2B · Ampère Law + Vector Potential (Full Solutions)

Q120182022×2

What is magnetic vector potential? (2018 adds: given $\vec{A}$, calculate $\vec{B}$.)

Recall

Steps

1. Define $\vec{A}$ as the field whose curl equals $\vec{B}$; unit T·m. Many $\vec{A}$ give the same $\vec{B}$ (gauge freedom).
2. Write components: $B_x = \partial_y A_z - \partial_z A_y$, $B_y = \partial_z A_x - \partial_x A_z$, $B_z = \partial_x A_y - \partial_y A_x$.
3. Case 1 (2018): $\vec{A} = x^2\hat{i}$ only, so $A_x = x^2$, $A_y = A_z = 0$.
4. Apply to Case 1: $B_y = \partial_z(x^2) - 0 = 0$, $B_z = 0 - \partial_y(x^2) = 0$, $B_x = 0 - 0 = 0$.
5. Case 2 (2018): $\vec{A} = (x^2+y^2+z^2)\hat{i}$, so $A_x = x^2+y^2+z^2$, $A_y = A_z = 0$.
6. Apply to Case 2: $B_x = 0$, $B_y = \partial_z A_x = 2z$, $B_z = -\partial_y A_x = -2y$.

Answer

Exam tip

Q2201820192023×3

State Ampère's circuital law; mathematical form / show $\nabla\times\vec{B} = \mu_0\vec{J}$.

Recall

Steps

1. Statement: $\oint\vec{B}\cdot d\vec{l} = \mu_0 I_{in}$ — line integral of $\vec{B}$ round any closed loop equals $\mu_0$ times the enclosed current.
2. Proof for a long wire: take an Amperian circle of radius $r$ around it. $B = \mu_0 I/2\pi r$ is tangent to the circle and constant on it.
3. Integrate: $\oint\vec{B}\cdot d\vec{l} = B(2\pi r) = \mu_0 I$. By superposition this holds for any loop shape and any set of steady currents.
4. Get the differential form: by Stokes' theorem $\oint\vec{B}\cdot d\vec{l} = \int(\nabla\times\vec{B})\cdot d\vec{S}$ and $I_{in} = \int\vec{J}\cdot d\vec{S}$.
5. Since the surface is arbitrary, the integrands must match: $\nabla\times\vec{B} = \mu_0\vec{J}$.

Answer

Exam tip

Q32022

What is the meaning of $\nabla\cdot\vec{B} = 0$? Using Biot–Savart law prove $\nabla\cdot\vec{B} = 0$.

Recall

Steps

1. State the meaning in three points: (a) no magnetic monopoles — field lines never start or end; (b) lines close on themselves (circles round currents); (c) net flux through any closed surface is zero, $\oint\vec{B}\cdot d\vec{S} = 0$.
2. Biot–Savart integrand is a curl: write $d\vec{l}\times\hat{r}/r^2$ as the curl (with respect to the field point) of $d\vec{l}/r$.
3. Hence the total $\vec{B}$ equals $\nabla\times\vec{A}$ for some vector potential $\vec{A}$ (the vector potential from Q1).
4. Apply the identity: divergence of any curl is identically zero, $\nabla\cdot(\nabla\times\vec{A}) = 0$, because mixed partials cancel ($\partial_x\partial_y = \partial_y\partial_x$).

Answer

Exam tip