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

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsMagnetic Field due to a Current5 Marks
Question
A solid wire of radius 10cm carries a current 5.0A distributed uniformly over its cross-section. Find the magnetic field B at a point at a distance (a) 2cm (b) 10cm and (c) 20cm away from the axis. Sketch a graph of B versus x for 0 < x < 20cm.
✓

Answer

  1. r = 10cm = 10 × 10-2m

x = 2 × 10-2m,

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. 10cm 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 = 20cm

$\text{B}\times\pi(20\times10^{-2})^2=\mu_0\times5$

$\Rightarrow\text{B}=\frac{\mu_0\times5}{\pi\times(20\times10^{-2})^2}=\frac{4\pi10^{-7}\times5}{\pi\times400\times10^{-4}}=5\times10^{-5}$

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