Problem 9.3.1

Statement

9.3.1. Using the formula given in problem 9.2.1, determine the magnetic induction near a uniformly charged plate that moves with velocity $$v$$ along its plane. The surface charge density of the plate is $\sigma$ .

Solution

The electric field of a uniformly charged plate is:

$E=\frac{\sigma}{2\varepsilon_{0}}$

The field is directed perpendicular to the plate. Problem 9.2.1 gives the formula:

$\vec{B}=\mu_{0}\varepsilon_{0}\left[ \vec{v} \times \vec{E}\right]$

Since the velocity is directed perpendicular to the electric field voltage vector, then for the magnitude of the magnetic induction vector we obtain:

$B=\mu_{0}\varepsilon_{0}vE \sin\frac{\pi}{2}=\mu_{0}\varepsilon_{0}vE$

$\vec{B}$ is directed parallel to the surface of the plate and perpendicular to the direction of its motion.

Substitute the value of $$E$$ into the formula:

$B=\mu_{0}\varepsilon_{0}v\frac{\sigma}{2\varepsilon_{0}}=\frac{\mu_{0}\sigma v}{2}$

Answer

$B=\frac{\mu_{0}\sigma v}{2}$

Formulas in this solution the whole sheet

$уповерхностипроводника$E = \frac{\sigma}{2\varepsilon_0}, \qquad E = \frac{\sigma}{\varepsilon_0}\ \text{(у поверхности проводника)}$$ Field of a charged plane 6.2 · in 44 more problems
$\vec B = \varepsilon_0\mu_0[\vec v\times\vec E] = \frac{\mu_0}{4\pi}\,\frac{q[\vec v\times\vec r]}{r^3}$ Field of a moving charge 9.2 · in 9 more problems
$внутрислояснаружислоя$B = \frac{\mu_0 i}{2}, \qquad B = \mu_0 j x\ \text{(внутри слоя)}, \qquad B = \frac{\mu_0 j d}{2}\ \text{(снаружи слоя)}$$ Field of a sheet current 9.3 · in 23 more problems

Contributed by emixter Edited by astrosander Last edited Mar 26, 2025 All edits

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9.3: The magnetic field from a current distributed over a surface of space

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