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(D) The magnetic induction $B$ is given by the formula $B = \mu_0(M_v + H)$,where $M_v$ is the intensity of magnetization and $H$ is the magnetic fiel

Vedclass · Accepted Answer

$6.28$

Vedclass · Answer

(D) The magnetic induction $B$ is given by the formula $B = \mu_0(M_v + H)$,where $M_v$ is the intensity of magnetization and $H$ is the magnetic field strength.First,calculate the volume $V$ of the rod: $V = 10 \, cm 	imes 0.5 \, cm 	imes 0.2 \, cm = 10^{-1} \, m 	imes 0.5 	imes 10^{-2} \, m 	imes 0.2 	imes 10^{-2} \, m = 10^{-6} \, m^3$.The intensity of magnetization $M_v$ is $M_v = \frac{M}{V} = \frac{5 \, A \cdot m^2}{10^{-6} \, m^3} = 5 	imes 10^6 \, A/m$.The magnetic field strength $H = 0.5 	imes 10^4 \, A/m = 5000 \, A/m$.Now,$B = 4\pi 	imes 10^{-7} 	imes (5 	imes 10^6 + 5000) = 4 	imes 3.14 	imes 10^{-7} 	imes (5,005,000) \approx 12.56 	imes 10^{-7} 	imes 5 	imes 10^6 = 6.28 \, T$.

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