Gauss' Law for magnetism, $\nabla \cdot \mathbf{B} = 0$ suggests we can generate a potential for $\mathbf{B}$. What form should the definition of this potential take ($\Phi$ and $\mathbf{A}$ are placeholder scalar and vector functions, respectively)? 1. $\mathbf{B} = \nabla \Phi$ 2. $\mathbf{B} = \nabla \times \Phi$ 3. $\mathbf{B} = \nabla \cdot \mathbf{A}$ 4. $\mathbf{B} = \nabla \times \mathbf{A}$ 5. Something else?! Note: * CORRECT ANSWER: D
## Announcements * Danny will be out Monday * Norman Birge will cover class
Consider a toroid, which is like a finite solenoid connected end to end. In which direction do you expect the B field to point? <img src="./images/toroid.png" align="right" style="width: 300px";/> 1. Azimuthally ($\hat{\phi}$ direction) 2. Radially ($\hat{s}$ direction) 3. In the $\hat{z}$ direction (perp. to page) 4. Loops around the rim 5. Mix of the above... Note: * CORRECT ANSWER: A
<img src="./images/toroid_loops.png" align="right" style="width: 300px";/> Which Amperian loop would you draw to find B “inside” the Torus (region II)? 1. Large "azimuthal" loop 2. Smallish loop from region II to outside (where B=0) 3. Small loop in region II 4. Like A, but perp to page 5. Something entirely different Note: * CORRECT ANSWER: A
With $\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}$, we can write (in Cartesian coordinates): $$\nabla^2 A_x = -\mu_0 J_x$$ Does that also mean in spherical coordinates that $\nabla^2 A_r = -\mu_0 J_r$? 1. Yes 2. No Note: * CORRECT ANSWER: B
We can compute $\mathbf{A}$ using the following integral: $\mathbf{A}(\mathbf{r}) = \dfrac{\mu_0}{4\pi}\int \dfrac{\mathbf{J}(\mathbf{r}')}{\mathfrak{R}}d\tau'$ Can you calculate that integral using spherical coordinates? 1. Yes, no problem 2. Yes, $r'$ can be in spherical, but $\mathbf{J}$ still needs to be in Cartesian components 3. No. Note: * CORRECT ANSWER: B * It's subtle. Griffiths discusses this in a footnote, you can't solve for, say, the phi component of A by integrating the "phi component" of J (because the unit vectors in spherical coordinates themselves depend on position, and get differentiated by del squared too)
For a infinite solenoid of radius $R$, with current $I$, and $n$ turns per unit length, which is the current density $\mathbf{J}$? 1. $\mathbf{J} = nI\hat{\phi}$ 2. $\mathbf{J} = nI\delta(r-R)\hat{\phi}$ 3. $\mathbf{J} = \frac{I}{n}\delta(r-R)\hat{\phi}$ 4. $\mathbf{J} = \mu_0 nI\delta(r-R)\hat{\phi}$ 5. Something else?! Note: * CORRECT ANSWER: B