Exercise 120519

Problem 1 (on demand)

In a circular region, there is a uniform magnetic field ${\vec {B}}$ pointing into the page. An xy coordinate system has its origin at the circular region’s center. A free positive point charge $Q=+1.0\mu$C is initially at rest at a position $x=+10$ cm on the x axis. If the magnitude of the magnetic field is now decreased at a rate of $-0.10$ T/s, what force (magnitude and direction) will act on $+Q$?

Is this force conservative?

Solution

Remember the Lenz's law

${\mathcal {E}}=-{\frac {d\Phi _{B}}{dt}}$

Combining this with the definition of emf, $V_{ab}=\int _{a}^{b}{\vec {E}}\cdot d{\vec {l}}$

$\oint {\vec {E}}\cdot d{\vec {l}}=-{\frac {d\Phi _{B}}{dt}}$

i.e. a changing magnetic field produces an electric field.

Choose a circular path centered at the origin with radius 10 cm. By symmetry the electric field is uniform along this path and is parallel to the path. We then use Eq. 29-8 to calculate the electric field at each point on this path. From the electric field we calculate the force on the charged particle.

$\oint {\vec {E}}\cdot d{\vec {l}}=E(2\pi r)=-{\frac {d\Phi _{B}}{dt}}=-(\pi r^{2}){\frac {dB}{dt}}$

$F=QE=-Q{\frac {r}{2}}{\frac {dB}{dt}}=-(1.0\times 10^{-6}{\textrm {C}}){\frac {0.10{\textrm {m}}}{2}}(-0.10{\textrm {T/s}})=50{\textrm {N}}$

Since the magnetic field points into the page and is decreasing, Lenz’s law tells us that an induced circular current centered at the origin would flow in the clockwise direction. Therefore, the initial force on a positive charge along the positive x-axis would be down, or in the $-{\hat {j}}$ direction.

In the electrostatic case, the potential difference between two points is

$V_{ba}=V_{b}-V_{a}=-\int _{a}^{b}{\vec {E}}\cdot d{\vec {l}}$

In the case of a circle since the start and end points are the same,

$\oint {\vec {E}}\cdot d{\vec {l}}=0$

the relation above, $\oint {\vec {E}}\cdot d{\vec {l}}=0$, tells us that the work done per unit charge around any closed path is zero (or the work done between any two points is independent of path), which is a property only of a conservative force. But in the nonelectrostatic case, when the electric field is produced by a changing magnetic field, the integral around a closed path is not zero

$\oint {\vec {E}}\cdot d{\vec {l}}=-{\frac {d\Phi _{B}}{dt}}$

thus this force is not conservative.

Problem 2

Figure shows a parallel-plate capacitor of plate area $A=10.5$ cm 2 and plate separation $2d=7.12$ mm. The left half of the gap is filled with material of dielectric constant $\kappa _{1}=21.0$ ; the top of the right half is filled with material of dielectric constant $\kappa _{2}=42.0$ ; the bottom of the right half is filled with material of dielectric constant $\kappa _{3}=58.0$.What is the capacitance?

Solution

Let

$C_{1}=\epsilon _{0}(A/2)\kappa _{1}/2d=\epsilon _{0}A\kappa _{1}/4d$

$C_{2}=\epsilon _{0}(A/2)\kappa _{2}/d=\epsilon _{0}A\kappa _{2}/2d$

$C_{3}=\epsilon _{0}(A/2)\kappa _{3}/d=\epsilon _{0}A\kappa _{3}/2d$

Note that $C_{2}$ and $C_{3}$ are effectively connected in series, while $C_{1}$ is effectively connected in parallel with the $C_{2}--C_{3}$ combination. Thus,

$C=C_{1}+{\frac {C_{2}C_{3}}{C_{2}+C_{3}}}={\frac {\epsilon _{0}A}{4d}}\left(\kappa _{1}+{\frac {2\kappa _{2}\kappa _{3}}{\kappa _{2}+\kappa _{3}}}\right)$

Plugging in the values

$C={\frac {(8.85\times 10^{-12}{\textrm {C}}^{2}/{\textrm {N.m}}^{2})(1.05\times 10^{-3}{\textrm {m}}^{2})}{4(3.56\times 10^{-3}{\textrm {m}}}}\left(21+{\frac {2(42)(58)}{42+58}}\right)=4.55\times 10^{-11}{\textrm {F}}$

Problem 3

A group of $N$ identical batteries of emf ${\mathcal {E}}$ and internal resistance $r$ may be connected all in series or all in parallel and then across a resistor $R$. Show that both arrangements give the same current in $R$ if $R=r$

Solution

When all the batteries are connected in parallel, the emf is ${\mathcal {E}}$ and the equivalent resistance is $R_{\textrm {parallel}}=R+r/N$ so the current is

$i_{\textrm {parallel}}={\frac {\mathcal {E}}{R_{\textrm {parallel}}}}={\frac {\mathcal {E}}{R+r/N}}={\frac {N{\mathcal {E}}}{NR+r}}$

Similarly, when all the batteries are connected in series, the total emf is $N{\mathcal {E}}$ and the equivalent resistance is $R_{\textrm {series}}=R+Nr$ Therefore,

$i_{\textrm {series}}={\frac {N{\mathcal {E}}}{R_{\textrm {series}}}}={\frac {N{\mathcal {E}}}{R+Nr}}$

hence when $R=r$

$i_{\textrm {parallel}}=i_{\textrm {series}}={\frac {N{\mathcal {E}}}{(N+1)r}}$

Problem 4

Find an expression for the oscillation frequency of an electric dipole of dipole moment ${\vec {p}}$ and rotational inertia $I$ for small amplitudes of oscillation about its equilibrium position in a uniform electric field of magnitude $E$.

Solution

$\tau =-pE\sin \theta$

this is a restoring torque, trying to bring the tilted dipole back to its aligned equilibrium position. If the amplitude of the motion is small, $\sin \theta \approx \theta$ and

$\tau \approx -pE\theta$

since this exhibits a simple negative proportionality to the angle of rotation, the dipole oscillates in simple harmonic motion, like a torsional pendulum with torsion constant $\kappa =pE$.

$\theta (t)=\theta _{m}\cos(\omega t+\phi )$

${\frac {d}{dt}}\theta (t)=-\omega \theta _{m}\sin(\omega t+\phi )$

${\frac {d^{2}}{dt^{2}}}\theta (t)=-\omega ^{2}\theta _{m}\cos(\omega t+\phi )$

The angular frequency $\omega$ is given by

$\omega ^{2}={\frac {\kappa }{I}}={\frac {PE}{I}}$

where $I$ is the rotational inertia of the dipole. The frequency of oscillation is

$f={\frac {\omega }{2\pi }}={\frac {1}{2\pi }}{\sqrt {\frac {PE}{I}}}$