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The value of susceptibility of aluminium is $2.2 \times 10^{- 5}$ . The percentage increase in the magnetic field if space with in a current carrying toroid is filled with aluminium is $\frac{X}{10^{4}}$ then the value of “ $X$ ” is ______

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answer is 22.

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Detailed Solution

$B = μ_{0} μ_{r} H$

$= μ_{0} (1 + χ) H$

$B = B_{0} [1 + χ]$

$B - B_{0} = B_{0} χ$

$\frac{B - B_{0}}{B_{0}} = χ$

$\frac{B - B_{0}}{B_{0}} \times 100 = 100 χ$

$= 100 [22 \times 10^{- 5}]$

$= 22 \times 10^{- 3}$

$= 22 \times 10^{- 4}$

$= \frac{22}{10^{4}}$

$\frac{B - B_{0}}{B_{0}} \times 100 = {\frac{22}{10^{4}}}^{}$

Given it is $= \frac{X}{10^{4}}$

So, $X = 22$

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