Practice Exam 6

Magnetic Fields & Forces

Questions10

Time90 min

Points100

CalculatorAllowed

Use \(e = 1.60\times10^{-19}\) C, \(m_e = 9.11\times10^{-31}\) kg, \(m_p = 1.67\times10^{-27}\) kg, \(u = 1.66\times10^{-27}\) kg. The page is the \(xy\)-plane, with \(\hat{k}\) out of the page. Non-relativistic formulas throughout.

1Direction of the magnetic force8 pts

Give the direction of the magnetic force (or say it is zero): (a) an electron with \(\mathbf{v}\) along \(+\hat{\imath}\) in \(\mathbf{B}\) along \(+\hat{\jmath}\); (b) a proton with \(\mathbf{v}\) along \(+\hat{k}\) in \(\mathbf{B}\) along \(+\hat{k}\); (c) a proton with \(\mathbf{v}\) along \(-\hat{\jmath}\) in \(\mathbf{B}\) along \(-\hat{\imath}\); (d) an electron moving into the page in a field pointing to the right (\(+\hat{\imath}\)).

Solution

\[ \text{(a)}\ (-e)(\hat{\imath}\times\hat{\jmath}) = -e\,\hat{k}\;\Rightarrow\;\mathbf{-\hat{k}}\ \text{(into the page)}\qquad \text{(b)}\ \hat{k}\times\hat{k} = \mathbf{0} \] \[ \text{(c)}\ (+e)(-\hat{\jmath})\times(-\hat{\imath}) = e(\hat{\jmath}\times\hat{\imath}) = -e\,\hat{k}\;\Rightarrow\;\mathbf{-\hat{k}}\qquad \text{(d)}\ (-e)(-\hat{k})\times\hat{\imath} = e(\hat{k}\times\hat{\imath}) = e\,\hat{\jmath}\;\Rightarrow\;\mathbf{+\hat{\jmath}}\ \text{(up the page)} \]

In (c), the two minus signs from the vectors cancel; only the proton's positive charge remains. In (d), the minus from “into the page” and the minus from the electron's charge cancel. Track every sign explicitly — the right-hand rule alone loses one of them about half the time.

2Circular orbit10 pts

A 5.0 MeV proton moves perpendicular to a uniform 1.5 T field. Find its speed, the radius of its orbit, its period and its frequency. Would doubling its kinetic energy change the period?

Solution

\[ K = (5.0\times10^{6})(1.60\times10^{-19}) = 8.0\times10^{-13}\ \text{J},\qquad v = \sqrt{\frac{2K}{m_p}} = \sqrt{\frac{1.6\times10^{-12}}{1.67\times10^{-27}}} = \mathbf{3.1\times10^{7}\ m/s} \] \[ r = \frac{m_pv}{eB} = \frac{(1.67\times10^{-27})(3.10\times10^{7})}{(1.60\times10^{-19})(1.5)} = \mathbf{0.22\ m} \] \[ T = \frac{2\pi m_p}{eB} = \frac{2\pi(1.67\times10^{-27})}{(1.60\times10^{-19})(1.5)} = \mathbf{43.7\ ns},\qquad f = \frac1T = \mathbf{22.9\ MHz} \]

No: \(T\) contains no \(v\). Doubling \(K\) multiplies \(v\) and \(r\) by \(\sqrt2\) — the longer path is covered proportionally faster. (\(v\approx0.10c\), so the classical result is good to about 1 %.)

3Helical motion10 pts

An electron with speed \(3.0\times10^{6}\) m/s enters a uniform 2.0 mT field with its velocity at \(60^\circ\) to the field. Find the radius, the period and the pitch of the helix.

Solution

\[ v_\perp = v\sin60^\circ = 2.60\times10^{6}\ \text{m/s},\qquad v_\parallel = v\cos60^\circ = 1.50\times10^{6}\ \text{m/s} \] \[ r = \frac{m_ev_\perp}{eB} = \frac{(9.11\times10^{-31})(2.60\times10^{6})}{(1.60\times10^{-19})(2.0\times10^{-3})} = \mathbf{7.4\ mm} \] \[ T = \frac{2\pi m_e}{eB} = \mathbf{17.9\ ns},\qquad p = v_\parallel T = (1.50\times10^{6})(1.79\times10^{-8}) = \mathbf{2.7\ cm} \]

The angle is measured from the field, so \(v_\perp\) uses the sine. Swapping sine and cosine here gives \(r = 4.3\) mm and \(p = 4.6\) cm — both wrong.

4Velocity selector & mass spectrometer12 pts

Singly ionised neon passes through a velocity selector with \(E = 4.0\times10^{4}\) V/m and \(B = 0.080\) T, then enters a region with \(B_0 = 0.40\) T. (a) What speed is selected? Does it depend on the ion's mass? (b) Where do \(^{20}\)Ne and \(^{22}\)Ne land, measured from the entrance slit (\(m = 20u\) and \(22u\))? (c) An ion moving slightly faster than the selected speed — which way is it deflected in the selector?

Solution

\[ \text{(a)}\ v = \frac{E}{B} = \frac{4.0\times10^{4}}{0.080} = \mathbf{5.0\times10^{5}\ m/s}\quad(\text{independent of } m \text{ and } q) \] \[ \text{(b)}\ r_{20} = \frac{(20)(1.66\times10^{-27})(5.0\times10^{5})}{(1.60\times10^{-19})(0.40)} = 25.9\ \text{cm},\qquad r_{22} = \tfrac{22}{20}r_{20} = 28.5\ \text{cm} \] \[ 2r_{20} = \mathbf{51.9\ cm},\qquad 2r_{22} = \mathbf{57.1\ cm},\qquad \text{separation} = \mathbf{5.2\ cm} \]

(c) The magnetic force \(qvB\) grows with \(v\); the electric force \(qE\) does not. A faster ion is deflected toward the side the magnetic force pushes and hits the selector wall.

5Hall-effect sensor · EE application12 pts

A Hall element is a semiconductor plate of thickness \(t = 0.10\) mm along \(\mathbf{B}\), with \(n = 5.0\times10^{21}\) conduction electrons per m³. Its bias current is \(I = 2.0\) mA along \(+\hat{\imath}\) and the field is along \(+\hat{k}\). (a) Find its sensitivity in mV per mT. (b) It reads 5.0 mV: what is \(B\)? (c) Which edge, \(+\hat{\jmath}\) or \(-\hat{\jmath}\), is at the higher potential? (d) Why not just raise the bias current to 200 mA for more signal?

Solution

\[ \text{(a)}\ \frac{V_H}{B} = \frac{I}{net} = \frac{2.0\times10^{-3}}{(5.0\times10^{21})(1.60\times10^{-19})(1.0\times10^{-4})} = 0.025\ \text{V/T} = \mathbf{0.025\ mV/mT} \] \[ \text{(b)}\ B = \frac{V_H}{0.025\ \text{V/T}} = \frac{5.0\times10^{-3}}{0.025} = \mathbf{0.20\ T} \]

(c) Electrons drift along \(-\hat{\imath}\). Force: \((-e)(-v\,\hat{\imath})\times(B\,\hat{k}) = evB(\hat{\imath}\times\hat{k}) = -evB\,\hat{\jmath}\). Electrons pile up on the \(-\hat{\jmath}\) edge, so the \(+\hat{\jmath}\) edge is at the higher potential. (With positive carriers it would be the other way.)

(d) \(V_H\propto I\), but the element's resistance dissipates \(I^2R\): 100× the current means 10 000× the heat, which drifts \(n\) (it is a semiconductor) and ruins the calibration, or destroys the part. Real sensors use a modest bias and an on-chip amplifier instead.

6Force on a straight wire (vector)10 pts

A straight wire segment \(\mathbf{L} = (0.20\,\hat{\imath} + 0.10\,\hat{\jmath})\) m carries 4.0 A in the direction of \(\mathbf{L}\), in a uniform field \(\mathbf{B} = (0.30\,\hat{\imath} - 0.40\,\hat{k})\) T. Find the force vector and its magnitude, and verify it is perpendicular to both \(\mathbf{L}\) and \(\mathbf{B}\).

Solution

\[ \mathbf{L}\times\mathbf{B} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 0.20 & 0.10 & 0\\ 0.30 & 0 & -0.40\end{vmatrix} = \hat{\imath}(-0.040 - 0) - \hat{\jmath}(-0.080 - 0) + \hat{k}(0 - 0.030) \] \[ \mathbf{F} = I\,\mathbf{L}\times\mathbf{B} = 4.0(-0.040,\ 0.080,\ -0.030) = \mathbf{(-0.16\,\hat{\imath} + 0.32\,\hat{\jmath} - 0.12\,\hat{k})\ N} \] \[ |\mathbf{F}| = \sqrt{0.16^2 + 0.32^2 + 0.12^2} = \sqrt{0.1424} = \mathbf{0.38\ N} \] \[ \mathbf{F}\cdot\mathbf{L} = -0.032 + 0.032 + 0 = 0\ \checkmark\qquad \mathbf{F}\cdot\mathbf{B} = -0.048 + 0 + 0.048 = 0\ \checkmark \]

7Curved wire: derivation8 pts

A wire bent into a semicircle of radius \(R\) lies in the upper half of the \(xy\)-plane, carrying current \(I\) from \((-R, 0)\) over the top to \((R, 0)\), in a uniform field \(\mathbf{B} = B\hat{k}\). (a) By integrating \(I\,d\mathbf{l}\times\mathbf{B}\), find the force. (b) Evaluate it for \(I = 3.0\) A, \(R = 0.10\) m, \(B = 0.25\) T. (c) The circuit is closed by a straight wire along the diameter from \((R,0)\) back to \((-R,0)\). What is the net force on the whole loop?

Solution

Position \(\mathbf{r} = R(\cos\theta, \sin\theta, 0)\) with \(\theta\) running from \(\pi\) down to \(0\); \(d\mathbf{l} = R(-\sin\theta, \cos\theta, 0)\,d\theta\).

\[ d\mathbf{l}\times B\hat{k} = RB(\cos\theta,\ \sin\theta,\ 0)\,d\theta \] \[ \mathbf{F} = IRB\int_{\pi}^{0}(\cos\theta,\ \sin\theta,\ 0)\,d\theta = IRB\left(0,\ [-\cos\theta]_{\pi}^{0},\ 0\right) = IRB(0,\ -2,\ 0) = -2IRB\,\hat{\jmath} \] \[ \text{(b)}\ |\mathbf{F}| = 2(3.0)(0.10)(0.25) = \mathbf{0.15\ N},\ \text{along } -\hat{\jmath} \]

Same as a straight wire from \((-R,0)\) to \((R,0)\): \(I(2R\hat{\imath})\times B\hat{k} = -2IRB\,\hat{\jmath}\) ✔ — in a uniform field only the endpoints matter.

(c) The diameter carries current along \(-\hat{\imath}\): force \(+2IRB\,\hat{\jmath}\). Net force on the closed loop: zero, as for any closed loop in a uniform field.

8Torque & dipole energy10 pts

A circular coil of 50 turns, radius 2.0 cm, carries 1.5 A in a uniform 0.30 T field; its magnetic moment makes \(60^\circ\) with \(\mathbf{B}\). Find (a) \(\mu\); (b) the torque; (c) the potential energy; (d) the work the field does as the coil swings to \(\theta = 0\); (e) the work needed to turn it from \(\theta = 0\) to \(180^\circ\).

Solution

\[ \text{(a)}\ \mu = NI\pi r^2 = 50(1.5)\pi(0.020)^2 = \mathbf{0.0942\ A\cdot m^2} \] \[ \text{(b)}\ \tau = \mu B\sin60^\circ = (0.0942)(0.30)(0.866) = \mathbf{0.0245\ N\cdot m} \] \[ \text{(c)}\ U = -\mu B\cos60^\circ = \mathbf{-0.0141\ J} \] \[ \text{(d)}\ W_{\text{field}} = U(60^\circ) - U(0^\circ) = -0.0141 - (-0.0283) = \mathbf{+0.0141\ J} \] \[ \text{(e)}\ W = U(180^\circ) - U(0^\circ) = 2\mu B = \mathbf{0.0565\ J} \]

The field does positive work as the coil swings toward alignment — the energy goes into rotational kinetic energy (or into whatever load the shaft drives).

9Designing a DC motor · EE application10 pts

A small DC motor has a single 200-turn coil, 3.0 cm × 2.0 cm, in a 0.25 T field, with a commutator. (a) What current gives a peak torque of 0.030 N·m? (b) What is the average torque over a revolution? (c) The coil resistance is 4.0 Ω. How much power is dissipated in the coil when the motor is held stalled at that current? (d) Why do real motors use several coils at different angles?

Solution

\[ \text{(a)}\ \tau_{\max} = NIAB\;\Rightarrow\; I = \frac{0.030}{(200)(6.0\times10^{-4})(0.25)} = \frac{0.030}{0.030} = \mathbf{1.0\ A} \] \[ \text{(b)}\ \bar\tau = \frac{1}{\pi}\int_0^{\pi}\tau_{\max}\sin\theta\,d\theta = \frac{2}{\pi}\tau_{\max} = \mathbf{0.019\ N\cdot m} \] \[ \text{(c)}\ P = I^2R = (1.0)^2(4.0) = \mathbf{4.0\ W} \]

(b) The commutator makes the torque \(\tau_{\max}|\sin\theta|\), whose average is \(2/\pi\) of the peak.

(d) One coil gives a torque that drops to zero twice a revolution (the motor can stall at a dead spot and vibrates). Several coils offset in angle fill in each other's zeros, giving a nearly constant torque and guaranteed self-starting. (c) also shows why a stalled motor overheats: all the electrical input becomes heat, and with no back EMF (Unit 8) nothing limits the current except \(R\).

10The cyclotron: derivation10 pts

(a) Show that a charged particle in a uniform field has an orbital period independent of its speed, and explain why that makes the cyclotron possible. (b) Show that a particle leaving a cyclotron of outer radius \(R\) has kinetic energy \(K = q^2B^2R^2/(2m)\). (c) For protons, \(B = 1.2\) T, \(R = 0.50\) m: find the required oscillator frequency and the final kinetic energy in MeV.

Solution

\[ \text{(a)}\ qvB = \frac{mv^2}{r}\;\Rightarrow\; r = \frac{mv}{qB},\qquad T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \]

\(v\) cancels. As the particle gains energy at each gap crossing, it moves to a larger radius but still takes the same time per half-turn, so a fixed-frequency AC voltage across the gap stays in step and kicks it forward every time.

\[ \text{(b)}\ v_{\max} = \frac{qBR}{m}\;\Rightarrow\; K = \tfrac12mv_{\max}^2 = \frac{q^2B^2R^2}{2m} \] \[ \text{(c)}\ f = \frac{eB}{2\pi m_p} = \frac{(1.60\times10^{-19})(1.2)}{2\pi(1.67\times10^{-27})} = \mathbf{18.3\ MHz} \] \[ K = \frac{\left[(1.60\times10^{-19})(1.2)(0.50)\right]^2}{2(1.67\times10^{-27})} = 2.76\times10^{-12}\ \text{J} = \mathbf{17\ MeV} \]

Final speed \(v = 5.7\times10^{7}\) m/s \(\approx 0.19c\): the mass correction is about 2 %, enough that the particle slowly falls out of step — which is why higher-energy machines vary the frequency (synchrocyclotron) or the field.

Scoring

Score	Where you are
90–100	Solid. On to the sources of magnetic field.
75–89	Physics right, signs slipping (Q1, Q5c) or sine/cosine swapped (Q3, Q8). Redo the unit-vector products in Section 1 and the angle convention in Section 7.
60–74	Device questions lost (Q4, Q5, Q9)? Re-read Section 4, Section 5 and Section 8.
< 60	Start from Section 1 and Section 2 with the lab open: \(\mathbf{F} = q\mathbf{v}\times\mathbf{B}\) and \(r = mv/(\|q\|B)\) must be automatic before the devices make sense. Then Section 6.

Where the points usually go

Losing the electron's minus sign (Q1, Q5); using the full speed instead of \(v_\perp\) or \(v_\parallel\) (Q3); quoting the radius when the question asks for the landing point \(2r\) (Q4); measuring \(\theta\) from the plane of the loop instead of its normal (Q8); and forgetting the \(2/\pi\) average for a commutated coil (Q9).