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A solenoid has a core of a material with relative permeability $\frac{800}{\pi}$. The windings of the solenoid are insulated from the core and carry current of $2 \mathrm{~A}$. If the number of turns is 1000 per metre, find the magnetic field $B$.
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The correct answer is:
$640 \mathrm{mT}$
Given, relative permeability, $\mu_r=\frac{800}{\pi}$
Current, $I=2 \mathrm{~A} \Rightarrow n=1000 \mathrm{~m}^{-1}$
As we know that,
$H=n I=1000 \times 2=2 \times 10^3 \mathrm{Am}^{-1}$
Since, $\quad B=\mu H=\mu_0 \mu_r H \quad\left(\because \mu=\mu_0 \mu_r\right)$
$=4 \pi \times 10^{-7} \times \frac{800}{\pi} \times 2 \times 10^3$
$=0.64 \mathrm{~T}=640 \mathrm{mT}$
Current, $I=2 \mathrm{~A} \Rightarrow n=1000 \mathrm{~m}^{-1}$
As we know that,
$H=n I=1000 \times 2=2 \times 10^3 \mathrm{Am}^{-1}$
Since, $\quad B=\mu H=\mu_0 \mu_r H \quad\left(\because \mu=\mu_0 \mu_r\right)$
$=4 \pi \times 10^{-7} \times \frac{800}{\pi} \times 2 \times 10^3$
$=0.64 \mathrm{~T}=640 \mathrm{mT}$
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