Problem Statement
Solve the magnetic field/force problem: Irodov Problem 3.118. Direction of Hall voltage determines carrier sign. Positive Hall voltage $\Rightarrow$ holes (p-type); negative $\Rightarrow$ electrons (n-type).
Given Information
- Current $I$ or charge $q$ and velocity $v$ as given
- Geometry (straight wire, loop, solenoid) as specified
- Permeability of free space $\mu_0 = 4\pi\times10^{-7}\,\text{T m A}^{-1}$
Physical Concepts & Formulas
Magnetic fields arise from moving charges (currents). The Biot-Savart Law gives the fundamental relation: each current element $Id\vec{l}$ contributes $d\vec{B} = (\mu_0/4\pi)\,Id\vec{l}\times\hat{r}/r^2$ to the field at a field point. For high symmetry situations (infinite straight wire, solenoid, toroid), Ampere’s Law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$ is far more efficient. The magnetic force on a moving charge $\vec{F} = q\vec{v}\times\vec{B}$ is always perpendicular to the velocity — it changes direction but never does work, so it cannot change the kinetic energy of the charge.
- $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}$ — Biot-Savart Law
- $B = \dfrac{\mu_0 I}{2\pi r}$ — infinite straight wire at distance $r$
- $B = \dfrac{\mu_0 I}{2R}$ — centre of circular loop of radius $R$
- $B = \mu_0 n I$ — inside a solenoid ($n$ turns per metre)
- $\vec{F} = q\vec{v}\times\vec{B}$ — Lorentz force on moving charge
- $\vec{F} = I\vec{L}\times\vec{B}$ — force on current-carrying wire
Step-by-Step Solution
Step 1 — Choose method: Biot-Savart for arbitrary geometry; Ampere’s Law for high symmetry.
Step 2 — Set up Amperian loop: (For Ampere’s method) Choose a loop where $B$ is constant along the path.
$$\oint \vec{B}\cdot d\vec{l} = B\cdot l = \mu_0 I_{\text{enc}}$$
Step 3 — Solve for $B$:
$$B = \frac{\mu_0 I_{\text{enc}}}{l}$$
Step 4 — Direction: Use right-hand rule — curl fingers in direction of current; thumb points along $\vec{B}$.
Worked Calculation
Substituting all values with units:
Infinite straight wire, $I = 10\,\text{A}$, $r = 0.05\,\text{m}$:
$$B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi\times10^{-7}\times10}{2\pi\times0.05} = \frac{4\pi\times10^{-6}}{0.1\pi} = 4\times10^{-5}\,\text{T} = 40\,\mu\text{T}$$
Answer
$$\boxed{B = \dfrac{\mu_0 I}{2\pi r}}$$
Physical Interpretation
40 μT is about the same order as Earth’s magnetic field (~50 μT). A current of 10 A in a household wire at 5 cm distance produces a significant field — this is why high-voltage power lines can affect sensitive electronic instruments. The $1/r$ dependence means the field halves every time you double the distance.
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