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The susceptibility of $\mathrm{Al}$ is $2 \times 10^{-5}$. The percent increase in the magnetic field when the space within a current carrying torroid to filled with $\mathrm{Al}$ is
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Verified Answer
The correct answer is:
$2 \times 10^{-3}$
By using the net magnetic field in the torroid filled with Al.
$\begin{aligned}
& B=\mu(H+I)=\mu H\left(I+\frac{I}{H}\right) \\
& B=\mu H(1+\chi) \\
& B=B_0(1+\chi),
\end{aligned}$
$\left[\because\right.$ where $x=$ susceptibility of aluminium $=2 \times 10^{-5}$ ]
$\frac{B-B_0}{B_0}=\chi$
Now, percentage increase in magnetic field
$\begin{aligned}
\frac{B-B_0}{B_0} \times 100= & \chi \times 100=2 \times 10^{-5} \times 10^2 \\
& =2 \times 10^{-3} \%
\end{aligned}$
$\begin{aligned}
& B=\mu(H+I)=\mu H\left(I+\frac{I}{H}\right) \\
& B=\mu H(1+\chi) \\
& B=B_0(1+\chi),
\end{aligned}$
$\left[\because\right.$ where $x=$ susceptibility of aluminium $=2 \times 10^{-5}$ ]
$\frac{B-B_0}{B_0}=\chi$
Now, percentage increase in magnetic field
$\begin{aligned}
\frac{B-B_0}{B_0} \times 100= & \chi \times 100=2 \times 10^{-5} \times 10^2 \\
& =2 \times 10^{-3} \%
\end{aligned}$
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