Suppose you have a circuit driven by a voltage:
$$V(t)=V_0\cos(\omega t)$$
You observe the resulting current is:
$$I(t) = I_0\cos(\omega t-\pi/4)$$
Would you say the current is
1. leading
2. lagging
the voltage by 45 degrees?
Note:
* Correct Answer: B
## Announcements
* Quiz 3 (Friday 2/14) - RLC circuits
* Solve a circuit problem using the phasor method
* Discuss limits on the response and how it might act as a filter
Two LR circuits driven by an AC power supply are shown below.
<img src="./images/two_RL_circuits.png" align="center" style="width: 800px";/>
Which circuit is a low pass filter?
1. The left circuit
2. The right circuit
3. Both circuits
4. Neither circuit
Note:
* Correct Answer: B
Two RC circuits driven by an AC power supply are shown below.
<img src="./images/two_RC_circuits.png" align="center" style="width: 800px";/>
Which circuit is a high pass filter?
1. The left circuit
2. The right circuit
3. Both circuits
4. Neither circuit
Note:
* Correct Answer: A
Ampere's Law relates the line integral of B around some closed path, to a current flowing through a surface bounded by the chosen closed path.
$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I$
By calling it a "Law", we expect that:
1. It is neither correct nor useful.
2. It is sometimes correct and sometimes easy to use.
3. It is correct and sometimes easy to use.
4. It is correct and always easy to use.
5. None of the above.
Note:
* Correct Answer: C
Take the divergence of the curl of any (well-behaved)
vector function $\mathbf{F}$, what do you get?
$$\nabla \cdot \left(\nabla \times \mathbf{F}\right) = ???$$
1. Always 0
2. A complicated partial differential of $\mathbf{F}$
3. The Laplacian: $\nabla^2 \mathbf{F}$
4. Wait, this vector operation is ill-defined!
Note:
* Correct Answer: A
Take the divergence of both sides of Faraday's law:
$$\nabla \times \mathbf{E} = -\dfrac{\partial \mathbf{B}}{\partial t}$$
What do you get?
1. 0 = 0 (is this interesting?)
2. A complicated partial differential equation (perhaps a wave equation of some sort ?!) for $\mathbf{B}$
3. Gauss’ law!
4. ???
Note:
* Correct Answer: A
Take the divergence of both sides of Ampere's law:
$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$$
**According to this**, the divergence of $\mathbf{J}$ is:
1. $-\partial \rho/\partial t$
2. A complicated partial differential of $\mathbf{B}$
3. Always 0
4. ???
Note:
* Correct Answer: C
Ampere's Law relates the line integral of $\mathbf{B}$ around some closed path, to a current flowing through a surface bounded by the chosen closed path.
$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I$
The **path** can be:
1. Any closed path
2. Only circular paths
3. Only sufficiently symmetrical paths
4. Paths that are parallel to the B-field direction.
5. None of the above.
Note:
* Correct Answer: A
Ampere's Law relates the line integral of $\mathbf{B}$ around some closed path, to a current flowing through a surface bounded by the chosen closed path.
$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I$
The **surface** can be:
1. Any closed bounded surface
2. Any open bounded surface
3. Only surfaces perpendicular to $\mathbf{J}$.
4. Only surfaces tangential to the B-field direction.
5. None of the above.
Note:
* Correct Answer: B
Rank order $\int \mathbf{J} \cdot d\mathbf{A}$ (over blue surfaces) where $\mathbf{J}$ is uniform, going left to right:
<img src="./images/current_surfaces.png" align="center" style="width: 600px";/>
1. iii > iv > ii > i
2. iii > i > ii > iv
3. i > ii > iii > iv
4. Something else!!
5. Not enough info given!!
Note:
* CORRECT ANSWER: D
* They are all the same!
We are interested in $\mathbf{B}$ on the dashed "Amperian loop", and plan to use $\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_t$ to figure it out. What is $I_t$ here?
<img src="./images/wire_w_cap.png" align="center" style="width: 700px";/>
1. $I$
2. $I/2$
3. 0
4. Something else
Note:
* Correct Answer: A
We are interested in $\mathbf{B}$ on the dashed "Amperian loop", and plan to use $\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_t$ to figure it out. What is $I_t$ here? *The surface over which we integrate $\mathbf{J}\cdot d\mathbf{A}$ is shown
in blue.*
<img src="./images/cap_w_wire_highlighted.png" align="center" style="width: 700px";/>
1. $I$
2. $I/2$
3. 0
4. Something else
Note:
* Correct Answer: A
We are interested in $\mathbf{B}$ on the dashed "Amperian loop", and plan to use $\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_t$ to figure it out. What is $I_t$ here? *The surface over which we integrate $\mathbf{J}\cdot d\mathbf{A}$ is shown
in blue.*
<img src="./images/cap_soap_bubble.png" align="center" style="width: 700px";/>
1. $I$
2. $I/2$
3. 0
4. Something else
Note:
* Correct Answer: C
The complete differential form of Ampere's Law is now argued to be:
$$\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \epsilon_0\dfrac{\partial \mathbf{E}}{\partial t}$$
The integral form of this equation is:
1. $\iint \mathbf{B}\cdot d\mathbf{A} = \mu_0 I + \mu_0 \epsilon_0 \frac{d}{dt}\oint \mathbf{E}\cdot d\mathbf{l}$
2. $\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I + \mu_0 \epsilon_0 \frac{d}{dt}\oint \mathbf{E}\cdot d\mathbf{l}$
3. $\iint \mathbf{B}\cdot d\mathbf{A} = \mu_0 I + \mu_0 \epsilon_0 \frac{d}{dt}\iint \mathbf{E}\cdot d\mathbf{A}$
4. $\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I + \mu_0 \epsilon_0 \frac{d}{dt}\iint \mathbf{E}\cdot d\mathbf{A}$
5. Something else/???
Note:
* Correct Answer: D
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