MHB Electromagnetic Induction: Problem 30.85 in Halliday and Resnick and Walker

[Ackbach]

$\newcommand{\uvec}[1]{\boldsymbol{\hat{\textbf{#1}}}}$
There is a uniform magnetic field $\mathbf{B}$ directed along the axis, and confined to a cylindrical volume of radius $R$. The magnitude of $\mathbf{B}$ is decreasing at a constant rate of $10$ mT/s. In unit-vector notation, what is the initial acceleration of an electron released at (a) point $a$ (radial distance $r=5\times 10^{-2} \, \text{m}$, (b) point $b$ (at the center), and (c) point $c$ (distance $r$ down)?

So, if you look at a cross-section of the cylinder, you see that the $\mathbf{B}$ field is directed away from you. Define $x$ to be positive to the right, $y$ positive up, and $z$ positive out. Then point $a$ is at $r\uvec{j}$ and point $b$ is at the origin, and point $c$ is at $-r\uvec{j}$.

Now then: I tried two avenues. One was that $\mathbf{F}=m\mathbf{a}=q \mathbf{v}\times\mathbf{B},$ so $\dot{\mathbf{F}}=q (\mathbf{a}\times\mathbf{B}+\mathbf{v}\times\dot{\mathbf{B}})=m\dot{\mathbf{a}}.$ But while I'm told what $\dot{\mathbf{B}}$ is, I don't know what $\mathbf{B}$ is.

The next thing I tried was Faraday's Law: $V=-\dfrac{\Phi_B}{dt}$. The problem here is that I'm not very clear on what the current loop would be. It's not stated in the problem that the cylindrical volume even has a conducting surface, so I'm not sure I can say that the cylinder itself is conducting. Moreover, it's not clear to me what surface I should use to define $\Phi_B$. Presumably it would have a constant area $A$, so that I could say $\Phi_B=BA$, and $\dot{\Phi}_B=\dot{B}A$. But how would I relate this surface/boundary combination with points $a, b,$ and $c$?

 

[I like Serena]

Your derivative of the Lorentz formula would give you the "jerk" instead of the acceleration.
And indeed, it requires the initial velocity.
However, if I understand the problem statement correctly, the initial velocity is zero.
So there would not be any Lorentz force.

Instead you may want to try the Maxwell-Faraday equation (from which Faraday's law can be derived).
In integral form it is:
$$\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$
You would be supposed to use the cylindrical symmetries to find $\mathbf{E}$.

 

[Ackbach]

$\newcommand{\uvec}[1]{\boldsymbol{\hat{\textbf{#1}}}}$

Your derivative of the Lorentz formula would give you the "jerk" instead of the acceleration.

Right: $\dot{\mathbf{a}}=\dfrac{d\mathbf{a}}{dt}=\mathbf{j}$, which is what I had there. Perhaps you didn't see the dot?

And indeed, it requires the initial velocity.
However, if I understand the problem statement correctly, the initial velocity is zero.
So there would not be any Lorentz force.

Instead you may want to try the Maxwell-Faraday equation (from which Faraday's law can be derived).
In integral form it is:
$$\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$
You would be supposed to use the cylindrical symmetries to find $\mathbf{E}$.

So if I let $\Sigma$ be a circle of radius $s$ concentric with a cross-section of the cylinder; we'll traverse the circular boundary $\partial\Sigma$ clockwise such that $\mathbf{A}=-\pi s^{2}\uvec{k}$. Then
$$-\int_{\Sigma}\frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{A}
=-A\,\frac{dB}{dt}=-\pi s^2 \frac{dB}{dt}.$$

Next, we have to compute
$$\int_{\partial\Sigma}\mathbf{E}\cdot d\mathbf{l}.$$
Surely $\mathbf{E}$ is constant on $\partial\Sigma$, by symmetry. There can't be any $z$-dependence, by symmetry, and there can't be any angular dependence. It could depend on $s$, the distance from the axis of the cylinder. There must be an azimuthal component, because if there weren't, the line integral would be zero. I want to argue that there can't be a radial or $z$ component, but I'm not entirely sure how to argue that. So far, I have that
\begin{align*}
\mathbf{E}&=E_s \uvec{s}+E_{\varphi} \hat{\mathbf{\varphi}}+E_z \uvec{k} \\
E_s&=E_s(s) \\
E_{\varphi}&=E_{\varphi}(s) \\
E_z&=E_z(s).
\end{align*}

Suppose $E_{\varphi}\not=0$, and $E_s=E_z=0$. Then
$$\int_{\partial\Sigma}\mathbf{E} \cdot d\mathbf{l}=E_{\varphi}(2\pi s).$$
It follows that
$$E_{\varphi}=\frac{-\pi s^2 dB/dt}{2\pi s}=-\frac{s}{2} \frac{dB}{dt}.$$
From this, we can compute $F=qE=ma$, so that
$$a=\frac{qE}{m}=-\frac{q}{m}\frac{s}{2} \frac{dB}{dt}.$$
The minus sign indicates (because we are taking about azimuthal direction here) that we are in the negative $\varphi$ direction, or the clockwise direction. Hence,
$$\mathbf{a}=-\frac{q}{m}\frac{s}{2} \frac{dB}{dt}\hat{\varphi}.$$

 

[I like Serena]

$\newcommand{\uvec}[1]{\boldsymbol{\hat{\textbf{#1}}}}$Right: $\dot{\mathbf{a}}=\dfrac{d\mathbf{a}}{dt}=\mathbf{j}$, which is what I had there. Perhaps you didn't see the dot?

Ah well, I only meant that the problem statement asks for acceleration, not jerk.

So if I let $\Sigma$ be a circle of radius $s$ concentric with a cross-section of the cylinder; we'll traverse the circular boundary $\partial\Sigma$ clockwise such that $\mathbf{A}=-\pi s^{2}\uvec{k}$.

I'm somewhat confused by your choices for orientation.
Since point c is supposed to be down and B is pointing the in z-direction, it seems to me that positive z should be up.
And your point c in the OP seems to refer to the wrong unit vector.

Anyway, since you are looking in the $\uvec{k}$ direction, I believe a clockwise boundary would have a positive sign.

I want to argue that there can't be a radial or $z$ component, but I'm not entirely sure how to argue that.

Perhaps you can use another equation from Maxwell's equations to argue that.

Hence,
$$\mathbf{a}=-\frac{q}{m}\frac{s}{2} \frac{dB}{dt}\hat{\varphi}.$$

Looks good!

 

[I like Serena]

Perhaps you can use another equation from Maxwell's equations to argue that.

To elaborate, from:
$$\bigcirc \!\!\!\!\!\!\!\! \iint_{\partial \Omega} \mathbf{E}\cdot\mathrm{d}\mathbf{S} = \frac{1}{\varepsilon_0} \iiint_\Omega \rho \,\mathrm{d}V$$
we can get that $E_r=0$.

And from another application of:
$$\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$
over a different contour (perpendicular to the one you have), we get that $E_z=0$.

 

[Ackbach]

To elaborate, from:
$$\bigcirc \!\!\!\!\!\!\!\! \iint_{\partial \Omega} \mathbf{E}\cdot\mathrm{d}\mathbf{S} = \frac{1}{\varepsilon_0} \iiint_\Omega \rho \,\mathrm{d}V$$
we can get that $E_r=0$.

And from another application of:
$$\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$
over a different contour (perpendicular to the one you have), we get that $E_z=0$.

Or how about just a straight Gauss's Law application?

 

[I like Serena]

Or how about just a straight Gauss's Law application?

What do you mean?

 

[Ackbach]

Well, if you draw a Gaussian cylinder inside the physical one, there'd be no enclosed charge (other than the test charge that's going to experience the force), and hence there can be no electric flux going out of the Gaussian surface or into it. But the area vectors there are precisely the $z$ and $r$ directions. Hence, there can be no $z$ or $r$ component of the electric field inside the cylinder.

 

[I like Serena]

Well, if you draw a Gaussian cylinder inside the physical one, there'd be no enclosed charge (other than the test charge that's going to experience the force), and hence there can be no electric flux going out of the Gaussian surface or into it. But the area vectors there are precisely the $z$ and $r$ directions. Hence, there can be no $z$ or $r$ component of the electric field inside the cylinder.

From symmetry we get that $E_r(a)$ would be constant on a cylinder with radius $a$.
And $E_z(s)$ would be the same at both caps of the cylinder.

The contribution of $E_z(s)$ on both caps would cancel, since at one end it is positive and at the other end it is negative.
As a consequence $E_r(a)$ must be zero.

However, this does not tell us anything about $E_z(s)$ yet.