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Magnetic Field due to a Current — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMagnetic Field due to a Current5 Marks
Question
A solid wire of radius $10\ cm$ carries a current $5.0\ A$ 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. $r = 10cm = 10 \times 10^{-2}m$
$x = 2 \times 10^{-2}m,$
$i = 5A$
i in the region of radius 2cm
$\frac{5}{\pi(10\times\times10^{-2})^2}\times\pi(2\times10^{-2})^2=0.2\text{A}$
$\text{B}\times\pi(2\times10^{-2})^2=\mu_0(0-2)$
$\Rightarrow\text{B}=\frac{4\pi\times10^{-7}\times0.2}{\pi\times4\times10^{-4}}=\frac{0.2\times10^{-7}}{10^{-4}}=2\times10^{-4}$
  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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