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(A) Given: Volume $V = 30\, cm^3 = 30 	imes 10^{-6} \, m^3$,Magnetising field $H = \frac{1250}{\pi} \, A/m$,Magnetic moment $M = 6 \, A\cdot m^2$.Fir

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(A) Given: Volume $V = 30\, cm^3 = 30 	imes 10^{-6} \, m^3$,Magnetising field $H = \frac{1250}{\pi} \, A/m$,Magnetic moment $M = 6 \, A\cdot m^2$.First,calculate the intensity of magnetisation $I$:$I = \frac{M}{V} = \frac{6}{30 	imes 10^{-6}} = 0.2 	imes 10^6 = 2 	imes 10^5 \, A/m$.Next,calculate the magnetic susceptibility $\chi$:$\chi = \frac{I}{H} = \frac{2 	imes 10^5}{1250/\pi} = \frac{2 	imes 10^5 	imes \pi}{1250} = 160\pi$.Calculate relative permeability $\mu_r$:$\mu_r = 1 + \chi = 1 + 160\pi \approx 1 + 502.65 = 503.65$.Finally,calculate magnetic induction $B$:$B = \mu_0 \mu_r H = (4\pi 	imes 10^{-7}) 	imes (1 + 160\pi) 	imes \frac{1250}{\pi}$.$B = 4 	imes 10^{-7} 	imes (1 + 160\pi) 	imes 1250 = 5000 	imes 10^{-7} 	imes (1 + 160\pi)$.$B = 5 	imes 10^{-4} 	imes (1 + 502.65) = 5 	imes 10^{-4} 	imes 503.65 \approx 0.2518 \, T$.Rounding to the nearest option,the value is $0.25 \, T$.

By putting a magnetic substance of volume $30\, cm^3$ in a magnetising field of strength $\frac{1250}{\pi} \, A/m$,the value of the produced magnetic moment is $6\, A\cdot m^2$. Calculate the value of magnetic induction produced in the substance in $T$.

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