Investigate Current and Magnetic Fields - Grade 11 Physics

1. How does the magnetic field inside solenoid 2 compare with that inside solenoid 1?
   Answer: It has $0.75$ times the magnitude and the opposite axial direction. When comparing magnetic fields in solenoids, you need to evaluate two things separately: the magnitude ratio and the direction. The formula $B = \mu_0 \frac{N}{L} I$ tells you that field strength depends on turn density ($N/L$) and current. Calculate each field's magnitude. For solenoid 1: $\frac{N}{L} = \frac{600}{0.30} = 2000 \text{ turns/m}$, so $B_1 = \mu_0 (2000)(0.40) = 800\mu_0$. For solenoid 2: $\frac{N}{L} = \frac{900}{0.45} = 2000 \text{ turns/m}$, so $B_2 = \mu_0 (2000)(0.30) = 600\mu_0$. The ratio is $\frac{B_2}{B_1} = \frac{600}{800} = 0.75$. Now for direction: since solenoid 2's current runs opposite to solenoid 1's (when viewed from the same end), the right-hand rule gives the opposite axial direction. This confirms answer B. Choice A gets the direction right as "same" — which is incorrect — and also states 0.50 times the magnitude, which would only be true if the currents were in a 1:2 ratio or turn densities differed. Choice C arrives at the correct direction (opposite) but calculates 1.25 times the magnitude — this likely comes from inverting the ratio ($800/600$) instead of computing $B_2/B_1$. Choice D inverts both the magnitude ratio and gets the direction wrong, suggesting a systematic confusion about which solenoid is being compared to which. A useful strategy: always define your ratio as $B_2 / B_1$ explicitly before calculating, and treat magnitude and direction as two completely independent steps — they're tested separately here, and getting one right doesn't guarantee the other.
2. What current in loop 2 makes the net magnetic field at the common center equal to zero?
   Answer: $4.0\,\text{A}$ counterclockwise, because its radius is twice as large Whenever you see two current loops sharing a center, think about superposition of magnetic fields: each loop independently creates a field at the center, and you need them to cancel. The formula given, $B = \mu_0 I / (2R)$, tells you everything — field strength depends on both current and radius. For Loop 1: $B_1 = \mu_0 (2.0) / (2 \times 0.10) = 10\mu_0$. Loop 1 is clockwise, so by the right-hand rule its field points into the page at the center. For the net field to be zero, Loop 2 must produce an equal field pointing out of the page — which a counterclockwise current does. Setting $B_2 = B_1$: $\frac{\mu_0 I_2}{2(0.20)} = 10\mu_0 \implies I_2 = \frac{10\mu_0 \times 0.40}{\mu_0} = 4.0\,\text{A}$ Choice C is correct. Because Loop 2 has twice the radius, it's less efficient at producing field per amp, so you need twice the current (4.0 A) to match Loop 1's field. Choice A (1.0 A) makes the opposite error — it assumes a larger radius needs less current, perhaps confusing "larger loop" with "stronger field." Choice B (2.0 A) ignores radius entirely, treating the loops as equivalent just because equal currents were mentioned. Choice D (8.0 A) misapplies an inverse-square law — that's for point charges, not magnetic fields at a loop's center, where field scales as $1/R$, not $1/R^2$. Your study tip: always write out $B = \mu_0 I / (2R)$ for each loop and set them equal. Don't reason qualitatively — the math prevents you from mixing up $1/R$ and $1/R^2$.
3. How does the force between the wires change?
   Answer: It remains equal in magnitude and becomes repulsive. Whenever you see a question about forces between current-carrying wires, focus on two things: the magnitude (using the proportionality formula) and the direction rule (same-direction currents attract; opposite-direction currents repel). The force per unit length is proportional to $\frac{I_1 I_2}{d}$. In the original setup, the force is proportional to $\frac{I_1 I_2}{d}$. After the changes, wire 2's current becomes $\frac{I_2}{2}$ and the separation becomes $\frac{d}{2}$, so the new force is proportional to: $\frac{I_1 \cdot \frac{I_2}{2}}{\frac{d}{2}} = \frac{I_1 I_2}{2} \cdot \frac{2}{d} = \frac{I_1 I_2}{d}$ The magnitude is unchanged. Now for direction: reversing wire 2's current means the currents now flow in opposite directions, which makes the force repulsive. So the correct answer is D — equal magnitude, now repulsive. Choice A is wrong on both counts: the magnitude doesn't double, and the force doesn't stay attractive after the current reversal. Choice B gets the magnitude right but ignores the direction reversal — a classic trap of doing the math correctly while forgetting the physics of attraction vs. repulsion. Choice C correctly identifies the repulsive nature but makes an arithmetic error, perhaps forgetting that halving both the current and the distance causes the effects to cancel, not reduce the force by half. A quick study tip: always treat magnitude and direction as two separate checks. Do the algebra first, then ask yourself whether the currents are parallel or antiparallel to determine attraction or repulsion.
4. At which position on the east–west line through the wires is the net magnetic field zero?
   Answer: $0.15\,\text{m}$ west of wire A When two wires carry currents in opposite directions, their magnetic fields point in the same direction between the wires, so the fields can only cancel outside the pair. This immediately rules out any point between the wires ($x = 0$ to $x = 0.30\,\text{m}$). Next, you need to decide which side. The cancellation point must be closer to the weaker wire (Wire A, 2.0 A) because a weaker source needs less distance to match the stronger one. That means the zero-field point lies west of Wire A ($x < 0$), not east of Wire B. Setting the field magnitudes equal at distance $r_A$ from Wire A and $r_B$ from Wire B: $\frac{I_A}{r_A} = \frac{I_B}{r_B} \implies \frac{2.0}{r_A} = \frac{6.0}{r_B}$ Since the point is west of Wire A, $r_B = r_A + 0.30$. Substituting: $2.0(r_A + 0.30) = 6.0\,r_A \implies 0.60 = 4.0\,r_A \implies r_A = 0.15\,\text{m}$ So the zero-field point is 0.15 m west of Wire A — confirming answer A. Choice B places the point between the wires, where opposite-direction currents produce fields that add, not cancel. Choice C is also between the wires for the same reason. Choice D (0.45 m east, west of Wire B's far side) is on the wrong side entirely — there the weaker wire's field can never match the stronger one at any distance. Strategy tip: For antiparallel currents, always reason first about which side the null point must be on before solving numerically — elimination saves time and prevents algebraic mistakes.
5. Which statement correctly describes the initial magnetic forces on the two particles?
   Answer: Both forces point upward because reversing both charge and velocity preserves the force direction. When a charged particle moves through a magnetic field, the force it experiences depends on three things together: the charge sign, the velocity direction, and the field direction — via $\vec{F} = q\vec{v} \times \vec{B}$. Start with the proton. Its velocity points east ($+\hat{x}$) and the field points north ($+\hat{y}$). The cross product $\hat{x} \times \hat{y} = +\hat{z}$, which is upward. Since the proton's charge is positive, its force points upward. Now the electron. Its velocity points west ($-\hat{x}$) and the field is still north ($+\hat{y}$). The cross product gives $(-\hat{x}) \times \hat{y} = -\hat{z}$, pointing downward. But the electron carries negative charge, so multiplying by $q = -e$ flips the sign once more: $(-e)(-\hat{z}) = +\hat{z}$, which is upward. Reversing both the charge and the velocity each contribute a sign flip, and two negatives cancel — both particles experience an upward force. Answer A is correct. Answer B is wrong because it incorrectly claims the electron's negative charge alone controls both force directions, ignoring that the electron's velocity is also reversed relative to the proton's. Answer C incorrectly applies the proton's cross product result directly to the electron without accounting for the negative charge's sign reversal. Answer D gets the proton's direction backward entirely. A reliable strategy: treat the charge sign and velocity direction as separate factors in $q(\vec{v} \times \vec{B})$. When both flip simultaneously, their effects cancel — always track each piece independently before combining.
6. Which follow-up investigation would best test whether increasing current alone caused the stronger magnetic effect?
   Answer: Keep the turns and core fixed, vary the current, and measure the field at one fixed position. Whenever you see a question about electromagnets and magnetic field strength, the key concept is controlled variable testing — to isolate one cause, you must change only that variable while holding everything else constant. In the original experiment, both the number of turns and the current increased simultaneously (more coils + more batteries). Because two variables changed at once, you cannot determine which one — or whether both — caused the stronger effect. The follow-up investigation needs to isolate current as the single independent variable. Choice A is correct because it keeps the number of turns and the core material fixed, changes only the current (the independent variable), and measures the magnetic field at one consistent position. Every condition except current is controlled, so any change in magnetic effect can be attributed directly to the change in current. This is a properly designed controlled experiment. Choice B fails because it varies the number of turns rather than the current — this would test whether turns cause the effect, not current. It also measures the field at "several changing positions," introducing a second uncontrolled variable (measurement location), making results unreliable. Choice C increases both current and turns again, repeating the same flaw as the original experiment. You still cannot separate the individual contributions of each variable, so it answers nothing new. Choice D changes the core material (iron → steel), which introduces yet another variable entirely. This tests core material, not current, and is irrelevant to the follow-up question. Study tip: On questions about experimental design, immediately ask yourself, "How many variables are changing?" A valid test of one factor must change exactly that factor and freeze everything else.
7. What is the torque on the replacement coil?
   Answer: $0.087\,\text{N}\cdot\text{m}$, because the turn–current product is unchanged and the normal is $60^\circ$ from the field Whenever you see a magnetic torque problem, your first job is to correctly identify the angle θ in the formula $\tau = NIAB\sin\theta$. The critical detail: θ is measured from the field to the coil's normal, not to the plane of the coil. These two angles are complementary — if the plane makes 30° with the field, the normal makes 90° − 30° = 60° with the field. For the replacement coil: $N = 200$, $I = 0.10\,\text{A}$, $A = 0.010\,\text{m}^2$, $B = 0.50\,\text{T}$, and $\theta = 60°$. Plugging in: $\tau = (200)(0.10)(0.010)(0.50)\sin 60° = 0.10 \times \frac{\sqrt{3}}{2} \approx \mathbf{0.087\,\text{N}\cdot\text{m}}$ This confirms C is correct. Notice also that the turn–current product $NI = 200 \times 0.10 = 20$, identical to the original coil ($100 \times 0.20 = 20$), so any torque difference comes entirely from the angle. A is wrong because it ignores the angle change entirely and simply halves the original torque, treating the geometry as unchanged. B uses 30° directly in the formula — a classic trap. $\sin 30° = 0.50$, giving 0.050 N·m, but 30° is the plane angle, not the angle between the normal and the field. D correctly notices the equal turn–current product but wrongly assumes the torque equals the original (which used $\sin 90° = 1$ since the plane was parallel to the field), ignoring that the new angle changes the sine factor. Study tip: Always sketch the coil and field, label the normal vector, and ask yourself: "Is my angle measured from the normal or the plane?" That one habit will save you from the most common trap in torque problems.
8. If the current is doubled and the wire is rotated until it is perpendicular to the field, what is the new magnetic-force magnitude?
   Answer: $0.60\,\text{N}$, because doubling the current and changing $\sin30^\circ$ to $\sin90^\circ$ each contribute a factor of two Whenever you see a question about magnetic force on a current-carrying wire, your instinct should be to write out the full formula — $F = BIL\sin\theta$ — and track every variable that changes, not just the most obvious one. Here, both the current and the angle change simultaneously. The original force is $F_0 = (0.40)(3.0)(0.25)\sin30° = (0.40)(3.0)(0.25)(0.5) = 0.15\,\text{N}$. Now the current doubles to $6.0\,\text{A}$, and the wire rotates to $90°$, where $\sin90° = 1$. The new force is $F = (0.40)(6.0)(0.25)(1) = 0.60\,\text{N}$. Notice that doubling the current multiplied the force by 2, and replacing $\sin30° = 0.5$ with $\sin90° = 1$ also multiplied the force by 2 — giving an overall factor of 4 from the original $0.15\,\text{N}$. This confirms C is correct. Choice A ignores both changes entirely, freezing the situation at the original $0.15\,\text{N}$ — a trap for students who don't re-evaluate after the problem states conditions have changed. Choice B accounts for the doubled current but forgets that rotating the wire to perpendicular changes $\sin\theta$ dramatically — from $0.5$ to $1$, a factor of two. Choice D overcounts, suggesting a third doubling that doesn't exist in the formula. Your study tip: whenever a problem changes multiple variables, substitute the new values into the formula fresh rather than trying to reason in percentages or multipliers mentally — you're far less likely to miss a factor.
9. What is the magnitude of the average induced current in the coil?
   Answer: $0.50\,\text{A}$, using the full change in signed magnetic flux When a coil experiences a changing magnetic flux, Faraday's Law tells you the induced emf depends on three things: the number of turns $N$, the total change in flux $\Delta\Phi$, and the time interval $\Delta t$. The key phrase is "total change in signed flux" — you must subtract the final value from the initial value, respecting the signs. Here, $\Delta\Phi = \Phi_f - \Phi_i = (-1.0\times10^{-4}) - (+3.0\times10^{-4}) = -4.0\times10^{-4}\,\text{Wb}$. The magnitude is $4.0\times10^{-4}\,\text{Wb}$. Plugging into Faraday's Law: $|\varepsilon| = N\frac{|\Delta\Phi|}{\Delta t} = 200 \times \frac{4.0\times10^{-4}}{0.020} = 4.0\,\text{V}$. The induced current is then $I = \varepsilon/R = 4.0/8.0 = 0.50\,\text{A}$, confirming B is correct. Choice A is a common trap: it only accounts for the flux dropping from $+3.0\times10^{-4}$ to zero (half the actual change), giving $|\Delta\Phi| = 3.0\times10^{-4}\,\text{Wb}$ and $I = 0.25\,\text{A}$. This ignores the flux continuing to reverse sign. Choice C forgets to multiply by $N = 200$, massively underestimating the emf and yielding $2.0\,\text{A}$ through faulty arithmetic. Choice D skips Ohm's Law entirely — $4.0\,\text{V}$ is the induced emf, not the current; dividing by resistance is a required step. Study tip: When flux changes sign, always calculate $\Delta\Phi$ using both endpoints with their signs — the reversal doubles the effective change. Then remember the two-step process: Faraday's Law gives emf, Ohm's Law gives current.
10. Approximately where will the compass point after the current is changed?
    Answer: $56^\circ$ west of north, because the wire's field reverses and doubles When a compass sits in two perpendicular magnetic fields, it aligns with their vector sum — so you need to track both the magnitude and direction of every field component carefully. Initially, Earth contributes $40\,\mu\text{T}$ north and the wire contributes $30\,\mu\text{T}$ east. After the current is reversed and doubled, the wire's field flips direction (east → west) and its magnitude doubles: $2 \times 30 = 60\,\mu\text{T}$ west. Now the two fields are $40\,\mu\text{T}$ north and $60\,\mu\text{T}$ west. The compass angle west of north is: $\theta = \arctan\!\left(\frac{60}{40}\right) = \arctan(1.5) \approx 56^\circ \text{ west of north}$ That confirms D is correct. A is wrong because it claims reversing the current doesn't change field direction — it absolutely does. By the right-hand rule, flipping current direction flips the magnetic field direction. B correctly identifies that the direction reverses (east → west), but ignores that the current also doubles the field magnitude. Using the original $30\,\mu\text{T}$ gives $\arctan(30/40) \approx 37^\circ$, which is outdated once the current doubles. C correctly doubles the wire's field but keeps the direction east, ignoring the reversal. Doubling without reversing would give $56^\circ$ east — the right angle, wrong direction. Study tip: On vector-sum compass problems, always update both magnitude and direction before computing your angle. Changing current simultaneously changes both, and exam distractors are specifically designed to trap students who only handle one change at a time.

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