Homework 4 · Magnetic Force and Current-Generated Fields

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New geometry diagrams for Ch. 8, Ex. 3 (left), and Ch. 9, Ex. 9 (right), from the supplied exercises; figure record.

Ch. 8, Ex. 3 · A loop partly in a magnetic field

Problem. A field \(\mathbf B=B\hat{\mathbf z}\) exists only in the region \(y\geq a\) of the \(xy\) plane. A circular loop of radius \(R\), centered at the origin, carries current \(I\) counterclockwise seen from \(+z\), with \(0\leq a<R\). Find its net force.

Method: Force on a current element.

Solution. Only the upper arc is in the field. Parameterize it by \(\mathbf r=R(\cos\theta,\sin\theta,0)\), with \(\alpha=\arcsin(a/R)\) and \(\alpha\leq\theta\leq\pi-\alpha\). Since \(\mathbf B\) is constant on that arc,

\[ \mathbf F=I\int_{\rm arc}d\boldsymbol\ell\times\mathbf B =I(\mathbf r_{\rm end}-\mathbf r_{\rm start})\times B\hat{\mathbf z}. \]

The displacement along the current is \(-2\sqrt{R^2-a^2}\hat{\mathbf x}\), so

\[ \boxed{\mathbf F=2IB\sqrt{R^2-a^2}\hat{\mathbf y}.} \]

It reduces to \(2IBR\hat{\mathbf y}\) for the upper semicircle (\(a=0\)), and vanishes as \(a\to R\). 不能把整圆代入 \(I\oint d\boldsymbol\ell=0\)；下半部分没有磁场。

Ch. 8, Ex. 4 · Torque on a rotating charged disk

Problem. A plastic disk of radius \(R\) has uniform surface charge density \(\sigma\) and rotates about its axis with angular velocity \(\boldsymbol\omega\). A uniform field \(\mathbf B\) is perpendicular to the rotation axis. Find the torque magnitude.

Method: Current-loop moment and torque.

Solution. Split the disk into rings. For \(\omega=|\boldsymbol\omega|\),

\[ dq=2\pi\sigma r\,dr,\qquad dI=\frac{\omega}{2\pi}dq=\sigma\omega r\,dr, \qquad d\mu=\pi r^2dI. \]

Integrating gives the signed vector moment

\[ \boldsymbol\mu=\frac{\pi\sigma R^4}{4}\boldsymbol\omega, \qquad \boldsymbol\tau=\boldsymbol\mu\times\mathbf B, \qquad \boxed{|\boldsymbol\tau|=\frac{\pi|\sigma|\omega BR^4}{4}.} \]

The moment follows \(\boldsymbol\omega\) for positive charge and opposes it for negative charge. The disk is insulating: charges rotate with the material rather than redistributing as on a conductor.

Ch. 8, Ex. 7 · Motion from rest in crossed fields

Problem. Uniform fields are \(\mathbf E=E\hat{\mathbf x}\) and \(\mathbf B=B\hat{\mathbf z}\), \(E,B>0\). A particle of mass \(m\) and charge \(q\ne0\) starts at rest at the origin. Ignore gravity. (a) Find the mean velocity over one cyclotron period \(T=2\pi m/(|q|B)\) and show it is independent of \(q,m\). (b) Find the largest \(|x|\) and the sign of \(q\) for which the particle moves toward \(+x\).

Method: Lorentz force and crossed fields.

Solution. (a) First derive the motion using the signed frequency \(\omega_s=qB/m\). The equations are

\[ \dot v_x=\omega_s(v_y+E/B),\qquad \dot v_y=-\omega_sv_x,\qquad v_z=0. \]

With zero initial velocity, the shifted vector \((v_x,v_y+E/B)\) rotates uniformly:

\[ \boxed{v_x=\frac EB\sin\omega_st,\qquad v_y=\frac EB(\cos\omega_st-1),\qquad v_z=0.} \]

Integrating with \(\mathbf r(0)=0\),

\[ \boxed{x=\frac{E}{B\omega_s}(1-\cos\omega_st),\qquad y=\frac{E}{B\omega_s}\sin\omega_st-\frac EB t,\qquad z=0.} \]

The positive period and drift are

\[ \boxed{T=\frac{2\pi m}{|q|B},\qquad \langle\mathbf v\rangle=\frac{\mathbf r(T)-\mathbf r(0)}T =-\frac EB\hat{\mathbf y} =\frac{\mathbf E\times\mathbf B}{B^2}.} \]

The drift does not depend on \(q\) or \(m\), although the cyclotron motion does.

(b) Since \(0\leq1-\cos\omega_st\leq2\),

\[ \boxed{|x|_{\max}=\frac{2mE}{|q|B^2}.} \]

Positive charges stay on the \(x\geq0\) side; negative charges stay on \(x\leq0\). 漂移方向对两种电荷相同，横向偏转方向相反。

Ch. 9, Ex. 2 · Field of a rotating charged spherical shell

Problem. A spherical surface of radius \(R\) carries uniformly distributed total charge \(Q\). It rotates about \(z\) with angular speed \(\omega\), counterclockwise seen from \(+z\). Find the magnetic field magnitude on the axis at \((0,0,z)\) for \(|z|<R\) and \(|z|>R\).

Method: Biot–Savart superposition and circular-loop axial field.

Solution. Write \(\sigma=Q/(4\pi R^2)\) and build the shell from rings at polar angle \(\theta\). Each ring has radius \(R\sin\theta\), height \(R\cos\theta\), and current

\[ dI=\frac{\omega}{2\pi}(\sigma\,2\pi R^2\sin\theta\,d\theta) =\sigma\omega R^2\sin\theta\,d\theta. \]

Its axial field contribution gives

\[ B_z=\frac{\mu_0\sigma\omega R^4}{2} \int_0^\pi\frac{\sin^3\theta\,d\theta} {(R^2+z^2-2Rz\cos\theta)^{3/2}}. \]

To evaluate, set \(s=|z|\), \(u=\cos\theta\), \(A=R^2+s^2\), \(D=2Rs\). For \(s>0\),

\[ \begin{aligned} \mathcal I&=\int_{-1}^1\frac{1-u^2}{(A-Du)^{3/2}}du\\ &=\frac2{D^3}\left[-\frac{D^2-A^2}{w}+2Aw-\frac{w^3}{3}\right]_{|R-s|}^{R+s} =\frac4{3\max(R,s)^3}, \end{aligned} \]

where \(w=\sqrt{A-Du}\). At \(s=0\) the same result follows directly from \(\int_0^\pi\sin^3\theta\,d\theta=4/3\). Thus

\[ \boxed{\mathbf B(0,0,z)=\begin{cases} \dfrac{\mu_0Q\omega}{6\pi R}\hat{\mathbf z},&|z|<R,\\[3pt] \dfrac{\mu_0Q\omega R^2}{6\pi|z|^3}\hat{\mathbf z},&|z|>R. \end{cases}} \]

For the requested magnitude replace \(Q\) by \(|Q|\). The interior axial field is constant. The exterior result agrees with a dipole \(\mu=Q\omega R^2/3\). This is a surface charge distribution, not a uniformly charged solid ball.

Ch. 9, Ex. 7 · Ampère paths with winding numbers

Problem. Wire \(A\) carries \(i_1>0\) out of the page; wire \(B\) carries \(i_2>0\) into it. Find \(\oint\mathbf B\cdot d\boldsymbol\ell\) for (a) a path winding twice counterclockwise around \(A\), never around \(B\); (b) a figure-eight whose side-by-side lobes meet at one genuine crossing, with the lobe about \(A\) traversed counterclockwise and the other surrounding \(B\).

Method: Ampère's law and orientation.

Solution. For an oriented planar path,

\[ \oint\mathbf B\cdot d\boldsymbol\ell =\mu_0\sum_j n_jI_{j,z}, \]

where \(n_j\) is its signed winding number and \(I_{j,z}\) is positive out of the page.

(a) \(n_A=2,n_B=0\), hence \(\boxed{\oint\mathbf B\cdot d\boldsymbol\ell=2\mu_0i_1}\).

(b) A genuine crossing makes the other lobe clockwise, so \(n_A=1,n_B=-1\), while \(I_{B,z}=-i_2\). Therefore

\[ \boxed{\oint\mathbf B\cdot d\boldsymbol\ell=\mu_0[i_1+(-1)(-i_2)] =\mu_0(i_1+i_2).} \]

The two signs for the second lobe cancel. 自交环路按各叶片的有向绕数计算，不能只看几何上圈住了哪些电流。

Ch. 9, Ex. 9 · Force across a toroid's finite width

Problem. A toroid has \(N\) turns, current \(I\), major radius \(R\), and circular cross-section radius \(a<R\). Its mid-plane is \(xy\) and its internal field circulates counterclockwise seen from \(+z\). A straight test wire runs along the positive \(x\) axis from \(R-a\) to \(R+a\), carrying current \(i\) toward \(+x\). (a) Find the force on this segment. (b) Compare its magnitude with \(2aiB_0\), where \(B_0=\mu_0NI/(2\pi R)\); show which is larger.

Method: Toroid field and wire force.

Solution. (a) At \(x>0\), \(\mathbf B(x)=\mu_0NI\hat{\mathbf y}/(2\pi x)\). Thus

\[ \boxed{\mathbf F=i\int_{R-a}^{R+a}dx\,\hat{\mathbf x}\times\mathbf B(x) =\frac{\mu_0NIi}{2\pi}\ln\frac{R+a}{R-a}\hat{\mathbf z}.} \]

(b) Put \(u=a/R\in(0,1)\). Then

\[ \frac{F}{2aiB_0}=\frac1{2u}\ln\frac{1+u}{1-u} =1+\frac{u^2}{3}+\frac{u^4}{5}+\cdots>1. \]

Hence \(\boxed{F>2aiB_0}\). The center-field approximation underestimates the force, tending to equality only as \(a/R\to0\).