The susceptibility of Al is 2 times 10^{-5}. | vedclass

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(B) The magnetic field $B$ inside a toroid filled with a magnetic material is given by $B = \mu_0(H + M) = \mu_0 H(1 + \chi)$, where $\chi$ is the mag

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$2 	imes 10^{-3}$

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(B) The magnetic field $B$ inside a toroid filled with a magnetic material is given by $B = \mu_0(H + M) = \mu_0 H(1 + \chi)$, where $\chi$ is the magnetic susceptibility.Since the initial magnetic field in vacuum is $B_0 = \mu_0 H$, we have $B = B_0(1 + \chi)$.The increase in the magnetic field is $\Delta B = B - B_0 = B_0 \chi$.The fractional increase is $\frac{\Delta B}{B_0} = \chi$.The percentage increase is $\frac{\Delta B}{B_0} 	imes 100 = \chi 	imes 100$.Given $\chi = 2 	imes 10^{-5}$, the percentage increase is $(2 	imes 10^{-5}) 	imes 100 = 2 	imes 10^{-3} \%$.

The susceptibility of $Al$ is $2 \times 10^{-5}$. The percent increase in the magnetic field when the space within a current-carrying toroid is filled with $Al$ is:

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