Practice 3: Magnetic Fields

10 MCQ + 5 FRQ · Covers Magnetic Forces, Biot-Savart Law, Ampere's Law, Solenoids, and Torque

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Multiple Choice Questions

Q1. A proton moves with velocity \(\vec{v}\) in a uniform magnetic field \(\vec{B}\) directed into the page, tracing a circle of radius \(r\). If replaced by an alpha particle (charge \(+2e\), mass \(4m_p\)) at the same speed, what is the new radius?

Correct! \(r = mv/(qB)\). For alpha: \(r_\alpha = (4m_p)v/(2eB) = 2(m_p v/eB) = 2r\). The radius doubles because mass quadruples but charge only doubles.
Incorrect. The correct answer is (D) \(2r\). \(r = mv/(qB)\). Alpha: \(m \to 4m_p\), \(q \to 2e\), so \(r_\alpha = 4m_p v/(2eB) = 2r\).

Q2. An electron moves north in a region where \(\vec{B}\) points east. The magnetic force on the electron is directed:

Correct! \(\vec{F} = q\vec{v} \times \vec{B}\). RHR: \(\vec{v}\) (north) × \(\vec{B}\) (east) = down. But \(q = -e\), so force is reversed: up (out of ground).
Incorrect. The correct answer is (A) Up. \(\vec{v}\) north × \(\vec{B}\) east = down by RHR, but electron has negative charge → force is up.

Q3. A straight wire of length \(L\) carries current \(I\) at angle \(\theta\) to a uniform field \(\vec{B}\). The magnitude of the magnetic force is:

Correct! \(\vec{F} = I\vec{L} \times \vec{B}\), magnitude \(F = ILB\sin\theta\) where \(\theta\) is the angle between the wire direction and \(\vec{B}\).
Incorrect. The correct answer is (B) \(ILB\sin\theta\). Cross product: \(|\vec{F}| = I|\vec{L}||\vec{B}|\sin\theta\).

Q4. Two long parallel wires carry equal currents \(I\) in the same direction. The force between them is:

Correct! Parallel currents in the same direction attract. Wire 1's B field at wire 2: using \(\vec{F} = I_2\vec{L} \times \vec{B}_1\) gives force toward wire 1. Opposite currents repel.
Incorrect. The correct answer is (A) Attractive. Same direction → attract; opposite directions → repel. \(F/L = \mu_0 I^2/(2\pi d)\).

Q5. A circular loop of wire (radius \(R\)) carries current \(I\). The magnetic field at the center is:

Correct! Biot-Savart: \(dB = \frac{\mu_0}{4\pi}\frac{I d\vec{\ell} \times \hat{r}}{r^2}\). All \(d\vec{\ell} \perp \hat{r}\) at center, integrate around \(2\pi R\): \(B = \frac{\mu_0 I}{4\pi R^2} \cdot 2\pi R = \frac{\mu_0 I}{2R}\).
Incorrect. The correct answer is (D) \(\mu_0 I/(2R)\). Integrate Biot-Savart around the circle.

Q6. Ampere's Law applied to a toroidal solenoid with \(N\) turns and current \(I\) gives, at distance \(r\) within the toroid:

Correct! Circular Amperian loop of radius \(r\): \(\oint \vec{B} \cdot d\vec{\ell} = B(2\pi r) = \mu_0 NI\). So \(B = \mu_0 NI/(2\pi r)\). Outside the toroid, \(B = 0\).
Incorrect. The correct answer is (A) \(\mu_0 NI/(2\pi r)\). Amperian loop encloses \(N\) turns: \(B(2\pi r) = \mu_0 NI\).

Q7. For an ideal infinitely long solenoid with \(n\) turns per unit length and current \(I\), the magnetic field outside the solenoid is:

Correct! For an ideal solenoid, Ampere's Law on a rectangular loop shows \(B = 0\) outside and \(B = \mu_0 nI\) inside (uniform).
Incorrect. The correct answer is (B) \(0\). Ideal solenoid: uniform \(B = \mu_0 nI\) inside, \(B = 0\) outside.

Q8. A current loop in a uniform magnetic field experiences maximum torque when its plane is:

Correct! \(\vec{\tau} = \vec{\mu} \times \vec{B} = I\vec{A} \times \vec{B}\). \(\tau = \mu B\sin\phi\) where \(\phi\) = angle between \(\vec{\mu}\) (normal to plane) and \(\vec{B}\). Max at \(\phi = 90^\circ\) — area vector \(\perp \vec{B}\) — which means the plane is parallel to \(\vec{B}\).
Incorrect. The correct answer is (C) Parallel to \(\vec{B}\). \(\tau = \mu B\sin\phi\), max when \(\vec{\mu} \perp \vec{B}\), meaning the plane (perpendicular to \(\vec{\mu}\)) is parallel to \(\vec{B}\).

Q9. A charged particle passes undeflected through crossed \(\vec{E}\) and \(\vec{B}\) fields at speed \(v = E/B\). This device is a:

Correct! Crossed \(\vec{E} \perp \vec{B}\) fields: \(qE = qvB\) when \(v = E/B\), so only particles at that speed pass undeflected. Used as the first stage in mass spectrometers.
Incorrect. The correct answer is (E) Velocity selector. Crossed \(\vec{E}\) and \(\vec{B}\): \(qE = qvB\) → \(v = E/B\).

Q10. The field \(B = \frac{\mu_0 I}{2\pi r}\) around a long straight wire is most directly derived from:

Correct! Ampere's Law: \(\oint \vec{B} \cdot d\vec{\ell} = B(2\pi r) = \mu_0 I\), so \(B = \mu_0 I/(2\pi r)\). The cylindrical symmetry makes Ampere's Law the simplest derivation.
Incorrect. The correct answer is (C). Ampere's Law: \(B(2\pi r) = \mu_0 I\) → \(B = \mu_0 I/(2\pi r)\).

Free Response Questions

FRQ

F1: Cyclotron Motion

A proton (\(m = 1.67\times10^{-27}\ \text{kg}\), \(q = 1.6\times10^{-19}\ \text{C}\)) enters a uniform \(B = 0.50\ \text{T}\) (into page) with \(v = 2.0\times10^6\ \text{m/s}\) perpendicular to \(\vec{B}\).

  1. Determine the radius of the proton's circular path.
  2. Calculate the period of the motion.
  3. Find the work done by the magnetic force over one revolution. Explain.
  4. The proton is replaced by a deuteron (same charge, twice the mass) at the same speed. Find the new period.
Show Solution
(a): \(r = \frac{mv}{qB} = \frac{(1.67\times10^{-27})(2.0\times10^6)}{(1.6\times10^{-19})(0.50)} = 4.18\times10^{-2}\ \text{m} = 4.18\ \text{cm}\).
(b): \(T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} = \frac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.50)} = 1.31\times10^{-7}\ \text{s}\). Note: period is independent of speed!
(c): \(W = 0\). Magnetic force is always perpendicular to velocity (\(\vec{F} = q\vec{v} \times \vec{B} \perp \vec{v}\)), so it does zero work. It changes direction but not speed.
(d): \(T = 2\pi m/(qB)\). With \(m \to 2m\): \(T_{\text{new}} = 2T = 2.62\times10^{-7}\ \text{s}\).
FRQ

F2: Biot-Savart — Semicircular Arc

A wire carries current \(I\) and is bent into a semicircular arc of radius \(R\) in the \(xy\)-plane, with straight segments extending radially outward from the center.

  1. Use the Biot-Savart Law to derive \(\vec{B}\) at the center of curvature.
  2. What is the direction of \(\vec{B}\)?
  3. If the arc is extended to a full circle, compare the field to part (a).
Show Solution
(a): Biot-Savart: \(d\vec{B} = \frac{\mu_0}{4\pi}\frac{I d\vec{\ell} \times \hat{r}}{r^2}\). Arc: \(d\ell = R\,d\theta\), \(d\vec{\ell} \times \hat{r}\) magnitude = \(R\,d\theta\). \(dB = \frac{\mu_0 I}{4\pi R^2}R\,d\theta = \frac{\mu_0 I}{4\pi R}d\theta\). Integrate \(\theta: 0 \to \pi\): \(B = \frac{\mu_0 I}{4R}\). Straight segments: \(d\vec{\ell} \parallel \hat{r}\) → contribution = 0.
(b): RHR: current in \(+\theta\) direction, \(d\vec{\ell} \times \hat{r}\) points into page (\(-z\) direction).
(c): Full circle: integrate \(0 \to 2\pi\): \(B = \frac{\mu_0 I}{2R}\), exactly twice the semicircle result.
FRQ

F3: Ampere's Law — Coaxial Cable

A coaxial cable: inner solid cylinder (radius \(a\)) carries \(I\) out of page; outer thin cylindrical shell (radius \(b\)) carries \(I\) into page. Both currents uniformly distributed.

  1. Use Ampere's Law to find \(B(r)\) for: (i) \(r < a\), (ii) \(a < r < b\), (iii) \(r > b\).
  2. Sketch \(B(r)\) from \(r = 0\) to \(r = 2b\).
  3. Where is the magnetic field maximum?
Show Solution
(a)(i) \(r < a\): \(I_{\text{enc}} = I(\pi r^2)/(\pi a^2) = Ir^2/a^2\). \(B(2\pi r) = \mu_0 Ir^2/a^2\) → \(B = \frac{\mu_0 I r}{2\pi a^2}\).
(ii) \(a < r < b\): \(I_{\text{enc}} = I\). \(B = \frac{\mu_0 I}{2\pi r}\).
(iii) \(r > b\): \(I_{\text{enc}} = I - I = 0\). \(B = 0\).
(b): \(B\) increases linearly from 0 to \(\frac{\mu_0 I}{2\pi a}\) at \(r = a\), then decreases as \(1/r\) for \(a < r < b\), then drops to 0 at \(r = b\) and stays zero.
(c): Maximum at \(r = a\) (surface of inner conductor): \(B_{\text{max}} = \frac{\mu_0 I}{2\pi a}\).
FRQ

F4: Current Loop Torque

A rectangular loop has sides \(a = 0.10\ \text{m}\), \(b = 0.15\ \text{m}\), carries \(I = 2.0\ \text{A}\), and is in \(B = 0.30\ \text{T}\). The area vector makes a 30° angle with \(\vec{B}\).

  1. Calculate the magnetic dipole moment (magnitude and direction).
  2. Find the torque on the loop.
  3. Find the potential energy of the loop.
  4. If released from rest, describe the subsequent motion.
Show Solution
(a): \(\mu = NIA = (1)(2.0)(0.10 \times 0.15) = 0.030\ \text{A}\cdot\text{m}^2\). Direction: perpendicular to plane by RHR (curl fingers with current).
(b): \(\tau = \mu B\sin\phi = (0.030)(0.30)\sin(30^\circ) = 0.0045\ \text{N}\cdot\text{m}\).
(c): \(U = -\vec{\mu}\cdot\vec{B} = -\mu B\cos\phi = -(0.030)(0.30)\cos(30^\circ) = -7.79 \times 10^{-3}\ \text{J}\).
(d): Torque rotates loop to align \(\vec{\mu}\) with \(\vec{B}\) (minimum energy). The loop oscillates about this equilibrium due to inertia (simple harmonic for small angles).
FRQ

F5: Mass Spectrometer

Singly ionized atoms (\(q = +e\)) pass through a velocity selector (\(E = 5000\ \text{V/m}\), \(B_1 = 0.20\ \text{T}\)) then enter \(B_2 = 0.80\ \text{T}\) where they follow semicircular paths of radius \(r\).

  1. What speed passes the velocity selector undeflected?
  2. Derive an expression for mass in terms of \(r\), \(e\), \(B_1\), \(B_2\), and \(E\).
  3. Two isotopes strike the detector 4.0 mm apart. If the lighter one has \(r = 10.0\ \text{cm}\), find the mass difference in atomic mass units.
Show Solution
(a): \(v = E/B_1 = 5000/0.20 = 2.5 \times 10^4\ \text{m/s}\).
(b): In \(B_2\): \(r = mv/(eB_2)\), with \(v = E/B_1\): \(m = \frac{eB_2 r}{v} = \frac{eB_1 B_2 r}{E}\).
(c): \(m \propto r\), so \(\Delta m/m = \Delta r/r = 0.004/0.10 = 0.04\). Lighter: \(m_1 = \frac{(1.6\times10^{-19})(0.20)(0.80)(0.10)}{5000} = 5.12\times10^{-25}\ \text{kg}\). \(\Delta m = 0.04 \times 5.12\times10^{-25} = 2.05\times10^{-26}\ \text{kg} \times \frac{1\ \text{u}}{1.66\times10^{-27}\ \text{kg}} = 12.3\ \text{u}\).

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