Understand Electric & Magnetic Fields - IB Physics

1. A conducting rod of length $L$ rotates about one end with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation. What is the potential difference between the ends of the rod, and which end is at higher potential if the rod carries positive charge carriers?
   Answer: $\frac{1}{2}B\omega L^2$, with the free end at higher potential When you encounter a rotating conductor in a magnetic field, you're dealing with motional EMF caused by the magnetic force on moving charge carriers. The key insight is that different parts of the rod move at different speeds, creating a varying force distribution. As the rod rotates, each point at distance $r$ from the pivot moves with velocity $v = \omega r$. When positive charges move through the magnetic field $B$, they experience a magnetic force $F = qvB = q\omega rB$ directed radially outward (using the right-hand rule). This force pushes positive charges toward the free end of the rod. To find the total potential difference, you must integrate the electric field created by this charge separation. The electric field at distance $r$ is $E = \omega rB$, and the potential difference is: $V = \int_0^L E \, dr = \int_0^L \omega rB \, dr = \omega B \int_0^L r \, dr = \omega B \cdot \frac{L^2}{2} = \frac{1}{2}B\omega L^2$ Since positive charges accumulate at the free end, it becomes positively charged and at higher potential. Option A uses the wrong coefficient (missing the factor of $\frac{1}{2}$ from integration) but correctly identifies the higher potential end. Option B has the correct magnitude but incorrectly places higher potential at the pivot end. Option D combines both errors—wrong coefficient and wrong polarity. Remember: in motional EMF problems involving rotation, always integrate over the varying velocity, and use the right-hand rule to determine which end accumulates positive charge.
2. Three identical positive point charges, $+q$, are brought from infinity and fixed at the vertices of an equilateral triangle of side length $r$. What is the total electric potential energy of this system of three charges?
   Answer: $k\frac{3q^2}{r}$ The total electric potential energy of a system of charges is the sum of the potential energies for every pair of charges in the system. For three charges, there are three pairs: (1,2), (1,3), and (2,3). The potential energy of a single pair of charges $q$ separated by distance $r$ is $U = k\frac{q^2}{r}$. Since all charges are identical and all distances are $r$, the energy for each of the three pairs is the same. Therefore, the total potential energy is $U_{total} = U_{12} + U_{13} + U_{23} = 3 \times \left(k\frac{q^2}{r}\right) = k\frac{3q^2}{r}$.
3. Four point charges are fixed at the vertices of a square of side length $s$. Charges $+Q$ are at two diagonally opposite vertices, and charges $-Q$ are at the other two. What is the magnitude of the net electric field at the geometric center of the square?
   Answer: $\frac{4\sqrt{2}kQ}{s^2}$ The distance from each corner to the center is $r = s/\sqrt{2}$, so $r^2 = s^2/2$. The electric field from one $+Q$ charge and the diagonally opposite $-Q$ charge both point in the same direction (along the diagonal). Their combined magnitude is $E_{diag} = k\frac{Q}{r^2} + k\frac{Q}{r^2} = \frac{2kQ}{s^2/2} = \frac{4kQ}{s^2}$. The fields from the other pair of charges also combine to a magnitude of $\frac{4kQ}{s^2}$ along the other diagonal. These two resultant vectors are perpendicular. The net field is the vector sum, with magnitude $E_{net} = \sqrt{E_{diag}^2 + E_{diag}^2} = \sqrt{2} E_{diag} = \sqrt{2} \frac{4kQ}{s^2} = \frac{4\sqrt{2}kQ}{s^2}$.
4. Two large, parallel conducting plates are separated by a distance $d$ and connected to a battery of potential difference $V$. An electron with charge $-e$ is located midway between the plates. What is the magnitude of the electrostatic force on the electron?
   Answer: $eV/d$ The electric field $E$ between two large parallel plates is uniform and is given by $E = V/d$. The force $F$ on a charge $q$ in an electric field $E$ is given by $F = qE$. For an electron, the charge magnitude is $e$. Therefore, the magnitude of the force is $F = eE = e(V/d) = eV/d$. The force is constant at all points between the plates, including the midway point.
5. Two concentric spherical conducting shells have radii $R$ and $2R$. The inner shell carries charge $+Q$ and the outer shell carries charge $-Q$. A point charge $+q$ is placed at distance $3R$ from the center. What is the magnitude of the electric field at distance $1.5R$ from the center?
   Answer: $\frac{kQ}{(1.5R)^2}$ When analyzing electric fields with conductors, you need to understand how charges distribute and how conductors shield electric fields. This question tests your knowledge of electrostatic shielding and charge distribution on conducting surfaces. The key insight is that electric fields inside conductors are always zero, and charges on conductors reside only on surfaces. When the inner shell carries charge +Q, this charge distributes uniformly on its outer surface. The outer shell carries -Q total charge, but this charge redistributes due to the external point charge +q at distance 3R. At the point 1.5R from center, you're between the two conducting shells. Here, only the charge on the inner conductor (+Q) contributes to the electric field, because you're outside the inner shell but inside the outer conductor. The outer shell's charge doesn't create any field in the region between the shells - this is a consequence of electrostatic shielding. The external point charge +q also doesn't affect the field here because you're "shielded" by the outer conductor. Using Gauss's law, the electric field magnitude is $\frac{kQ}{(1.5R)^2}$. Choice A incorrectly assumes you're inside a conductor (you're between conductors). Choice B wrongly includes both the inner charge and external point charge, ignoring shielding effects. Choice C incorrectly suggests only the external charge matters, when it's actually shielded out. Remember: conductors create "Faraday cages" - charges outside a conductor don't create fields inside it, and you must carefully identify which region you're analyzing.
6. A charged particle enters a region with crossed electric and magnetic fields ($\vec{E} \perp \vec{B}$) and travels in a straight line. If the electric field strength is doubled while keeping the magnetic field constant, what type of motion will the particle exhibit?
   Answer: Helical motion with both circular and linear velocity components Initially, for straight-line motion: $qE = qvB$, so $v = E/B$. When $E$ is doubled: the electric force becomes $2qE$ while the magnetic force remains $qvB$. The net force is no longer zero. The component of electric force parallel to the initial velocity will accelerate the particle in that direction, while the perpendicular component of the net force will cause circular motion. This results in helical motion. Choice A assumes purely circular motion (ignoring the parallel acceleration). Choice C treats it like projectile motion (ignoring the magnetic force). Choice D assumes equilibrium is re-established at a different velocity.
7. In a uniform electric field $\vec{E}$, two point charges $+2q$ and $-q$ are held fixed at positions that form an equilateral triangle with a third point P. If the electric field due to the two charges at point P has the same magnitude as the external field $\vec{E}$ but points in the opposite direction, what is the net electric field at point P?
   Answer: Zero, because all field contributions cancel exactly The problem states that the electric field due to the two charges at point P has the same magnitude as the external field but points in the opposite direction. Therefore, $\vec{E}_{charges} = -\vec{E}$. The net field is $\vec{E}_{net} = \vec{E} + \vec{E}_{charges} = \vec{E} + (-\vec{E}) = 0$. The fields cancel exactly. Choice B incorrectly adds the magnitudes. Choice C assumes incomplete cancellation. Choice D incorrectly assumes the fields add rather than cancel.
8. An equipotential surface is a surface of constant electric potential. Which statement correctly describes the work done on a charge and the orientation of the electric field as the charge is moved along such a surface?
   Answer: The work done is zero, and the electric field is perpendicular to the surface. The work done in moving a charge $q$ between two points is $W = q \Delta V$. Since the potential $V$ is constant everywhere on an equipotential surface, the potential difference $\Delta V$ for any movement along the surface is zero. Therefore, the work done is zero. If work is zero for displacement, the force component along that displacement must be zero. This means the electric force, and thus the electric field, must be perpendicular to the equipotential surface at all points.
9. Three point charges are at the vertices of a right-angled isosceles triangle. A charge of $+Q$ is at (0, a), another $+Q$ is at (a, 0), and a charge of $-2Q$ is at the origin (0, 0). What is the magnitude of the net electric force on the charge at the origin?
   Answer: $\frac{2\sqrt{2}kQ^2}{a^2}$ The force on the $-2Q$ charge from the $+Q$ at (0, a) is attractive, with magnitude $F_y = k\frac{|(+Q)(-2Q)|}{a^2} = \frac{2kQ^2}{a^2}$ in the +y direction. The force from the $+Q$ at (a, 0) is also attractive, with magnitude $F_x = k\frac{|(+Q)(-2Q)|}{a^2} = \frac{2kQ^2}{a^2}$ in the +x direction. The net force is the vector sum of these two perpendicular forces. The magnitude is found using the Pythagorean theorem: $F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{(\frac{2kQ^2}{a^2})^2 + (\frac{2kQ^2}{a^2})^2} = \sqrt{2(\frac{2kQ^2}{a^2})^2} = \sqrt{2} \frac{2kQ^2}{a^2} = \frac{2\sqrt{2}kQ^2}{a^2}$.
10. A solenoid with $n$ turns per unit length carries current $I$. A small circular loop of radius $r$ and resistance $R$ is placed coaxially inside the solenoid. If the current in the solenoid decreases linearly from $I$ to zero in time $t$, what is the magnitude of the induced current in the loop?
    Answer: $\frac{\mu_0 n \pi r^2 I}{Rt}$ When you encounter electromagnetic induction problems involving solenoids and loops, you need to apply Faraday's law: the induced EMF equals the negative rate of change of magnetic flux through the loop. Start by finding the magnetic field inside the solenoid. For a solenoid with $n$ turns per unit length carrying current $I$, the magnetic field is $B = \mu_0 n I$. This field is uniform and parallel to the solenoid's axis. Next, calculate the magnetic flux through the circular loop: $\Phi = B \cdot A = \mu_0 n I \cdot \pi r^2$. Since the current decreases linearly from $I$ to zero in time $t$, the rate of change is $\frac{dI}{dt} = -\frac{I}{t}$. Using Faraday's law: $|\mathcal{E}| = \left|\frac{d\Phi}{dt}\right| = \left|\frac{d}{dt}(\mu_0 n I \pi r^2)\right| = \mu_0 n \pi r^2 \left|\frac{dI}{dt}\right| = \frac{\mu_0 n \pi r^2 I}{t}$ Finally, apply Ohm's law to find the induced current: $i = \frac{|\mathcal{E}|}{R} = \frac{\mu_0 n \pi r^2 I}{Rt}$, which is answer C. Let's examine the wrong answers: A includes an extra factor of $n$, suggesting confusion about the solenoid field formula. B omits $\pi r$ from the area calculation, treating the loop as if it had area $r^2$ instead of $\pi r^2$. D incorrectly places $n$ in the denominator, fundamentally misunderstanding how solenoid field strength depends on turn density. Remember: solenoid problems always involve $B = \mu_0 n I$, and circular areas always include $\pi r^2$. Double-check these fundamental relationships in your calculations.
11. At a point P, the electric field is directed due east. If a small negative test charge is placed at P, in which direction will the electrostatic force on it act?
    Answer: West By definition, the direction of the electric field at a point is the direction of the force that would be exerted on a positive test charge placed at that point. The force on a negative charge ($F = qE$) is in the opposite direction to the electric field because the charge $q$ is negative. Since the field is directed east, the force on a negative charge will be directed west.
12. The Millikan oil drop experiment, which balanced gravitational and electric forces on tiny charged oil drops, provided the first conclusive experimental evidence for which fundamental concept?
    Answer: The quantization of electric charge. The most significant conclusion from Millikan's experiment was that the electric charge on the oil drops was always an integer multiple of a fundamental unit of charge, $e \approx 1.6 \times 10^{-19} \, C$. This demonstrated that electric charge is quantized, meaning it exists in discrete packets rather than as a continuous quantity.
13. Which statement provides the best physical reason why two electric field lines can never cross?
    Answer: The electric field vector at any point must have a unique direction. The tangent to an electric field line at any point gives the direction of the electric field vector at that point. The electric field, which represents the net force per unit charge, can only have one resultant direction at any single point in space. If two lines crossed, it would imply two different directions for the electric field at the intersection point, which is physically impossible.
14. A proton is accelerated from rest by a uniform electric field between two points with a potential difference of 500 V. What is the kinetic energy gained by the proton in joules? (elementary charge $e = 1.60 \times 10^{-19} \, C$)
    Answer: $8.00 \times 10^{-17} \, J$ The work done on a charge $q$ when it moves through a potential difference $V$ is given by $W = qV$. By the work-energy theorem, this work is equal to the change in kinetic energy of the charge. The charge of a proton is $+e$. Therefore, $\Delta E_k = W = eV = (1.60 \times 10^{-19} \, C)(500 \, V) = 8.00 \times 10^{-17} \, J$.
15. In a certain region of space, the electric potential is found to increase uniformly in the positive x-direction. What can be deduced about the direction of the electric field in this region?
    Answer: It points in the negative x-direction. The electric field vector points in the direction of the steepest decrease in electric potential. The relationship in one dimension is $E_x = -dV/dx$. Since the potential $V$ increases uniformly in the positive x-direction, its derivative $dV/dx$ is a positive constant. Therefore, $E_x$ must be a negative constant, meaning the electric field points in the negative x-direction.
16. A rectangular loop of wire with dimensions $a \times b$ moves with constant velocity $v$ perpendicular to a uniform magnetic field $B$. The loop is initially outside the field region and enters the field completely before exiting on the other side. Which statement correctly describes the induced EMF during the three phases of motion?
    Answer: EMF is $Bav$ while entering, zero while completely inside, and $Bav$ while exiting, but opposite directions for entering and exiting Using Faraday's law, EMF = $-\frac{d\Phi}{dt}$. While entering: flux increases at rate $Bav$ (assuming the loop enters along dimension $a$), so EMF = $Bav$. While completely inside: flux is constant, so EMF = 0. While exiting: flux decreases at rate $Bav$, so EMF = $Bav$ but in opposite direction to the entering phase due to Lenz's law. Choice A ignores the direction change. Choice C uses wrong dimension and assumes EMF while completely inside. Choice D incorrectly calculates EMF while completely inside.
17. A particle has a net charge of $-2.4 \times 10^{-18} \, C$. Given that the elementary charge is $e = 1.6 \times 10^{-19} \, C$, what is the number of excess electrons on the particle?
    Answer: 15 The total charge $Q$ is the number of excess electrons $n$ multiplied by the charge of a single electron ($-e$). So, $Q = n(-e)$. The number of excess electrons is $n = Q/(-e) = (-2.4 \times 10^{-18} \, C) / (-1.6 \times 10^{-19} \, C) = 15$.
18. Two parallel infinite conducting plates are separated by distance $d$ and maintain a potential difference $V$. A small charged particle with charge $+q$ and mass $m$ is released from rest at the positive plate. When the particle reaches the negative plate, what is the ratio of its final kinetic energy to the work done by the electric field if air resistance causes the particle to lose 25% of the energy gained from the field?
    Answer: $\frac{3qV}{4qV} = \frac{3}{4}$ The work done by the electric field is $W_{field} = qV$ as the particle moves from high to low potential. If air resistance causes the particle to lose 25% of this energy, then the final kinetic energy is $KE_{final} = W_{field} - 0.25W_{field} = 0.75W_{field} = 0.75qV$. Therefore, the ratio is $\frac{KE_{final}}{W_{field}} = \frac{0.75qV}{qV} = \frac{3}{4}$. Choice B incorrectly assumes the final KE is 25% of the work done. Choice C inverts the ratio. Choice D assumes the final KE is three times the work done, which violates energy conservation.
19. A point charge of $+4Q$ is located at $x=0$ and a point charge of $-Q$ is located at $x=L$. At which point on the x-axis, other than at infinity, is the net electric field equal to zero?
    Answer: $x = 2L$ The electric field can only be zero in a region where the fields from the two charges oppose each other. This occurs on the x-axis outside the charges, on the side of the smaller magnitude charge ($-Q$). Let this point be at position $x > L$. The field from $+4Q$ is $E_{4Q} = k\frac{4Q}{x^2}$ and the field from $-Q$ is $E_{-Q} = k\frac{Q}{(x-L)^2}$. Setting the magnitudes equal: $\frac{4Q}{x^2} = \frac{Q}{(x-L)^2}$. This simplifies to $4(x-L)^2 = x^2$, so $2(x-L) = \pm x$. The solution $2x-2L=x$ gives $x=2L$, which is valid. The solution $2x-2L=-x$ gives $x=2L/3$, which is between the charges where the fields add, so it is not a solution.
20. A positive charge $+Q$ and a negative charge $-Q$ are separated by a distance $d$. The magnitude of the attractive force between them is $F$. A third charge, also with positive charge $+Q$, is now placed exactly halfway between them. What is the new magnitude of the net electrostatic force on the original positive charge $+Q$?
    Answer: 3F The original force on the $+Q$ charge from the $-Q$ charge is $F = kQ^2/d^2$, and it is attractive (directed towards $-Q$). The new charge $+Q$ is placed at a distance of $d/2$ from the original $+Q$ charge. It exerts a repulsive force on the original $+Q$ charge, directed away from $-Q$. The magnitude of this new force is $F_{new} = kQ^2/(d/2)^2 = 4kQ^2/d^2 = 4F$. Since the original attractive force and the new repulsive force are in opposite directions, the net force is the difference between their magnitudes: $F_{net} = F_{new} - F = 4F - F = 3F$.

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