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Magnetic Field due to a Current — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsMagnetic Field due to a Current5 Marks
Question
A solid wire of radius $10\ cm$ carries a current $5.0A$ distributed uniformly over its cross$-$section. Find the magnetic field $B$ at a point at a distance $(a) 2\ cm\ (b) 10\ cm$ and $(c) 20\ cm$ away from the axis. Sketch a graph of $B$ versus $x$ for $0 < x < 20\ cm.$
✓

Answer

  1. $r = 10\ cm = 10 \times 10^{-2}m$
  2. $x = 2 \times 10^{-2}m,$$i = 5A$
  3. $i$ in the region of radius $2\ cm$
$\frac{5}{\pi(10\times\times10^{-2})^2}\times\pi(2\times10^{-2})^2=0.2\text{A}$
$\text{B}\times\pi(2\times10^{-2})^2=\mu_0(0-2)$
$\Rightarrow\text{B}=\frac{4\pi\times10^{-7}\times0.2}{\pi\times4\times10^{-4}}=\frac{0.2\times10^{-7}}{10^{-4}}=2\times10^{-4}$
  1. $10\ cm$ radius
$\text{B}\times\pi(10\times10^{-2})^2=\mu_0\times5$
$\Rightarrow\text{B}=\frac{4\pi\times10^{-7}\times5}{\pi\times10^{-2}}=20\times10^{-5}$
  1. $x = 20\ cm$
$\text{B}\times\pi(20\times10^{-2})^2=\mu_0\times5$
$\Rightarrow\text{B}=\frac{\mu_0\times5}{\pi\times(20\times10^{-2})^2}=\frac{4\pi10^{-7}\times5}{\pi\times400\times10^{-4}}=5\times10^{-5}$

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