on the field difference across the loop's width.
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?
Answer: The net force is zero and the net torque is non-zero. 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.
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?
Answer: $v$ 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.
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 $R'/R$ ?
Answer: 6 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.
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?
Answer: North 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.
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?
Answer: 1/2 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: $(2e/e) \times (m_p/4m_p) = 2 \times (1/4) = 1/2$ .

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