The value of aluminum's susceptibility is $2.2 \times 10^{- 5}$ . The percentage increase in the magnetic field, if space within a current carrying toroid, is filled with aliminum, is $\frac{x}{10^{4}}$ . Then, the value of $x$ is

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Solution

$B = μ (H + 1)$
$B = μ . H (1 + \frac{1}{H})$
$B = B_{0} (1 + χ)$
$B - B_{0} = B_{0} χ$
$\frac{B - B_{0}}{B_{0}} = χ$
$\frac{B - B_{0}}{B_{0}} \times 100 = 100 χ$
$= 2.2 \times 10^{- 3} = \frac{22}{10^{4}}$
$\Rightarrow x = 22$

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The value of aluminum's susceptibility is 2.2×10−5. The percentage increase in the magnetic field, if space within a current carrying toroid, is filled with aliminum, is x104. Then, the value of x is

B=μ(H+1) B=μ.H(1+1H) B=B0(1+χ) B−B0=B0χ B−B0B0=χ B−B0B0×100=100 χ =2.2×10−3=22104 ⇒ x=22

The value of aluminum's susceptibility is $2.2 \times 10^{- 5}$ . The percentage increase in the magnetic field, if space within a current carrying toroid, is filled with aliminum, is $\frac{x}{10^{4}}$ . Then, the value of $x$ is

$B = μ (H + 1)$
$B = μ . H (1 + \frac{1}{H})$
$B = B_{0} (1 + χ)$
$B - B_{0} = B_{0} χ$
$\frac{B - B_{0}}{B_{0}} = χ$
$\frac{B - B_{0}}{B_{0}} \times 100 = 100 χ$
$= 2.2 \times 10^{- 3} = \frac{22}{10^{4}}$
$\Rightarrow x = 22$