Model Electric and Magnetic Fields - Grade 11 Physics

1. At what speed will the electron pass through the fields without deflection, and why?
   Answer: $4.0\times10^5\,m/s$, because setting the electric force magnitude equal to the magnetic force magnitude and solving $qE=qvB$ for $v$ gives $E/B$ When a charged particle moves through crossed electric and magnetic fields, it travels undeflected only when the two forces on it are equal in magnitude and opposite in direction — this is called a velocity selector. Your job is to identify both force directions and set their magnitudes equal. Here's the physics: the electron moves east. The electric field points south, so the electric force on the electron (charge negative) points north — opposite the field. The magnetic field points into the tabletop (downward). Using the right-hand rule for velocity (east) crossed with B (down), the magnetic force on a positive charge would point north; since the electron is negative, the magnetic force points south. So the two forces point in opposite directions and can cancel — confirming undeflected motion is possible. Setting magnitudes equal: $qE = qvB$, so $v = \frac{E}{B} = \frac{1.2 \times 10^3}{3.0 \times 10^{-3}} = 4.0 \times 10^5 \, \text{m/s}$. Answer A is correct. Answer B gets the speed right but the direction wrong — it claims both forces point north, which would mean they add together and deflect the electron, not cancel. The reasoning is flawed even though the number matches. Answer C inverts the ratio, computing $B/E$ instead of $E/B$. Dimensionally, $B/E$ gives units of $\text{s/m}$, not $\text{m/s}$ — a useful red flag. Answer D multiplies $E \times B$ instead of dividing, producing a quantity with completely wrong units and an unreasonable result. Study tip: Always check your formula's units. For a velocity selector, $v = E/B$ has units $(\text{N/C})/(\text{T}) = \text{m/s}$ — if your ratio is flipped or multiplied, the units will betray the mistake immediately.
2. Two fixed positive charges lie on the x-axis. A charge of $+4.0\,\mu C$ is at $x=0$, and a charge of $+1.0\,\mu C$ is at $x=3.0\,m$. Which statement correctly describes the electric field and electric potential at $x=2.0\,m$?
   Answer: The net electric field is zero, but the electric potential is positive. When a question asks about both electric field and electric potential at a single point, treat them as completely independent quantities — they follow different rules and can behave very differently at the same location. The electric field at $x = 2.0\,\text{m}$ is a vector, so direction matters. The +4.0 µC charge at $x = 0$ pushes rightward on a test charge at $x = 2.0\,\text{m}$, while the +1.0 µC charge at $x = 3.0\,\text{m}$ pushes leftward. Using Coulomb's law, the field from the 4.0 µC charge is $E_1 = k\frac{4.0\times10^{-6}}{(2.0)^2} = 9000\,\text{N/C}$ (rightward), and from the 1.0 µC charge: $E_2 = k\frac{1.0\times10^{-6}}{(1.0)^2} = 9000\,\text{N/C}$ (leftward). These are equal and opposite, so the net electric field is exactly zero. Electric potential, however, is a scalar — you simply add contributions without worrying about direction. Both positive charges produce positive potentials, so the total potential $V = k\frac{4.0\times10^{-6}}{2.0} + k\frac{1.0\times10^{-6}}{1.0}$ is strictly positive. This confirms A is correct. Choice B is wrong because while the field is indeed zero, potential is never zero when positive charges are present — scalars add without cancellation from direction. Choice C is wrong because the fields cancel, leaving zero net field, not a rightward one. Choice D is wrong on both counts: the field is zero, and positive charges always produce positive potential. Remember: zero electric field does not imply zero potential. These are fundamentally different quantities — one is a vector sum, the other a scalar sum.
3. A charge of $+2.0\,\mu C$ is at coordinates $(0,0)$, and a charge of $-2.0\,\mu C$ is at $(0,\,0.30\,m)$. Point P is at $(0.40\,m,\,0)$. Use $k=9.0\times10^9\,N\cdot m^2/C^2$.
   Answer: $7.0\times10^4\,N/C$, about $38^\circ$ north of east When a question asks for the electric field at a point due to multiple charges, your job is to find each field vector separately, then add them as vectors — never just add the magnitudes. Start by finding the distances. Charge 1 ($+2.0\,\mu C$) sits at the origin, so its distance to P at $(0.40, 0)$ is simply $r_1 = 0.40\,\text{m}$. Charge 2 ($-2.0\,\mu C$) is at $(0, 0.30)$, so its distance to P is $r_2 = \sqrt{0.40^2 + 0.30^2} = 0.50\,\text{m}$. Now compute each field magnitude: $E_1 = \frac{kq}{r_1^2} = \frac{9\times10^9 \times 2\times10^{-6}}{0.16} = 1.125\times10^5\,\text{N/C}$, directed east (away from $+$ charge). $E_2 = \frac{9\times10^9 \times 2\times10^{-6}}{0.25} = 7.2\times10^4\,\text{N/C}$, directed toward the negative charge (northwest from P). The unit vector from P toward charge 2 points in the $(-0.40, +0.30)/0.50$ direction, giving components $(-0.80,\,+0.60)$. So: $E_{2x} = -5.76\times10^4$, $E_{2y} = +4.32\times10^4$. Total: $E_x = 1.125\times10^5 - 5.76\times10^4 \approx 5.49\times10^4$, $E_y \approx 4.32\times10^4$. Magnitude: $\sqrt{(5.49)^2+(4.32)^2}\times10^4 \approx 7.0\times10^4\,\text{N/C}$. Angle: $\tan^{-1}(4.32/5.49)\approx 38°$ north of east — confirming C. Choice A incorrectly adds magnitudes without proper vector decomposition, giving a falsely large result. Choice B has the correct magnitude but flips the direction entirely — a sign error on both components. Choice D ignores the contribution of the positive charge and assumes the field points straight toward the negative charge. Your study tip: always sketch the geometry, draw each field arrow at point P, and decompose into components before adding. Direction errors are the most common trap on superposition problems.
4. A compass sits above a horizontal wire carrying current east. Its N pole points:
   Answer: Southward Point your right thumb east along the current. Your fingers curl around the wire, and at a point above the wire they point south, so the magnetic field there is southward. The compass N pole aligns with that field, so it points south. The tempting wrong answer is northward because a compass usually points north, but here the wire's field overrides Earth's.
5. An electron moves east in a magnetic field pointing south. The initial magnetic force is:
   Answer: Upward The magnetic force on a charge comes from q v x B. With v east and B south, v x B points downward for a positive charge. But an electron has negative charge, so the force reverses and points upward. The tempting wrong choice is downward, which is the direction for a positive charge, not an electron.
6. A $+2q$ charge is at $(-3,0)$, a $-q$ at $(3,0)$. At $(0,4)$, the electric field points:
   Answer: Up-right From (0,4), both charges are 5 units away. The +2q charge pushes away from it, up and to the right with strength 2. The -q charge pulls toward it, which is down and to the right with strength 1. The vertical parts partly cancel but leave upward, while the rightward parts add. The tempting wrong answer is straight up because it forgets that the negative charge also pulls to the right, not left.
7. A wire carries current north; a proton above it moves east. Magnetic force on proton:
   Answer: Zero At a point above a wire carrying current north, the magnetic field points east. The proton also moves east, so its velocity and the magnetic field are parallel. Magnetic force is q(v x B), and the cross product of parallel vectors is zero. Upward is tempting if you expect a vertical deflection, but there is no perpendicular component to produce one.
8. An alpha particle enters a uniform magnetic field perpendicular to the field lines. It follows exactly half of a circular path before leaving the field, and its speed remains $v$. Which statement correctly gives the magnetic field's work on the particle, its change in kinetic energy, and the magnitude of its impulse?
   Answer: Work is zero, kinetic-energy change is zero, and impulse magnitude is $2mv$. Whenever you see a charged particle moving in a magnetic field, the first principle to anchor yourself to is this: the magnetic force is always perpendicular to the velocity. Because work equals force times displacement in the direction of the force, a purely perpendicular force does zero work. This means the particle's kinetic energy — which equals $\frac{1}{2}mv^2$ — cannot change, confirming the speed stays constant at $v$. That rules out C and D immediately, since both claim positive or negative work and kinetic-energy changes. Now the subtle part: even though energy is conserved, momentum is a vector, and it can change direction without changing magnitude. After half a circular path, the particle has reversed direction completely. Its initial momentum is $+mv$ and its final momentum is $-mv$ (pointing the opposite way). The impulse — defined as the change in momentum — is: $|\Delta \vec{p}| = |{-mv - (+mv)}| = 2mv$ So the impulse magnitude is $2mv$, not zero. This is what makes A wrong: A correctly identifies zero work and zero kinetic-energy change, but incorrectly concludes the impulse is also zero. Impulse depends on the change in the velocity vector, not its magnitude. C and D are doubly wrong because they misidentify the work and energy change and get the impulse wrong. That confirms B as the correct answer. Study tip: Always distinguish between scalar quantities (speed, kinetic energy) and vector quantities (velocity, momentum). A magnetic field can change a particle's direction — and therefore its momentum and impulse — without ever changing its speed or energy.
9. Which combination gives the approximate radius and initial direction of the proton's circular deflection?
   Answer: $3.2\,cm$, initially upward from the horizontal plane When a charged particle moves through a magnetic field, two things matter: the radius of circular motion and the direction of the magnetic force. Master both using energy conservation and the right-hand rule. Finding the speed: The proton gains kinetic energy equal to the work done by the electric field: $qV = \frac{1}{2}mv^2$. Solving for $v$: $v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2(1.60\times10^{-19})(2000)}{1.67\times10^{-27}}} \approx 6.2\times10^5 \,\text{m/s}$ Finding the radius: The magnetic force provides centripetal force, so $qvB = \frac{mv^2}{r}$, giving $r = \frac{mv}{qB}$: $r = \frac{(1.67\times10^{-27})(6.2\times10^5)}{(1.60\times10^{-19})(0.20)} \approx 0.032\,\text{m} = 3.2\,\text{cm}$ This eliminates C and D immediately. Finding the direction: Apply the right-hand rule to $\vec{F} = q\vec{v}\times\vec{B}$. Point your fingers east (velocity), curl them north (field), and your thumb points upward. So the initial force — and deflection — is upward, confirming answer A. Choice B gets the radius right but flips the direction. This is the classic right-hand rule error — students often forget the force is perpendicular to both $\vec{v}$ and $\vec{B}$, not just one. Choices C and D use an incorrect radius, likely from forgetting the factor of 2 inside the square root when solving for $v$. Study tip: Always tackle these in two steps — energy for speed, then right-hand rule for direction. Practice $\vec{v}\times\vec{B}$ with cardinal directions until it becomes automatic.
10. Between two equal positive charges, the field line pattern at the midpoint has:
    Answer: No lines, field is zero At the midpoint, each positive charge creates a field pointing away from it, so the two fields are equal in size and opposite in direction. They cancel exactly, making the net field zero, and no field lines pass through a zero-field point. The tempting wrong answer is 'lines closest together,' because closely spaced lines mean a strong field, but at the midpoint the field is weakest, not strongest.
11. At a certain point, a charge of $+3.0\,\mu C$ is $0.40\,m$ away, and a charge of $-2.0\,\mu C$ is $0.20\,m$ away. A charge of $-4.0\,nC$ is then placed at that point. Using $k=9.0\times10^9\,N\cdot m^2/C^2$, what is the electric potential energy of the newly placed charge?
    Answer: $+9.0\times10^{-5}\,J$, because the net electric potential at the point is negative When a question involves electric potential energy of a placed charge, your two-step process is: first find the net electric potential at the point (due to existing charges), then multiply by the new charge placed there. Step 1 — Net Electric Potential: Electric potential is a scalar, so you simply add contributions with their signs: $V = k\frac{q_1}{r_1} + k\frac{q_2}{r_2} = (9.0\times10^9)\frac{+3.0\times10^{-6}}{0.40} + (9.0\times10^9)\frac{-2.0\times10^{-6}}{0.20}$ $V = +67{,}500\,\text{V} + (-90{,}000\,\text{V}) = -22{,}500\,\text{V}$ The net potential is negative. Step 2 — Electric Potential Energy: Multiply the potential by the placed charge: $U = qV = (-4.0\times10^{-9}\,\text{C})(-22{,}500\,\text{V}) = +9.0\times10^{-5}\,\text{J}$ This confirms B — the potential energy is $+9.0\times10^{-5}\,\text{J}$ because the net potential at the point is negative. A gets the magnitude right but flips the sign, incorrectly claiming the net potential is positive — it's actually negative, making the potential energy positive when multiplied by the negative test charge. C ignores the signs of the source charges entirely, adding magnitudes rather than signed values, which gives a completely wrong potential and energy. D compounds the same magnitude-only error from C while also asserting a false rule that negative test charges always yield positive potential energy — sign depends on the actual potential, not just the charge's sign. A helpful habit: always keep track of signs for both the source charges and the placed charge separately — two negatives multiplied give a positive result, which is exactly the trap this question sets.
12. Two long, straight, parallel wires are separated by a fixed distance and carry equal currents in the same direction. Consider the point exactly halfway between the wires. Which statement correctly describes the magnetic field at the midpoint and the interaction between the wires?
    Answer: The fields cancel at the midpoint, and the wires attract each other. Whenever you see a question involving parallel current-carrying wires, you need to analyze two separate phenomena: the magnetic field at a specific point in space, and the force the wires exert on each other. These are related but distinct ideas. At the midpoint between two wires carrying current in the same direction, apply the right-hand rule to each wire separately. For two wires side by side with current flowing upward, the left wire produces a field pointing out of the page at the midpoint, while the right wire produces a field pointing into the page at the same location. These fields are equal in magnitude (same current, same distance) but opposite in direction — they cancel completely, giving a net magnetic field of zero at the midpoint. This confirms the "fields cancel" part of the answer. For the force between the wires, each wire sits inside the magnetic field created by the other. Using $\vec{F} = I\vec{L} \times \vec{B}$, the current in each wire interacts with the neighboring wire's field. When currents flow in the same direction, the resulting force pulls the wires toward each other — they attract. This makes D the correct answer. A and C are wrong because the fields don't add at the midpoint — they cancel due to opposing directions. B gets the field cancellation right but incorrectly states the wires repel; repulsion only happens when currents flow in opposite directions. A handy memory rule: same direction currents attract, opposite directions repel — and field cancellation/addition at the midpoint follows from careful right-hand rule application to each wire independently.
13. Which statement correctly models the induced current and the effect of increasing the magnet's speed?
    Answer: The current is counterclockwise, making the near face north; greater speed produces a larger current. Whenever you see a question about electromagnetic induction, your two-step framework is: (1) determine the direction of changing flux, then (2) apply Lenz's Law — the induced current will oppose that change. Here, the north pole approaches the loop, so the magnetic flux through the loop is increasing in the direction pointing toward the observer (away from the magnet). By Lenz's Law, the induced current must create a magnetic field that opposes this increase — meaning the induced field must point away from the observer (back toward the magnet). Using the right-hand rule, a field pointing away from you requires a counterclockwise current as seen from the observer's side. A counterclockwise current, viewed face-on, creates a north pole on the near face — which repels the approaching magnet, opposing the motion exactly as Lenz's Law demands. Now, when the magnet moves faster, the flux changes more rapidly. Since $\mathcal{E} = -\frac{d\Phi_B}{dt}$, a greater rate of flux change produces a larger induced EMF and therefore a larger induced current. This confirms answer D. Choice A is wrong twice: a clockwise current would create a south pole on the near face, and it incorrectly claims speed doesn't affect the current. Choice B correctly identifies a larger current with greater speed but gets the current direction and pole wrong — clockwise would attract the magnet, violating Lenz's Law. Choice C correctly identifies the counterclockwise direction and north pole, but claims greater speed produces a smaller current, which contradicts Faraday's law. Your study tip: always pair Lenz's Law with Faraday's law. Lenz's Law gives direction; Faraday's law gives magnitude. Faster motion → larger $|d\Phi/dt|$ → larger current, every time.
14. Which statement correctly predicts the droplet's motion and energy changes before it reaches a plate?
    Answer: It accelerates upward; its electric potential energy decreases while its kinetic energy increases. Whenever you see a charged particle in a uniform electric field, your first job is to identify the direction of the electric force — which depends on both the field direction and the sign of the charge. In a parallel-plate setup, the electric field points from high potential to low potential — here, downward (from the positive upper plate to the negative lower plate). For a negative charge, the electric force is opposite to the field direction, so the droplet experiences a force upward. Released from rest, it accelerates upward toward the upper (higher-potential) plate. Now for the energy changes. As the negative droplet moves toward higher electric potential, its electric potential energy actually decreases. This might feel counterintuitive, but remember: $U = qV$. Since q is negative, moving to higher V means U becomes more negative — a decrease. By conservation of energy (no gravity, no friction), this lost potential energy converts entirely into kinetic energy, so kinetic energy increases. That makes C the correct answer. A is wrong because it claims both energies increase — this violates conservation of energy. You can't gain kinetic energy without losing potential energy in a closed system. B is wrong about the direction. The field points down, but a negative charge is pushed opposite to the field — upward, not downward. D is a classic trap. Being equidistant from the plates has nothing to do with field strength. The field is uniform and nonzero everywhere between the plates. Study tip: For charged particles, always apply $F = qE$ carefully — the sign of q flips the force direction, and $U = qV$ means negative charges behave "backwards" on potential energy diagrams.

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