College Physics Quiz: Magnetic Fields
20 questions · exam conditions
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Magnetic FieldsQuestion 1 of 20

A proton moving with velocity v⃗=2.0×106 m/s\vec{v} = 2.0 \times 10^6 \text{ m/s} in the +x+x direction enters a uniform magnetic field B⃗=0.50 T\vec{B} = 0.50 \text{ T} in the +z+z direction. What is the magnitude of the magnetic force on the proton? (qp=1.6×10−19 Cq_p = 1.6 \times 10^{-19} \text{ C})

1.6×10−13 N1.6 \times 10^{-13} \text{ N}
3.2×10−13 N3.2 \times 10^{-13} \text{ N}
6.4×10−14 N6.4 \times 10^{-14} \text{ N}
8.0×10−14 N8.0 \times 10^{-14} \text{ N}
Zero, because the velocity and field are perpendicular
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College Physics Quiz

College Physics Quiz: Magnetic Fields

Practice Magnetic Fields in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Magnetic Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A proton moving with velocity v⃗=2.0×106 m/s\vec{v} = 2.0 \times 10^6 \text{ m/s} in the +x+x direction enters a uniform magnetic field B⃗=0.50 T\vec{B} = 0.50 \text{ T} in the +z+z direction. What is the magnitude of the magnetic force on the proton? (qp=1.6×10−19 Cq_p = 1.6 \times 10^{-19} \text{ C})

  1. 1.6×10−13 N1.6 \times 10^{-13} \text{ N} (correct answer)
  2. 3.2×10−13 N3.2 \times 10^{-13} \text{ N}
  3. 6.4×10−14 N6.4 \times 10^{-14} \text{ N}
  4. 8.0×10−14 N8.0 \times 10^{-14} \text{ N}
  5. Zero, because the velocity and field are perpendicular
Explanation: When a charged particle moves through a magnetic field, it experiences a magnetic force given by the Lorentz force law: F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B}. The magnitude of this force is F=qvBsin⁡θF = qvB\sin\theta, where θ\theta is the angle between the velocity and magnetic field vectors. In this problem, the proton moves in the +x+x direction while the magnetic field points in the +z+z direction. These directions are perpendicular to each other, so θ=90°\theta = 90° and sin⁡θ=1\sin\theta = 1. This simplifies our calculation to F=qvBF = qvB. Substituting the given values: F=(1.6×10−19 C)(2.0×106 m/s)(0.50 T)F = (1.6 \times 10^{-19} \text{ C})(2.0 \times 10^6 \text{ m/s})(0.50 \text{ T}). Working through the calculation: F=1.6×10−19×2.0×106×0.50=1.6×10−13 NF = 1.6 \times 10^{-19} \times 2.0 \times 10^6 \times 0.50 = 1.6 \times 10^{-13} \text{ N}. This confirms answer A is correct. Looking at the wrong answers: B (3.2×10−13 N3.2 \times 10^{-13} \text{ N}) represents doubling the correct value, possibly from using 2qvB2qvB instead of qvBqvB. C (6.4×10−14 N6.4 \times 10^{-14} \text{ N}) is exactly 2.5 times smaller than the correct answer, suggesting an error in decimal placement or coefficient manipulation. D (8.0×10−14 N8.0 \times 10^{-14} \text{ N}) is half the correct value, which could result from using 0.25 T0.25 \text{ T} instead of 0.50 T0.50 \text{ T} for the magnetic field. Remember: when velocity and magnetic field are perpendicular (common in physics problems), the magnetic force simplifies to F=qvBF = qvB. Always check that your vectors are truly perpendicular before using this shortcut.

Question 2

A wire carrying current I=5.0 AI = 5.0 \text{ A} in the +y+y direction is placed in a uniform magnetic field B⃗=0.30 T\vec{B} = 0.30 \text{ T} in the −z-z direction. If the length of wire in the field is L=0.25 mL = 0.25 \text{ m}, what is the direction of the magnetic force on the wire?

  1. +x+x direction
  2. −x-x direction (correct answer)
  3. +y+y direction
  4. −y-y direction
  5. +z+z direction
Explanation: When you encounter a problem involving magnetic force on a current-carrying wire, you're dealing with the interaction between moving charges and magnetic fields. The key tool here is the right-hand rule for the cross product F⃗=IL⃗×B⃗\vec{F} = I\vec{L} \times \vec{B}, where L⃗\vec{L} points in the direction of current flow. To find the force direction, align your right hand so your fingers point along the current direction (+y+y), then curl them toward the magnetic field direction (−z-z). Your thumb will point in the force direction. Starting with fingers pointing in the +y+y direction, you need to curl them toward −z-z. This requires rotating your hand so your thumb points in the −x-x direction. You can verify this algebraically: L⃗×B⃗=(+y)×(−z)=−(y^×z^)=−x^\vec{L} \times \vec{B} = (+y) \times (-z) = -(\hat{y} \times \hat{z}) = -\hat{x}, since y^×z^=x^\hat{y} \times \hat{z} = \hat{x} in the standard right-handed coordinate system. Choice (A) gives +x+x, which would result if you incorrectly used the left-hand rule or forgot the negative sign on the magnetic field. Choice (C) gives +y+y, which might come from confusing the force direction with the current direction. Choice (D) gives −y-y, which could result from incorrectly applying the cross product rules. The correct answer is (B) −x-x direction. Study tip: Always double-check your right-hand rule technique by practicing with standard coordinate axes. Remember that the cross product A⃗×B⃗\vec{A} \times \vec{B} is perpendicular to both vectors, and direction matters for the signs.

Question 3

A rectangular current loop with dimensions a=0.20 ma = 0.20 \text{ m} by b=0.15 mb = 0.15 \text{ m} carries current I=2.5 AI = 2.5 \text{ A}. The loop is placed in a uniform magnetic field B=0.80 TB = 0.80 \text{ T} such that the field is parallel to the plane of the loop. What is the maximum torque on the loop?

  1. 6.0×10−2 N⋅m6.0 \times 10^{-2} \text{ N⋅m} (correct answer)
  2. 3.0×10−2 N⋅m3.0 \times 10^{-2} \text{ N⋅m}
  3. 1.2×10−1 N⋅m1.2 \times 10^{-1} \text{ N⋅m}
  4. 4.8×10−2 N⋅m4.8 \times 10^{-2} \text{ N⋅m}
  5. 2.4×10−2 N⋅m2.4 \times 10^{-2} \text{ N⋅m}
Explanation: When you encounter a current loop in a magnetic field, you're dealing with magnetic torque. The key insight is that torque depends on the orientation between the magnetic field and the loop's magnetic dipole moment. The torque on a current loop is given by τ=μBsin⁡θ\tau = \mu B \sin \theta, where μ\mu is the magnetic dipole moment, BB is the magnetic field strength, and θ\theta is the angle between the field and the normal to the loop. The magnetic dipole moment equals μ=IA\mu = IA, where II is current and AA is the loop's area. First, calculate the area: A=ab=(0.20)(0.15)=0.030 m2A = ab = (0.20)(0.15) = 0.030 \text{ m}^2. Then find the magnetic dipole moment: μ=IA=(2.5)(0.030)=0.075 A⋅m2\mu = IA = (2.5)(0.030) = 0.075 \text{ A⋅m}^2. Since the field is parallel to the loop's plane, the angle between the field and the loop's normal is 90°, making sin⁡θ=1\sin \theta = 1. This gives maximum torque: τmax=μB=(0.075)(0.80)=0.060 N⋅m=6.0×10−2 N⋅m\tau_{max} = \mu B = (0.075)(0.80) = 0.060 \text{ N⋅m} = 6.0 \times 10^{-2} \text{ N⋅m}. Answer A (6.0×10−2 N⋅m6.0 \times 10^{-2} \text{ N⋅m}) is correct. Answer B (3.0×10−2 N⋅m3.0 \times 10^{-2} \text{ N⋅m}) results from incorrectly using sin⁡45°\sin 45° or halving the calculation. Answer C (1.2×10−1 N⋅m1.2 \times 10^{-1} \text{ N⋅m}) doubles the correct value, possibly from using diameter instead of side length. Answer D (4.8×10−2 N⋅m4.8 \times 10^{-2} \text{ N⋅m}) comes from calculation errors in the area or dipole moment. Remember: maximum torque occurs when the magnetic field is perpendicular to the magnetic dipole moment (parallel to the loop's plane). Always check that you're using the correct geometric relationship.

Question 4

A proton and an electron enter the same uniform magnetic field with the same speed perpendicular to the field lines. How do their circular path radii compare? (mp=1836mem_p = 1836 m_e)

  1. The proton's radius is 1836 times larger than the electron's radius (correct answer)
  2. The proton's radius is 1836 times smaller than the electron's radius
  3. The proton's radius is 1836\sqrt{1836} times larger than the electron's radius
  4. The radii are equal because they have the same charge magnitude and speed
  5. The proton's radius is 42.8 times larger than the electron's radius
Explanation: When charged particles move perpendicular to a magnetic field, they follow circular paths due to the magnetic force providing centripetal force. Understanding how particle properties affect this motion is crucial for analyzing cyclotron motion and particle accelerators. The key relationship comes from setting magnetic force equal to centripetal force: qvB=mv2rqvB = \frac{mv^2}{r}. Solving for radius gives us r=mvqBr = \frac{mv}{qB}. This shows that radius depends on mass, velocity, and charge. Since both particles enter with the same speed vv and experience the same magnetic field BB, and both have charge magnitude ee, the radius ratio becomes: rpre=mpme=1836\frac{r_p}{r_e} = \frac{m_p}{m_e} = 1836. The proton's much larger mass creates a correspondingly larger circular path. Looking at the wrong answers: Answer B incorrectly inverts the relationship, suggesting the heavier proton somehow has a smaller radius. Answer C uses 1836\sqrt{1836}, which might come from incorrectly applying kinetic energy relationships or confusing this with other physics formulas where square roots appear. Answer D falls into the trap of thinking equal charge magnitudes and speeds mean equal radii, but this ignores the crucial role of mass in determining the path. Remember: In magnetic field problems, radius is directly proportional to mass when charge and speed are constant. Heavier particles are "harder to turn" and need larger circular paths. Always check which quantities are held constant and which vary when comparing particle motion.

Question 5

A charged particle with charge q=+3.2×10−19 Cq = +3.2 \times 10^{-19} \text{ C} moves with velocity v⃗=(4.0i^+3.0j^)×105 m/s\vec{v} = (4.0\hat{i} + 3.0\hat{j}) \times 10^5 \text{ m/s} in a magnetic field B⃗=0.60k^ T\vec{B} = 0.60\hat{k} \text{ T}. What is the magnitude of the magnetic force on the particle?

  1. 9.6×10−14 N9.6 \times 10^{-14} \text{ N} (correct answer)
  2. 7.7×10−14 N7.7 \times 10^{-14} \text{ N}
  3. 5.8×10−14 N5.8 \times 10^{-14} \text{ N}
  4. 1.2×10−13 N1.2 \times 10^{-13} \text{ N}
  5. 3.8×10−14 N3.8 \times 10^{-14} \text{ N}
Explanation: When you encounter a charged particle moving in a magnetic field, you're dealing with the Lorentz force law: F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B}). The key insight is that the magnetic force depends on the cross product of velocity and magnetic field vectors, making the angle between them crucial. To find the magnitude, you need ∣F⃗∣=∣q∣∣v⃗∣∣B⃗∣sin⁡θ|\vec{F}| = |q||\vec{v}||\vec{B}|\sin\theta, where θ\theta is the angle between v⃗\vec{v} and B⃗\vec{B}. First, calculate the velocity magnitude: ∣v⃗∣=(4.0)2+(3.0)2×105=5.0×105 m/s|\vec{v}| = \sqrt{(4.0)^2 + (3.0)^2} \times 10^5 = 5.0 \times 10^5 \text{ m/s}. Since v⃗\vec{v} lies in the xy-plane and B⃗\vec{B} points along the z-axis, they're perpendicular (sin⁡90°=1\sin 90° = 1). Therefore: ∣F⃗∣=(3.2×10−19)(5.0×105)(0.60)(1)=9.6×10−14 N|\vec{F}| = (3.2 \times 10^{-19})(5.0 \times 10^5)(0.60)(1) = 9.6 \times 10^{-14} \text{ N} This confirms answer A is correct. Answer B (7.7×10−14 N7.7 \times 10^{-14} \text{ N}) likely results from using only one velocity component instead of the full magnitude. Answer C (5.8×10−14 N5.8 \times 10^{-14} \text{ N}) might come from incorrectly assuming a different angle or making calculation errors. Answer D (1.2×10−13 N1.2 \times 10^{-13} \text{ N}) is too large and suggests adding velocity components instead of using the Pythagorean theorem. Study tip: Always remember that magnetic force problems require vector magnitudes, not components. When velocity has multiple components, use ∣v⃗∣=vx2+vy2+vz2|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} and check if vectors are perpendicular—this often simplifies the cross product significantly.

Question 6

Three parallel wires carry currents I1=4.0 AI_1 = 4.0 \text{ A}, I2=2.0 AI_2 = 2.0 \text{ A}, and I3=6.0 AI_3 = 6.0 \text{ A} all in the same direction. Wire 2 is positioned midway between wires 1 and 3, which are separated by distance d=0.20 md = 0.20 \text{ m}. What is the direction of the net magnetic force on wire 2?

  1. Toward wire 1
  2. Toward wire 3 (correct answer)
  3. Zero force (wire 2 is in equilibrium)
  4. Perpendicular to the plane containing all three wires
  5. The force alternates direction due to interference between the fields
Explanation: When you encounter parallel current-carrying wires, remember that the magnetic force between them follows a simple rule: currents in the same direction attract, while opposite currents repel. The force magnitude depends on both currents and decreases with distance. Since wire 2 sits midway between wires 1 and 3, it's d/2=0.10 md/2 = 0.10 \text{ m} from each. Using the force equation F=μ0I1I22πrF = \frac{\mu_0 I_1 I_2}{2\pi r}, you can calculate the forces on wire 2. Wire 1 exerts force F12=μ0(4.0)(2.0)2π(0.10)F_{12} = \frac{\mu_0 (4.0)(2.0)}{2\pi(0.10)} pulling wire 2 toward wire 1. Wire 3 exerts force F32=μ0(6.0)(2.0)2π(0.10)F_{32} = \frac{\mu_0 (6.0)(2.0)}{2\pi(0.10)} pulling wire 2 toward wire 3. Since I3>I1I_3 > I_1, the force from wire 3 is stronger than from wire 1. The net force points toward wire 3, making (B) correct. (A) ignores that wire 3 carries more current than wire 1, creating a stronger attractive force. (C) assumes equilibrium occurs simply because wire 2 is centered, but equilibrium requires equal forces, which only happens when I1=I3I_1 = I_3. Here, the unequal currents create unequal forces. (D) misunderstands the geometry—magnetic forces between parallel wires always act in the plane perpendicular to the wires, either attracting or repelling along the line connecting them. Study tip: For parallel wire problems, always check current magnitudes and distances separately. Equal spacing doesn't guarantee equilibrium unless the currents are also equal.

Question 7

A rectangular coil with N=50N = 50 turns, width w=0.08 mw = 0.08 \text{ m}, and length l=0.12 ml = 0.12 \text{ m} rotates at f=60 Hzf = 60 \text{ Hz} in a uniform magnetic field B=0.15 TB = 0.15 \text{ T}. What is the maximum EMF induced in the coil?

  1. 2.7 V2.7 \text{ V} (correct answer)
  2. 5.4 V5.4 \text{ V}
  3. 1.4 V1.4 \text{ V}
  4. 0.68 V0.68 \text{ V}
  5. 4.1 V4.1 \text{ V}
Explanation: When you encounter a rotating coil in a magnetic field, you're dealing with electromagnetic induction and motional EMF. The key insight is that as the coil rotates, the magnetic flux through it changes sinusoidally, inducing an EMF according to Faraday's law. For a rectangular coil rotating in a uniform magnetic field, the maximum induced EMF is given by: εmax=NBAω\varepsilon_{max} = NBA\omega, where NN is the number of turns, BB is the magnetic field strength, AA is the coil's area, and ω\omega is the angular frequency. First, calculate the coil's area: A=w×l=0.08×0.12=0.0096 m2A = w \times l = 0.08 \times 0.12 = 0.0096 \text{ m}^2 Next, convert frequency to angular frequency: ω=2πf=2π×60=377 rad/s\omega = 2\pi f = 2\pi \times 60 = 377 \text{ rad/s} Now substitute into the formula: εmax=50×0.15×0.0096×377=2.7 V\varepsilon_{max} = 50 \times 0.15 \times 0.0096 \times 377 = 2.7 \text{ V} This confirms answer A is correct at 2.7 V2.7 \text{ V}. Answer B (5.4 V5.4 \text{ V}) would result from doubling the correct answer, possibly from incorrectly using 2ω2\omega instead of ω\omega. Answer C (1.4 V1.4 \text{ V}) represents roughly half the correct value, likely from using frequency ff instead of angular frequency ω\omega. Answer D (0.68 V0.68 \text{ V}) is about one-quarter the correct answer, possibly from confusing the area calculation or misapplying the frequency conversion. Remember: always convert frequency to angular frequency using ω=2πf\omega = 2\pi f for rotational EMF problems, and double-check your area calculation for rectangular geometries.

Question 8

A charged particle enters a region with both electric field E⃗=500 V/m\vec{E} = 500 \text{ V/m} (downward) and magnetic field B⃗=0.25 T\vec{B} = 0.25 \text{ T} (into the page). If the particle moves horizontally to the right with constant velocity, what must be the speed of the particle?

  1. 2000 m/s2000 \text{ m/s} (correct answer)
  2. 1250 m/s1250 \text{ m/s}
  3. 125 m/s125 \text{ m/s}
  4. 500 m/s500 \text{ m/s}
  5. 800 m/s800 \text{ m/s}
Explanation: When a charged particle moves through crossed electric and magnetic fields at constant velocity, you're dealing with a velocity selector configuration. The key insight is that the electric and magnetic forces must be perfectly balanced for the particle to travel in a straight line. The electric force on a charged particle is FE=qEF_E = qE, acting downward (same direction as the electric field for a positive charge). The magnetic force is FB=qvBF_B = qvB, and using the right-hand rule with velocity to the right and magnetic field into the page, this force acts upward on a positive charge. For constant velocity, these forces must be equal in magnitude: qE=qvBqE = qvB. Notice that the charge qq cancels out, giving us E=vBE = vB. Solving for velocity: v=EB=500 V/m0.25 T=2000 m/sv = \frac{E}{B} = \frac{500 \text{ V/m}}{0.25 \text{ T}} = 2000 \text{ m/s}. This confirms answer A is correct. Answer B (1250 m/s) might result from incorrectly calculating 5000.4\frac{500}{0.4} or misreading the magnetic field value. Answer C (125 m/s) comes from the calculation error 5004\frac{500}{4}, possibly from treating 0.25 as 4 instead of 0.25. Answer D (500 m/s) represents the misconception that velocity equals the electric field strength, ignoring the magnetic field entirely. Remember: in velocity selector problems, the particle's charge always cancels out, and the solution depends only on the field strengths. The formula v=E/Bv = E/B is your key relationship for crossed fields.

Question 9

An electron moves in a circular path of radius r=2.0 cmr = 2.0 \text{ cm} in a uniform magnetic field B=1.5×10−3 TB = 1.5 \times 10^{-3} \text{ T}. What is the speed of the electron? (me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg}, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C})

  1. 5.3×106 m/s5.3 \times 10^6 \text{ m/s} (correct answer)
  2. 2.6×106 m/s2.6 \times 10^6 \text{ m/s}
  3. 1.1×107 m/s1.1 \times 10^7 \text{ m/s}
  4. 8.8×105 m/s8.8 \times 10^5 \text{ m/s}
  5. 4.4×105 m/s4.4 \times 10^5 \text{ m/s}
Explanation: When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as the centripetal force, causing circular motion. This is a fundamental concept in electromagnetism that appears frequently in physics problems. For an electron moving in a circular path, the magnetic force FB=evBF_B = evB provides the centripetal force Fc=mv2rF_c = \frac{mv^2}{r}. Setting these equal: evB=mv2revB = \frac{mv^2}{r} Solving for velocity: v=eBrmv = \frac{eBr}{m} Substituting the given values: v=(1.6×10−19 C)(1.5×10−3 T)(0.02 m)9.1×10−31 kgv = \frac{(1.6 \times 10^{-19} \text{ C})(1.5 \times 10^{-3} \text{ T})(0.02 \text{ m})}{9.1 \times 10^{-31} \text{ kg}} v=4.8×10−249.1×10−31=5.3×106 m/sv = \frac{4.8 \times 10^{-24}}{9.1 \times 10^{-31}} = 5.3 \times 10^6 \text{ m/s} This confirms answer A is correct. Answer B (2.6×106 m/s2.6 \times 10^6 \text{ m/s}) would result from accidentally using half the radius or double the mass. Answer C (1.1×107 m/s1.1 \times 10^7 \text{ m/s}) comes from calculation errors, possibly forgetting to convert cm to meters properly. Answer D (8.8×105 m/s8.8 \times 10^5 \text{ m/s}) represents a magnitude error, likely from misplacing decimal points during the division. Remember the key relationship for charged particle motion in magnetic fields: v=qBrmv = \frac{qBr}{m}. Always double-check unit conversions (cm to m) and be careful with scientific notation arithmetic. This formula works for any charged particle in a uniform magnetic field.

Question 10

A solenoid with N=800N = 800 turns, length L=0.50 mL = 0.50 \text{ m}, and cross-sectional area A=2.0×10−3 m2A = 2.0 \times 10^{-3} \text{ m}^2 carries current I=1.5 AI = 1.5 \text{ A}. What is the magnetic field inside the solenoid? (μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \text{ T⋅m/A})

  1. 3.0×10−3 T3.0 \times 10^{-3} \text{ T} (correct answer)
  2. 1.5×10−3 T1.5 \times 10^{-3} \text{ T}
  3. 6.0×10−3 T6.0 \times 10^{-3} \text{ T}
  4. 2.4×10−3 T2.4 \times 10^{-3} \text{ T}
  5. 4.8×10−4 T4.8 \times 10^{-4} \text{ T}
Explanation: When you encounter solenoid problems, you're dealing with one of the most predictable magnetic field configurations in physics. The key insight is that a solenoid creates a uniform magnetic field inside that depends only on the current, number of turns, and length. The magnetic field inside a long solenoid is given by B=μ0nIB = \mu_0 n I, where nn is the turn density (turns per unit length). First, calculate the turn density: n=NL=8000.50=1600 turns/mn = \frac{N}{L} = \frac{800}{0.50} = 1600 \text{ turns/m}. Now apply the formula: B=μ0nI=(4π×10−7)(1600)(1.5)=9.6π×10−4 TB = \mu_0 n I = (4\pi \times 10^{-7})(1600)(1.5) = 9.6\pi \times 10^{-4} \text{ T} Since π≈3.14\pi \approx 3.14, this gives B≈3.0×10−3 TB \approx 3.0 \times 10^{-3} \text{ T}, which is answer A. Let's examine why the other answers are incorrect. Answer B (1.5×10−3 T1.5 \times 10^{-3} \text{ T}) likely comes from forgetting the factor of 4π4\pi in μ0\mu_0 and just using 10−710^{-7}. Answer C (6.0×10−3 T6.0 \times 10^{-3} \text{ T}) might result from doubling the correct answer or making an error with the π\pi approximation. Answer D (2.4×10−3 T2.4 \times 10^{-3} \text{ T}) could come from incorrectly including the cross-sectional area in the calculation, even though it doesn't affect the field strength inside a solenoid. Remember: the cross-sectional area is irrelevant for the magnetic field inside a solenoid—only current, turns, and length matter. Always double-check that you're using the complete value of μ0\mu_0 including the 4π4\pi factor.

Question 11

Two identical circular loops of radius R=0.15 mR = 0.15 \text{ m} carry currents I=2.0 AI = 2.0 \text{ A} in opposite directions. They are separated by a distance d=0.30 md = 0.30 \text{ m} along their common axis. At the midpoint between the loops, what is the direction of the net magnetic field?

  1. The net magnetic field is zero at the midpoint (correct answer)
  2. Along the axis, in the direction from the first loop to the second loop
  3. Along the axis, in the direction from the second loop to the first loop
  4. Perpendicular to the axis, in the plane of the loops
  5. The field direction alternates rapidly due to the opposing currents
Explanation: When you encounter problems involving magnetic fields from multiple current loops, you need to apply the principle of superposition while carefully considering the direction of each field using the right-hand rule. For a circular current loop, the magnetic field along the axis points in the direction determined by the right-hand rule: curl your fingers in the direction of current flow, and your thumb points in the direction of the magnetic field. Since these loops carry currents in opposite directions, their magnetic fields at any point along the axis will point in opposite directions. At the midpoint between the loops, you're equidistant from both loops (0.15 m0.15 \text{ m} from each). The magnetic field magnitude from a current loop along its axis is given by B=μ0IR22(R2+z2)3/2B = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}, where zz is the distance from the loop center. Since both loops are identical (same II and RR) and the midpoint is equidistant from both, the magnetic field magnitudes are equal. However, because the currents flow in opposite directions, these equal-magnitude fields point in opposite directions along the axis, completely canceling each other out. Choice B and C are wrong because they suggest a net field along the axis in one direction or the other, ignoring the cancellation. Choice D is incorrect because magnetic fields from current loops along their axis always point along that axis, never perpendicular to it. Remember: when identical current sources are arranged symmetrically but with opposite orientations, look for cancellation effects at symmetric points like the midpoint.

Question 12

A horizontal conducting rod of length L=0.40 mL = 0.40 \text{ m} moves with velocity v=6.0 m/sv = 6.0 \text{ m/s} perpendicular to a vertical magnetic field B=0.25 TB = 0.25 \text{ T}. What is the magnitude of the induced EMF across the rod?

  1. 0.60 V0.60 \text{ V} (correct answer)
  2. 1.20 V1.20 \text{ V}
  3. 0.30 V0.30 \text{ V}
  4. 2.40 V2.40 \text{ V}
  5. 0.15 V0.15 \text{ V}
Explanation: When you encounter a moving conductor in a magnetic field, you're dealing with motional EMF - one of the fundamental ways to induce voltage according to Faraday's law. The key insight is that when a conductor moves through a magnetic field, the magnetic force on charge carriers creates a separation of charge, resulting in an induced EMF. For a straight conductor moving perpendicular to a uniform magnetic field, the motional EMF is given by ε=BLv\varepsilon = BLv, where BB is the magnetic field strength, LL is the length of the conductor, and vv is the velocity perpendicular to the field. Substituting the given values: ε=(0.25 T)(0.40 m)(6.0 m/s)=0.60 V\varepsilon = (0.25 \text{ T})(0.40 \text{ m})(6.0 \text{ m/s}) = 0.60 \text{ V}. This confirms answer A is correct. Looking at the wrong answers: Answer B (1.20 V1.20 \text{ V}) represents doubling the correct answer, possibly from incorrectly using 2BLv2BLv or making an arithmetic error. Answer C (0.30 V0.30 \text{ V}) is exactly half the correct value, suggesting someone might have divided by 2 unnecessarily or used half of one of the given values. Answer D (2.40 V2.40 \text{ V}) is four times the correct answer, likely from multiplying by an extra factor or misplacing a decimal point. Remember the motional EMF formula ε=BLv\varepsilon = BLv for perpendicular motion - it's straightforward multiplication with no hidden factors. Always double-check your arithmetic, as physics problems often include distractors that are simple multiples or fractions of the correct answer.

Question 13

A long straight wire carries current I=5.0 AI = 5.0 \text{ A}. At a distance r=0.10 mr = 0.10 \text{ m} from the wire, the magnetic field strength is B1B_1. At what distance from the wire will the magnetic field strength be B2=0.25B1B_2 = 0.25 B_1?

  1. 0.40 m0.40 \text{ m} (correct answer)
  2. 0.25 m0.25 \text{ m}
  3. 0.20 m0.20 \text{ m}
  4. 0.05 m0.05 \text{ m}
  5. 0.80 m0.80 \text{ m}
Explanation: When you encounter problems about magnetic fields around current-carrying wires, you're dealing with Ampère's law, which tells us that the magnetic field strength varies inversely with distance from the wire. For a long straight wire, the magnetic field at distance rr is given by B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}. The key insight is that B∝1rB \propto \frac{1}{r} — as distance increases, field strength decreases proportionally. Since B1∝1r1B_1 \propto \frac{1}{r_1} and B2∝1r2B_2 \propto \frac{1}{r_2}, we can write: B2B1=r1r2\frac{B_2}{B_1} = \frac{r_1}{r_2} Given that B2=0.25B1B_2 = 0.25 B_1 and r1=0.10 mr_1 = 0.10 \text{ m}: 0.25B1B1=0.10r2\frac{0.25 B_1}{B_1} = \frac{0.10}{r_2} 0.25=0.10r20.25 = \frac{0.10}{r_2} r2=0.100.25=0.40 mr_2 = \frac{0.10}{0.25} = 0.40 \text{ m} This confirms answer A) 0.40 m0.40 \text{ m}. Let's check why the other options are wrong. Answer D) 0.05 m0.05 \text{ m} would place you closer to the wire, creating a stronger field (4B14 B_1), not weaker. Answer C) 0.20 m0.20 \text{ m} would give 0.5B10.5 B_1 — half the original strength, not one-quarter. Answer B) 0.25 m0.25 \text{ m} would yield 0.4B10.4 B_1, which is too strong. Study tip: For inverse relationships like B∝1rB \propto \frac{1}{r}, remember that to make the field one-quarter as strong, you need to be four times farther away. This 1r\frac{1}{r} dependence appears frequently in electromagnetic problems.

Question 14

Two parallel wires separated by distance d=0.10 md = 0.10 \text{ m} carry currents I1=3.0 AI_1 = 3.0 \text{ A} and I2=4.0 AI_2 = 4.0 \text{ A} in the same direction. What is the magnitude of the magnetic force per unit length between the wires? (μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \text{ T⋅m/A})

  1. 2.4×10−5 N/m2.4 \times 10^{-5} \text{ N/m} (correct answer)
  2. 1.2×10−5 N/m1.2 \times 10^{-5} \text{ N/m}
  3. 4.8×10−5 N/m4.8 \times 10^{-5} \text{ N/m}
  4. 6.0×10−6 N/m6.0 \times 10^{-6} \text{ N/m}
  5. 9.6×10−6 N/m9.6 \times 10^{-6} \text{ N/m}
Explanation: When you encounter parallel current-carrying wires, you're dealing with magnetic forces between conductors—a fundamental concept in electromagnetism. The key is recognizing that one wire creates a magnetic field, and the other wire experiences a force in that field. The magnetic force per unit length between two parallel wires is given by: F/L=μ0I1I22πdF/L = \frac{\mu_0 I_1 I_2}{2\pi d} Substituting the given values: F/L=(4π×10−7)(3.0)(4.0)2π(0.10)F/L = \frac{(4\pi \times 10^{-7})(3.0)(4.0)}{2\pi(0.10)} The π\pi terms cancel: F/L=(4×10−7)(12)2(0.10)=48×10−70.20=2.4×10−5 N/mF/L = \frac{(4 \times 10^{-7})(12)}{2(0.10)} = \frac{48 \times 10^{-7}}{0.20} = 2.4 \times 10^{-5} \text{ N/m} This confirms answer A is correct. Answer B (1.2×10−51.2 \times 10^{-5}) represents half the correct value—you might get this if you accidentally used 4π4\pi instead of 2π2\pi in the denominator. Answer C (4.8×10−54.8 \times 10^{-5}) is double the correct answer, possibly from forgetting the factor of 2 in the denominator entirely. Answer D (6.0×10−66.0 \times 10^{-6}) is one-fourth the correct value, likely from using 8π8\pi in the denominator instead of 2π2\pi. Study tip: Memorize the parallel wire force formula and watch the denominator carefully—it's 2πd2\pi d, not just πd\pi d or 4πd4\pi d. Also remember that currents in the same direction create attractive forces, while opposite currents repel.

Question 15

A straight wire carries a current of 8.0 A8.0 \text{ A}. At what distance from the wire is the magnetic field strength equal to 2.0×10−5 T2.0 \times 10^{-5} \text{ T}? (μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \text{ T⋅m/A})

  1. 8.0×10−2 m8.0 \times 10^{-2} \text{ m} (correct answer)
  2. 4.0×10−2 m4.0 \times 10^{-2} \text{ m}
  3. 1.6×10−1 m1.6 \times 10^{-1} \text{ m}
  4. 3.2×10−2 m3.2 \times 10^{-2} \text{ m}
  5. 6.4×10−2 m6.4 \times 10^{-2} \text{ m}
Explanation: When you encounter problems about magnetic fields around current-carrying wires, you're dealing with Ampère's law and the fundamental relationship between electric current and magnetism. The key formula here is the magnetic field around a long straight wire: B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}, where BB is the magnetic field strength, μ0\mu_0 is the permeability of free space, II is the current, and rr is the distance from the wire. To find the distance, rearrange the formula to solve for rr: r=μ0I2πBr = \frac{\mu_0 I}{2\pi B} Substituting the given values: r=(4π×10−7)(8.0)2π(2.0×10−5)r = \frac{(4\pi \times 10^{-7})(8.0)}{2\pi(2.0 \times 10^{-5})} The π\pi terms cancel: r=(4×10−7)(8.0)2(2.0×10−5)=3.2×10−64.0×10−5=8.0×10−2 mr = \frac{(4 \times 10^{-7})(8.0)}{2(2.0 \times 10^{-5})} = \frac{3.2 \times 10^{-6}}{4.0 \times 10^{-5}} = 8.0 \times 10^{-2} \text{ m} This confirms answer A is correct. Answer B (4.0×10−2 m4.0 \times 10^{-2} \text{ m}) represents exactly half the correct distance—likely from forgetting the factor of 2 in the denominator. Answer C (1.6×10−1 m1.6 \times 10^{-1} \text{ m}) is double the correct answer, possibly from an algebraic error. Answer D (3.2×10−2 m3.2 \times 10^{-2} \text{ m}) appears to come from incorrectly handling the π\pi cancellation or making an arithmetic mistake in the final calculation. Remember: magnetic field strength decreases with distance from the wire, and always double-check that your π\pi terms cancel properly in these circular field problems.

Question 16

A particle with charge q=−2.0×10−19 Cq = -2.0 \times 10^{-19} \text{ C} and mass m=6.0×10−31 kgm = 6.0 \times 10^{-31} \text{ kg} moves in a uniform magnetic field. If the magnetic force on the particle is F=3.2×10−14 NF = 3.2 \times 10^{-14} \text{ N} and the field strength is B=0.80 TB = 0.80 \text{ T}, what is the speed of the particle?

  1. 2.0×105 m/s2.0 \times 10^5 \text{ m/s} (correct answer)
  2. 4.0×105 m/s4.0 \times 10^5 \text{ m/s}
  3. 1.0×105 m/s1.0 \times 10^5 \text{ m/s}
  4. 8.0×104 m/s8.0 \times 10^4 \text{ m/s}
  5. 1.6×105 m/s1.6 \times 10^5 \text{ m/s}
Explanation: When you encounter a charged particle moving in a magnetic field, you're dealing with the Lorentz force law. The magnetic force on a moving charged particle is given by F=∣q∣vBsin⁡θF = |q|vB\sin\theta, where the sine term accounts for the angle between velocity and magnetic field. Since we're told this is a "uniform magnetic field" problem without mention of angles, we can assume the particle moves perpendicular to the field (sin⁡θ=1\sin\theta = 1), giving us F=∣q∣vBF = |q|vB. Notice we use the absolute value of charge since force magnitude is always positive. Solving for velocity: v=F∣q∣B=3.2×10−14(2.0×10−19)(0.80)=3.2×10−141.6×10−19=2.0×105 m/sv = \frac{F}{|q|B} = \frac{3.2 \times 10^{-14}}{(2.0 \times 10^{-19})(0.80)} = \frac{3.2 \times 10^{-14}}{1.6 \times 10^{-19}} = 2.0 \times 10^5 \text{ m/s} This confirms answer A is correct. Looking at the wrong answers: B (4.0×105 m/s4.0 \times 10^5 \text{ m/s}) would result if you forgot to use the absolute value of charge and somehow doubled the result. C (1.0×105 m/s1.0 \times 10^5 \text{ m/s}) might come from incorrectly using the full charge value −2.0×10−19-2.0 \times 10^{-19} instead of its magnitude, or making an arithmetic error. D (8.0×104 m/s8.0 \times 10^4 \text{ m/s}) could result from calculation mistakes in the division. Study tip: Always use the magnitude of charge in force calculations, and remember that for maximum magnetic force, the velocity must be perpendicular to the magnetic field. Practice rearranging F=∣q∣vBF = |q|vB to solve for any unknown variable.

Question 17

Two parallel wires carry currents I1=3.0I_1 = 3.0 A and I2=2.0I_2 = 2.0 A in opposite directions. The wires are separated by distance d=0.10d = 0.10 m. At what distance from wire 1 (measured perpendicular to the wires) is the magnetic field exactly zero?

  1. 0.040 m from wire 1, between the two wires
  2. The magnetic field cannot be zero at any finite distance from the wires
  3. 0.040 m from wire 1, on the side opposite to wire 2
  4. 0.060 m from wire 1, between the two wires (correct answer)
Explanation: When two parallel wires carry current, each creates a magnetic field that follows the right-hand rule. The key insight is that when currents flow in opposite directions, there's a point between the wires where their magnetic fields cancel out completely. To find where the magnetic field is zero, you need to set up the equation where the magnetic fields from both wires are equal in magnitude but opposite in direction. The magnetic field from a long straight wire is B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}, where rr is the distance from the wire. Let xx be the distance from wire 1. Since the zero point must be between the wires, the distance from wire 2 is (0.10−x)(0.10 - x). Setting the field magnitudes equal: μ0I12πx=μ0I22π(0.10−x)\frac{\mu_0 I_1}{2\pi x} = \frac{\mu_0 I_2}{2\pi (0.10 - x)} This simplifies to: 3.0x=2.00.10−x\frac{3.0}{x} = \frac{2.0}{0.10 - x} Cross-multiplying: 3.0(0.10−x)=2.0x3.0(0.10 - x) = 2.0x 0.30−3.0x=2.0x0.30 - 3.0x = 2.0x 0.30=5.0x0.30 = 5.0x x=0.060 mx = 0.060 \text{ m} Answer D is correct: 0.060 m from wire 1, between the wires. Answer A uses the wrong calculation or setup. Answer B is incorrect because cancellation always occurs between wires with opposite currents. Answer C places the zero point on the wrong side—with opposite currents, cancellation only happens between the wires, never outside. Study tip: For parallel wires with opposite currents, the magnetic field zero point always lies between the wires, closer to the wire with smaller current.

Question 18

A rectangular current loop with dimensions 0.20 m × 0.30 m carries a current of 2.5 A. The loop is placed in a uniform magnetic field of magnitude 0.40 T. What is the maximum possible magnetic torque that can act on this loop?

  1. 6.0×10−26.0 \times 10^{-2} N·m when the loop's normal is perpendicular to the field
  2. 3.0×10−23.0 \times 10^{-2} N·m when the loop's plane is perpendicular to the field
  3. 1.2×10−11.2 \times 10^{-1} N·m when the loop's normal makes a 45° angle with the field
  4. 6.0×10−26.0 \times 10^{-2} N·m when the loop's plane is parallel to the field (correct answer)
Explanation: When you encounter magnetic torque problems, remember that torque on a current loop depends on the magnetic dipole moment and the field orientation. The torque formula is τ=μBsin⁡θ\tau = \mu B \sin \theta, where μ\mu is the magnetic dipole moment, BB is the magnetic field strength, and θ\theta is the angle between the magnetic dipole moment vector and the field. The magnetic dipole moment equals μ=IA\mu = IA, where II is current and AA is the loop area. Here, A=0.20×0.30=0.06A = 0.20 \times 0.30 = 0.06 m², so μ=2.5×0.06=0.15\mu = 2.5 \times 0.06 = 0.15 A·m². Maximum torque occurs when sin⁡θ=1\sin \theta = 1, meaning θ=90°\theta = 90°. This happens when the magnetic dipole moment (which points normal to the loop) is perpendicular to the magnetic field. Geometrically, this means the loop's plane is parallel to the field. The maximum torque is τmax=μB=0.15×0.40=6.0×10−2\tau_{max} = \mu B = 0.15 \times 0.40 = 6.0 \times 10^{-2} N·m. Choice A incorrectly states maximum torque occurs when the normal is perpendicular to the field—this actually gives zero torque since sin⁡(0°)=0\sin(0°) = 0. Choice B gives the wrong magnitude and incorrect geometry. Choice C claims maximum torque occurs at 45°, but sin⁡(45°)=0.707<1\sin(45°) = 0.707 < 1, so this cannot be maximum. Additionally, its calculated value is double the correct maximum. Remember: maximum magnetic torque always occurs when the loop's plane is parallel to the magnetic field, making the dipole moment perpendicular to the field.

Question 19

A square current loop with side length a=0.080a = 0.080 m carries current I=2.5I = 2.5 A clockwise when viewed from above. The loop is placed in a uniform magnetic field B⃗=0.30k^\vec{B} = 0.30 \hat{k} T. If the loop is initially in the xy-plane, what is the net magnetic force on the loop and the initial direction of rotation?

  1. Zero net force; the loop rotates to align its magnetic moment antiparallel to the field (correct answer)
  2. Net force of 0.24 N upward; the loop rotates about its center with no preferred direction
  3. Zero net force; the loop rotates to align its magnetic moment parallel to the field
  4. Net force of 0.60 N in the xy-plane; the loop translates without rotating
Explanation: In a uniform magnetic field, the net force on a closed current loop is always zero because the forces on opposite sides cancel. However, there can be a net torque. The magnetic dipole moment of the loop is μ⃗=IAn^\vec{\mu} = IA\hat{n}, where n^\hat{n} is determined by the right-hand rule. For clockwise current viewed from +z, the magnetic moment points in the -z direction: μ⃗=−IAk^\vec{\mu} = -IA\hat{k}. The torque is τ⃗=μ⃗×B⃗=(−IAk^)×(Bk^)=0\vec{\tau} = \vec{\mu} \times \vec{B} = (-IA\hat{k}) \times (B\hat{k}) = 0 initially, but any small perturbation will cause the loop to rotate to minimize potential energy, which occurs when μ⃗\vec{\mu} and B⃗\vec{B} are antiparallel (the stable configuration). Choice B incorrectly calculates a net force. Choice C has the wrong alignment direction. Choice D misunderstands the uniform field condition.

Question 20

Two identical circular coils, each with N=50N = 50 turns and radius R=0.12R = 0.12 m, are placed parallel to each other and separated by a distance d=0.12d = 0.12 m (equal to their radius). This configuration is known as Helmholtz coils. If each coil carries the same current I=1.5I = 1.5 A in the same direction, what is the approximate magnetic field at the midpoint between the coils?

  1. 1.4×10−31.4 \times 10^{-3} T, more uniform than the field from a single coil (correct answer)
  2. 7.0×10−47.0 \times 10^{-4} T, exactly half the field from a single coil at its center
  3. 2.8×10−32.8 \times 10^{-3} T, twice the field from a single coil at its center
  4. 5.6×10−45.6 \times 10^{-4} T, less uniform than the field from a single coil
Explanation: For Helmholtz coils (separation equal to radius), the field at the midpoint is particularly uniform. Each coil contributes a field at the midpoint. The distance from each coil center to the midpoint is d/2=0.06d/2 = 0.06 m. Using the formula for field along the axis: B=μ0NIR22(R2+x2)3/2B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} where x=0.06x = 0.06 m. We get B=(4π×10−7)(50)(1.5)(0.12)22((0.12)2+(0.06)2)3/2=(4π×10−7)(50)(1.5)(0.0144)2(0.0180)3/2=7.0×10−4B = \frac{(4\pi \times 10^{-7})(50)(1.5)(0.12)^2}{2((0.12)^2 + (0.06)^2)^{3/2}} = \frac{(4\pi \times 10^{-7})(50)(1.5)(0.0144)}{2(0.0180)^{3/2}} = 7.0 \times 10^{-4} T per coil. Total field = 2×7.0×10−4=1.4×10−32 \times 7.0 \times 10^{-4} = 1.4 \times 10^{-3} T. The Helmholtz configuration provides exceptional field uniformity. Choice B forgets to add both contributions. Choice C uses wrong calculation. Choice D mischaracterizes the uniformity.