Test your understanding of electric fields, Coulomb's Law, magnetic fields, the Lorentz force, and charged particle motion. Covers Lessons 1–6.
1. Two point charges of $+3.0 \\mu$C and $+5.0 \\mu$C are placed 0.30 m apart in vacuum. What is the magnitude of the electrostatic force between them?
2. An electron is placed in a uniform electric field of 500 N/C directed to the right. What is the magnitude and direction of the force on the electron?
3. The electric field 0.20 m from a point charge of $+4.0 \\mu$C is:
4. Two parallel plates are separated by 4.0 cm with a potential difference of 200 V. What is the electric field between the plates?
5. A proton is accelerated from rest through a potential difference of 500 V. What is its final speed? ($m_p = 1.67\\times10^{-27}$ kg, $e = 1.60\\times10^{-19}$ C)
6. A current of 5.0 A flows vertically upward. What is the direction of the magnetic field 0.10 m to the east of the wire?
7. The magnetic field 0.050 m from a long straight wire carrying 8.0 A is: ($\\mu_0 = 4\\pi\\times10^{-7}$ T·m/A)
8. An electron moves at $2.0\\times10^6$ m/s perpendicular to a uniform magnetic field of 0.30 T. What is the magnitude of the magnetic force on the electron?
9. A charged particle enters a uniform magnetic field perpendicular to its velocity. The path of the particle will be:
10. A proton moves in a circular path of radius 0.080 m in a uniform magnetic field of 0.50 T. What is its speed? ($m_p = 1.67\\times10^{-27}$ kg, $e = 1.60\\times10^{-19}$ C)
11. An alpha particle (charge $+2e$, mass $6.6\\times10^{-27}$ kg) and a proton (charge $+e$, mass $1.67\\times10^{-27}$ kg) are accelerated through the same potential difference and then enter a uniform magnetic field perpendicular to their velocity. The ratio of the radius of the alpha particle's path to that of the proton is:
12. Which statement correctly compares electric and magnetic forces on charged particles?
13. An electron is projected horizontally into a uniform electric field directed vertically upward. The electron will follow a path that is:
14. Two parallel wires carry currents in the same direction. The force between them is:
15. In a velocity selector, electric and magnetic fields are perpendicular and crossed so that particles with a specific velocity pass through undeflected. If $E = 2.0\\times10^4$ N/C and $B = 0.10$ T, what is the selected velocity?
Question 1. Two point charges, $+2.0 \\mu$C and $-5.0 \\mu$C, are placed 0.40 m apart in vacuum. (a) Calculate the magnitude of the electrostatic force between them. (b) Is the force attractive or repulsive? Explain. (c) Calculate the electric field at the midpoint between the charges. (3 marks)
Question 2. An electron is accelerated from rest through a potential difference of 250 V between two parallel plates separated by 2.0 cm. (a) Calculate the electric field between the plates. (b) Calculate the final speed of the electron. (c) Calculate the time taken for the electron to travel between the plates. ($m_e = 9.11\\times10^{-31}$ kg, $e = 1.60\\times10^{-19}$ C) (3 marks)
Question 3. A proton enters a uniform magnetic field of 0.40 T perpendicular to its velocity of $5.0\\times10^6$ m/s. (a) Calculate the radius of the circular path. (b) Calculate the period of the motion. (c) Explain why the kinetic energy of the proton remains constant while it moves in the magnetic field. (d) If the magnetic field strength is doubled, what happens to the radius and period? (4 marks)
Question 4. In a mass spectrometer, ions are accelerated through a potential difference $V$ and then enter a uniform magnetic field $B$ where they follow a semicircular path of radius $r$. (a) Show that $r = \\dfrac{1}{B}\\sqrt{\\dfrac{2mV}{q}}$. (b) Two singly ionised isotopes with masses $m_1$ and $m_2$ are accelerated through the same $V$ and enter the same $B$. Find an expression for the separation of their impact points on the detector. (c) Explain why this device can separate isotopes but cannot separate ions with the same mass-to-charge ratio. (4 marks)
Question 5. Evaluate the statement: "A charged particle moving through a uniform magnetic field experiences a force that does no work, so its speed must remain constant. However, a charged particle in a uniform electric field experiences a force that does work, so its speed must change." Discuss both parts of the statement using vector analysis and the definitions of work and kinetic energy. (4 marks)
1. B — $F = k\\dfrac{q_1 q_2}{r^2} = 8.99\\times10^9 \\times \\dfrac{(3.0\\times10^{-6})(5.0\\times10^{-6})}{0.30^2} = 8.99\\times10^9 \\times \\dfrac{15\\times10^{-12}}{0.09} = 1.50$ N.
2. B — $F = qE = 1.6\\times10^{-19} \\times 500 = 8.0\\times10^{-17}$ N. The electron is negative, so the force is opposite to the field direction: to the left.
3. C — $E = kQ/r^2 = 8.99\\times10^9 \\times 4.0\\times10^{-6} / 0.20^2 = 8.99\\times10^5$ N/C. Positive charge: field points away.
4. C — $E = V/d = 200 / 0.040 = 5.0\\times10^3$ N/C. Remember to convert cm to m.
5. B — $\\Delta K = qV = 1.6\\times10^{-19} \\times 500 = 8.0\\times10^{-17}$ J. $v = \\sqrt{2K/m} = \\sqrt{2 \\times 8.0\\times10^{-17} / 1.67\\times10^{-27}} = 3.1\\times10^5$ m/s.
6. C — RH grip rule: thumb up (current), fingers curl anticlockwise when viewed from above. East of wire = to the right; field points north.
7. B — $B = \\mu_0 I / (2\\pi r) = (4\\pi\\times10^{-7} \\times 8.0) / (2\\pi \\times 0.050) = 3.2\\times10^{-5}$ T.
8. B — $F = qvB = 1.6\\times10^{-19} \\times 2.0\\times10^6 \\times 0.30 = 9.6\\times10^{-14}$ N.
9. C — Magnetic force is always perpendicular to velocity, so it provides centripetal force, producing circular motion.
10. B — $v = rqB/m = 0.080 \\times 1.6\\times10^{-19} \\times 0.50 / 1.67\\times10^{-27} = 3.8\\times10^6$ m/s.
11. B — From $qV = \\frac{1}{2}mv^2$, $v = \\sqrt{2qV/m}$. Then $r = mv/(qB) = \\sqrt{2mV/q}/B$, so $r \\propto \\sqrt{m/q}$. Ratio: $\\sqrt{(6.6/1.67) \\times (1/2)} \\approx \\sqrt{2}$.
12. B — Electric force $F_E = qE$ can have a component parallel to displacement, so it can do work. Magnetic force $F_B = qv\\times B$ is always perpendicular to velocity, so $W = F \\cdot d = 0$.
13. C — Field is upward; electron is negative, so force is downward. Horizontal velocity is constant (no horizontal force), so parabolic trajectory curving downward.
14. A — Parallel currents attract; antiparallel currents repel. This follows from the motor effect.
15. B — For no deflection: $qE = qvB$, so $v = E/B = 2.0\\times10^4 / 0.10 = 2.0\\times10^5$ m/s.
(a) Force (1 mark)
$F = k\\dfrac{q_1 q_2}{r^2} = 8.99\\times10^9 \\times \\dfrac{(2.0\\times10^{-6})(5.0\\times10^{-6})}{0.40^2} = 8.99\\times10^9 \\times \\dfrac{10\\times10^{-12}}{0.16} = 0.562$ N $\\approx$ 0.56 N
(b) Attractive (1 mark)
The charges are opposite ($+2.0 \\mu$C and $-5.0 \\mu$C), so the force is attractive.
(c) Field at midpoint (1 mark)
$E_+ = kQ/r^2$ points away from $+2.0 \\mu$C (to the right, toward negative charge).
$E_- = kQ/r^2$ points toward $-5.0 \\mu$C (to the right).
$E_{total} = E_+ + E_- = 8.99\\times10^9 \\times \\left(\\dfrac{2.0\\times10^{-6}}{0.20^2} + \\dfrac{5.0\\times10^{-6}}{0.20^2}\\right) = 8.99\\times10^9 \\times \\dfrac{7.0\\times10^{-6}}{0.04} = 1.57\\times10^6$ N/C to the right (toward the negative charge).
(a) Electric field (1 mark)
$E = V/d = 250 / 0.020 = 1.25\\times10^4$ N/C
(b) Final speed (1 mark)
$\\Delta K = qV = 1.6\\times10^{-19} \\times 250 = 4.0\\times10^{-17}$ J
$v = \\sqrt{2K/m} = \\sqrt{2 \\times 4.0\\times10^{-17} / 9.11\\times10^{-31}} = \\sqrt{8.78\\times10^{13}} = 9.37\\times10^6$ m/s $\\approx$ $9.4\\times10^6$ m/s
(c) Time (1 mark)
$a = F/m = qE/m = 1.6\\times10^{-19} \\times 1.25\\times10^4 / 9.11\\times10^{-31} = 2.20\\times10^{15}$ m/s²
$s = ut + \\frac{1}{2}at^2$ with $u = 0$: $0.020 = \\frac{1}{2} \\times 2.20\\times10^{15} \\times t^2$
$t = \\sqrt{2 \\times 0.020 / 2.20\\times10^{15}} = 4.3\\times10^{-9}$ s $\\approx$ 4.3 ns
(a) Radius (1 mark)
$r = mv/(qB) = 1.67\\times10^{-27} \\times 5.0\\times10^6 / (1.6\\times10^{-19} \\times 0.40) = 0.130$ m $\\approx$ 13 cm
(b) Period (1 mark)
$T = 2\\pi r / v = 2\\pi \\times 0.130 / 5.0\\times10^6 = 1.6\\times10^{-7}$ s $\\approx$ 0.16 $\\mu$s
Alternatively: $T = 2\\pi m / (qB) = 2\\pi \\times 1.67\\times10^{-27} / (1.6\\times10^{-19} \\times 0.40) = 1.6\\times10^{-7}$ s
(c) Constant KE (1 mark)
The magnetic force is always perpendicular to the velocity ($F \\perp v$). Since work $W = F \\cdot d = Fd\\cos 90° = 0$, no work is done. By the work-energy theorem, $\\Delta K = W = 0$, so kinetic energy remains constant.
(d) Doubled B-field (1 mark)
Radius halves ($r \\propto 1/B$): new $r = 6.5$ cm. Period halves ($T \\propto 1/B$): new $T = 0.08 \\mu$s.
(a) Derivation (1 mark)
$qV = \\frac{1}{2}mv^2$ gives $v = \\sqrt{2qV/m}$. In magnetic field: $qvB = mv^2/r$, so $r = mv/(qB) = \\dfrac{m}{qB}\\sqrt{\\dfrac{2qV}{m}} = \\dfrac{1}{B}\\sqrt{\\dfrac{2mV}{q}}$.
(b) Separation (2 marks)
$r_1 = \\dfrac{1}{B}\\sqrt{\\dfrac{2m_1V}{q}}$, $r_2 = \\dfrac{1}{B}\\sqrt{\\dfrac{2m_2V}{q}}$
Diameter of semicircle = $2r$, so separation = $2(r_2 - r_1) = \\dfrac{2}{B}\\sqrt{\\dfrac{2V}{q}}(\\sqrt{m_2} - \\sqrt{m_1})$.
(c) Same mass-to-charge ratio (1 mark)
Since $r \\propto \\sqrt{m/q}$, ions with the same $m/q$ have the same radius and cannot be separated. The mass spectrometer separates by $m/q$, not by $m$ or $q$ individually.
Magnetic field part (2 marks)
The statement is correct. The magnetic force on a charged particle is $\\vec{F}_B = q(\\vec{v} \\times \\vec{B})$. By the definition of the cross product, $\\vec{F}_B$ is always perpendicular to $\\vec{v}$. The infinitesimal work done is $dW = \\vec{F}_B \\cdot d\\vec{s} = \\vec{F}_B \\cdot \\vec{v}\\,dt = 0$ since $\\vec{F}_B \\perp \\vec{v}$. Therefore $W = 0$ and $\\Delta K = 0$, so speed remains constant. Only the direction of velocity changes.
Electric field part (2 marks)
The statement is correct. The electric force is $\\vec{F}_E = q\\vec{E}$. If the particle moves parallel to $\\vec{E}$, then $\\vec{F}_E$ has a component parallel to displacement, so $W = qE\\,d \\neq 0$. By the work-energy theorem, $\\Delta K = W \\neq 0$, so the speed changes. The particle accelerates in the direction of $\\vec{F}_E$ (for positive $q$) or opposite (for negative $q$).
Key distinction: Magnetic force changes direction only (circular motion); electric force changes speed (acceleration along field lines).
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