The percentage increase in magnetic field (B) | vedclass

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(A) The magnetic field inside an air-cored solenoid is $B_0 = \mu_0 n I$.When the core is filled with a material of relative permeability $\mu_r$,the

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$1.2 	imes 10^{-3} \%$

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(A) The magnetic field inside an air-cored solenoid is $B_0 = \mu_0 n I$.When the core is filled with a material of relative permeability $\mu_r$,the new magnetic field is $B = \mu n I = \mu_0 \mu_r n I$.The change in magnetic field is $\Delta B = B - B_0 = \mu_0 (\mu_r - 1) n I$.The percentage increase is given by $\frac{\Delta B}{B_0} 	imes 100 \% = (\mu_r - 1) 	imes 100 \%$.Since magnetic susceptibility $\chi = \mu_r - 1$,the percentage increase is $\chi 	imes 100 \%$.Given $\chi_{mg} = 1.2 	imes 10^{-5}$,the percentage increase is $(1.2 	imes 10^{-5}) 	imes 100 \% = 1.2 	imes 10^{-3} \%$.

The percentage increase in magnetic field $(B)$ when the space within a current-carrying solenoid is filled with magnesium (magnetic susceptibility $\chi_{mg} = 1.2 \times 10^{-5}$) is:

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