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Selected Problems on DC-circuit

Nerdy August 21st, 2008

From Problems and Solutions on Electromagnetism by Lim Yung-Kuo

3012

3013

3014

Solution : 3012

Solution : 3013

Solution : 3014

Tags: IPST, Physics, Problems, Teaching, Triam Udom

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Selected AC-circuit problems from Feynman’s Lectures on Physics Exercise Vol 2

Nerdy August 18th, 2008

Update : Solutions added in comments, 20/08/08

22-3

22-6

Tags: IPST, Physics, Problems, Teaching, Triam Udom

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B9 IB Adv P1 2007

Nerdy August 13th, 2008

For time-varying fields, the generalisation of Amperes law is

$\oint_{}^{}\mathbf{H}\cdot d\mathbf{l}=\int \mathbf{J}\cdot d\mathbf{S}+\int \frac{\partial \mathbf{D} }{\partial t}\cdot d\mathbf{S}$
where,
$\mathbf{H}$ is the magnetic field strength, $\mathbf{B}=\mu_{0} \mu_{r} \mathbf{H}$
$\mathbf{J}$ is the current density.
$\mathbf{D}$ is the displacement or electric flux density, $\mathbf{D}=\epsilon_{0} \epsilon_{r} \mathbf{E}$
A parallel-plate capacitor is formed from two large circular coaxial metal discs of radius a and separation d (d âª a). The capacitor is initially charged. A thin straight wire of high resistance R is connected between the centres of the discs and the capacitor discharges slowly.

Draw a labelled diagram showing the configuration of the electric and magnetic fields and Poynting vector in the space between the discs as the capacitor discharges.

Show that when the potential difference between the plates is V the
magnetic field at a distance r (r<<a) from the wire is given by

$H=\frac{V}{2 \pi r R}\left(1-\frac{r^2}{a^2} \right)$
Calculate the Poynting vector and the total power crossing a cylindrical
surface of radius r between the plates.
Discuss the energy flow and dissipation in this situation.

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Poynting vector

Nerdy August 13th, 2008

an excerpt from the note by Prof S. Withington, 2007

The magnitude of P gives the power ﬂow per unit area, and the direction of P gives the direction in which the power is ﬂowing.

The Poynting vector,

$\mathbf{P}(r)=\mathbf{E}(r) \times \mathbf{H}(r)$
is a vector ﬁeld that gives the direction, and energy ﬂux, of an electromagnetic ﬁeld.
Notice that it is not linear in the ﬁeld, and thereforediﬀerent Poynting ﬁelds cannot be superposed.
Consider a plane electromagnetic wave travelling in thez direction; say
$\mathbf{E}=\hat{\mathbf{i}} E_{x}$
$\mathbf{H}=\hat{\mathbf{j}} H_{y}$
giving
$\mathbf{P}=\hat{\mathbf{k}} E_{x} H_{y} = \hat{\mathbf{k}} E_{x}^2 \sqrt{\frac{\epsilon_{0}}{\mu_{0}}}$
Also
$\left|\mathbf{P} \right|= E_{x} H_{y} = E_{x}^2 \sqrt{\frac{\epsilon_{0}}{\mu_{0}}} = \epsilon_{0} E_{x}^2 c =Uc$
where $U=\epsilon_{0} E_{x}^2$ is the energy desnsity
Again we see that the Poynting vector gives the rate of ﬂow of energy per unit area.

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B8 IB Adv P1 2005

Nerdy August 13th, 2008

A pair of thin coaxial coils A and B, of radii a and b, and with $N_{a}$ and $N_{B}$ turns, respectively are separated by a distance L. Show that, if b<<a the mutual inductance is

$M= \frac{\mu _{0} N_{A} N_{B} \pi a^2 b^2}{2(a^2+L^2)^\frac{3}{2}}$
What is the magnetic dipole moment of coil B when a current $I_{B}$ flows in it?

In a simple nuclear magnetic resonance experiment, atoms confined to a small sample chamber each possess a magnetic dipole moment of fixed magnitude $m_{0}$ whose direction precesses about the z-axis at angular frequency $\omega$ , at constant angle $\theta$ to that axis. At a distance $x_{0}$ from the sample chamber is a single-turn coil of radius a, with its axis on the x-axis. Write down an expression for the Cartesian components of the magnetic dipole moment of an atom at time t. Derive an expression for the r.m.s. voltage per atom induced in the coil.

In an experiment to map the human lungs, nuclear magnetic resonance is performed on xenon gas that is inhaled by a subject. Each xenon nucleus has a magnetic dipole moment of magnitude $3 \times 10^{-26} Am^2$ and precesses at a frequency of 400kHz. Stating carefully any assumptions you made, estimate the smallest volume of gas that could be detected by a single-turn coil with a voltage sensitivity of $1 \mu V$

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B7 IB Adv P1 2004

Nerdy August 13th, 2008

Update : Solution added in the comment. 18/08/08

In a magnetic braking system, a small, square-shaped wire loop of side a and resistance R is attached to the outer edge of an insulating cylinder rotating about its own axis with angular velocity $\omega$ , as shown in the diagram. On each rotation, the wire loop passes through a region containing a uniform magnetic field B directed perpendicular to the plane of the loop. The field region has a square cross section slightly larger than the size of the loop. The velocity $u=b \omega$ of the loop decreases by only a small amount during each pass through the magnetic field.

Sketch the time dependence of the total force F acting on the loop during each pass through the magnetic field. Determine an impulse exerted on the loop on each pass.

Show that the reduction in the total rotational kinetic energy of the system on each pass due to the impulse exerted on the loop is equal to the energy dissipated by the current flowing around the loop.

[The self-inductance of the wire loop can be neglected.]

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Completeness of Maxwell’s equations

Nerdy August 13th, 2008

an excerpt from the note by Prof S. Withington, 2007

We have studied the behaviour of electrostatic and magnetostatic ﬁelds, and have considered how time-varying magnetic ﬁelds lead to voltages.
We have discussed a number of equations that seem to solve adequately all of the physical problems encountered.
In particular, we have derived a set of equations that describe electrical circuit theory.
The most important equations are
$\bigtriangledown \cdot \mathbf{D}=\rho$
$\bigtriangledown \cdot \mathbf{B}=0$
$\bigtriangledown \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$
$\bigtriangledown \times \mathbf{H}=\mathbf{J}$
and we might be content having reached this point, but there is a big problem!
This set of ﬁeld equations is not consistent with the conservation of charge.
We know that for charged to be conserved

$\oint_{} d\tau \bigtriangledown \cdot \mathbf{J}+\int d\tau \frac{\partial \rho }{\partial t}=0$
but because this must be true for all volumes, we have the continuity equation
$\bigtriangledown \cdot \mathbf{J}+\frac{\partial \rho }{\partial t}=0$
In the steady state, the charge density does not change anywhere with time, and
$\bigtriangledown \cdot \mathbf{J}=0$
showing that currents form closed circulating loops.
Using the last of the ﬁeld equations to evaluate the continuity equation gives
$\bigtriangledown \cdot \mathbf{J}=\bigtriangledown \cdot (\bigtriangledown \times \mathbf{H}) = 0$
because the divergence of the curl of every vector ﬁeld is zero, but this means that the continuity equation cannot be satisﬁed for anything other than the steady state magnetostatics.
Oh dear, the whole of the scheme seems to have fallen apart just when we were doing so well!
All of the best physics comes out of unforseen problems, so what is going on?
Maxwell suggested that the last of the ﬁeld equations should be modiﬁed by adding an additional term ∂ D/∂ t,
$\bigtriangledown \times \mathbf{H} =\mathbf{J}+\frac{\partial \mathbf{D}}{\partial t}$
which he called the displacement current.
The continuity equation then becomes
$\bigtriangledown \cdot (\bigtriangledown \times \mathbf{H}) =\bigtriangledown \cdot (\mathbf{J}+\frac{\partial \mathbf{D}}{\partial t})$
$\bigtriangledown \cdot (-\frac{\partial \mathbf{D}}{\partial t})+\frac{\partial \rho }{\partial t}=0$
$-\frac{\partial \bigtriangledown \cdot \mathbf{D}}{\partial t}+\frac{\partial \rho }{\partial t}=0$
which is consistent with the ﬁrst of the ﬁeld equations, and so the equations are self consistent even when charge is accumulating or decreasing within some region.
The adjustment seems rather simple, but was one of Maxwell’s greatest contributions, and led to a
revolution in physics.
Indeed, we have ﬁnally arrived at Maxwell’s equations:
$\bigtriangledown \cdot \mathbf{D}=\rho$
$\bigtriangledown \cdot \mathbf{B}=0$
$\bigtriangledown \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$
$\bigtriangledown \times \mathbf{H}=\mathbf{J}+\frac{\partial \mathbf{D}}{\partial t}$
Maxwell’s equations are self consistent, and describe every possible physical phenomena in classical electromagnetism.

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