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  4. A magnetising field of 2 × 103 A m-1 produces a magnetic flux density of 8 π T in an iron rod. The relative permeability of the rod will be

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Q. A magnetising field of $2 \times 10^3\, A\, m^{-1}$ produces a magnetic flux density of $8\, \pi\, T$ in an iron rod. The relative permeability of the rod will be

Magnetism and Matter

Solution:

Here, $H = 2 \times 10^3\, A\,m^{-1}$,
$B = 8\pi \,T$, $\mu_{0} = 4\,\pi \times 10^{-7}$
Since $\mu_{r} = \frac{\mu}{\mu_{0}} = \frac{\mu H}{\mu_{0}H} = \frac{B}{\mu _{0}H}$
$ = \frac{8\pi}{4\pi \times 10^{-7}\times 2 \times 10^{3}} = 10^{4}$