How amazing is that $\dfrac{1}{\sqrt{\epsilon_0 \mu_0}} = 3\times10^8 \mathrm{m/s}$? 1. OMGBBQPIZZA, so amazing! 2. It's pretty cool 3. Meh 4. Whatever
## Correct Answer OMGBBQPIZZA, so amazing!
What do you want to do today? 1. Clickers and lecture 2. Tutorial *Either way, we are covering the same example.*
Consider a large parallel plate capacitor as shown, charging so that $Q = Q_0+\beta t$ on the positively charged plate. Assuming the edges of the capacitor and the wire connections to the plates can be ignored, what is the direction of the magnetic field $\mathbf{B}$ halfway between the plates, at a radius $r$? <img src="./images/charging_cap_dEdt.png" align="left" style="width: 400px";/> 1. $\pm \hat{\phi}$ 2. 0 3. $\pm \hat{z}$ 4. $\pm \hat{s}$ 5. ??? Note: * Correct Answer: A
Same capacitor with $Q = Q_0+\beta t$ on the positively charged plate. What is the direction of the magnetic field $\mathbf{B}$ halfway between the plates, at a radius $r$? <img src="./images/charging_cap_dEdt.png" align="left" style="width: 400px";/> 1. $+\hat{\phi}$ 2. $-\hat{\phi}$ 3. Not sure how to tell Note: * Correct Answer: A
Same capacitor with $Q = Q_0+\beta t$ on the positively charged plate. What kind of amperian loop can be used between the plates to find the magnetic field $\mathbf{B}$ halfway between the plates, at a radius $r$? <img src="./images/charging_cap_dEdt_loop.png" align="center" style="width: 500px";/> D) A different loop E) Not enough symmetry for a useful loop Note: * Correct Answer: B
Same capacitor with $Q = Q_0+\beta t$ on the positively charged plate. What is the magnitude of the magnetic field $\mathbf{B}$ halfway between the plates, at a radius $r$? <img src="./images/charging_cap_dEdt_smaller.png" align="left" style="width: 400px";/> 1. $\frac{\mu_0 \beta}{2 \pi r}$ 2. $\frac{\mu_0 \beta r}{2 d^2}$ 3. $\frac{\mu_0 \beta d}{2 a^2}$ 4. $\frac{\mu_0 \beta a}{2 \pi r^2}$ 5. None of the above Note: * Correct Answer: E
<img src="./images/cap_plates_boxed.png" align="right" style="width: 400px";/> Consider the surface of an imaginary volume (dashed lines, at right) that partly encloses the left capacitor plate. For this closed surface, is the total flux of the current density, $\iint \mathbf{J} \cdot d\mathbf{A}$ positive, negative or zero? 1. Positive 2. Negative 3. Zero Note: * Correct Answer: B * The charge density is increasing, so the total charge in the region is increasing with time, thus the flux of the current density must be negative (conservation of charge).
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 3**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: A * There is no current or charge inthe region
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 2**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: D * One way to think about this is that the charge on the plate is increasing (so the density is increasingly more positive - positive charge). So the divergence of the current density is negative (jives with previous CQ also).
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 4**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: E * One way to think about this is that the charge on the plate is increasing (so the density is increasingly more negative - negative charge). So the divergence of the current density is positive.
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 1**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: A * The divergence of the curl of the magnetic field is zero and there's no electric field out there, so the divergence of J must vanish.
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> At each location, we will evaluate the sign of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$. **At location 5**, the signs of $\partial \rho/\partial t$ and $\nabla \cdot \mathbf{J}$ are: 1. both zero 2. both negative 3. both positive 4. $\partial \rho/\partial t$ is positive and $\nabla \cdot \mathbf{J}$ is negative 5. $\partial \rho/\partial t$ is negative and $\nabla \cdot \mathbf{J}$ is positive *Recall that charge is conserved locally!* Note: * Correct Answer: A * The divergence of the curl of the magnetic field is zero and there's no electric field out there, so the divergence of J must vanish.
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> Suppose the original Ampere's law $\nabla \times \mathbf{B} = \mu_0\mathbf{J}$ were correct without any correction from Maxwell (it’s not, but suppose for a moment that it is). What would this imply about $\nabla \cdot \mathbf{J}$ at points 2 and 4 in the diagram? 1. The remain unchanged 2. They swap signs 3. They become zero 4. ??? Note: * Correct Answer: C
<img src="./images/5_locations_charging_cap.png" align="left" style="width: 400px";/> Let's continue with the (incomplete) definition of Ampere's Law: $\nabla \times \mathbf{B} = \mu_0\mathbf{J}$. What does this form tell you about the signs of $(\nabla \times \mathbf{B})_x$ at locations 1, 3, and 5? 1. All positive 2. All negative 3. Positive at 1 and 5, zero at 3 4. Negative at 1 and 5, zero at 3 5. Something else Note: * Correct Answer: C * All we need here is the direction of Jx, which is zero at 3