Tutor profile: Marc V.

There is an infinite line current along the $$z$$ axis with current $$I$$. Find the magnetic field $$B$$

The magnetic field $$B$$ can be found using Amperes law. For this we have $( \oint_C \vec{B} \cdot \mathrm{d} l = \mu_0 \iint_S \vec{J} \cdot \mathrm{d}\vec{s}. $) By selecting polar coordinates $$(r,\phi,z)$$ and $$C$$ as the circle of radius $$r$$ enclosing the line current going counterclockwise, we see that the double integral of the current density is just the current on the wire, $$I$$. This is because the normal is in the positive $$z$$ direction and the integral of the current density $\vec{J}$ through the surface is the total current $I$. Then, the line integral on the left is just $$2 \pi |\vec{B}| $$, by symmetry. Thus, using the right-hand rule we can see that the magnetic field must be $( \vec{B}=\frac{\mu_0 I}{2 \pi r} \hat{\phi} $)

Given $$A$$ and $$B$$ are $$n\times n$$ matrices and $$A^{-1}$$ is the inverse of $$A$$, state whether or not the following hold: 1. $$AB=BA$$. 2. $$\det (ABA^{-1})=\det(B)$$. 3. $$A$$ has nonzero eigenvalues. 4. $$B$$ has an orthonormal set of eigenvectors.

1. False, unlike scalar multiplication, matrix multiplication is not commutative in general. 2. True. This can be seen by noting that $$\det (ABA^{-1})=\det(A)\det(B)\det(A^{-1})$$. This then reduces to $$\frac{\det (A)}{\det (A)} \det(B)=\det (B)$$. 3. True. Since $$A$$ is invertible and $$\det (A)\neq 0$$, and its determinant is the product of its eigenvalues, no eigenvalue can be zero. 4. False. We cannot state this in general. As a counter example, consider $$B=\begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}$$. It is simple to calculate its eigenvectors as $$v_\pm=[\pm 1/\sqrt{2},1]^T$$, which have $$v_+\cdot v_- \neq 0$$.

Given a function $$f(x)$$ is continuous on the domain $$(a, b)$$, does $$f(x)$$ satisfy the following? Explain or provide examples. 1. $$f(x)$$ is is differentiable on $$(a,b)$$. 2. $$f(x)$$ is integrable $$a,b$$. 3. The limit $$\lim_{x\rightarrow c} f(x)$$ exists for any $$a< c < b$$.

1. No. Differentiability requires continuity, by definition, but not vice-versa. See, for example, $$f(x)=|x|$$, which is continuous but not differentiable at $$x=0$$ 2. No. For example, if we select $$f(x)=1/x$$ on the interval $$(0,1)$$, we satisfy out continuity requirements, but the integral $$\int_0^1f(x)\mathrm{dx}$$ does not exist. 3. This is true. By definition, if $$f(x)$$ is continuous on $$(a,b)$$, then limit for $$a<c<b$$ must exist.