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A rod with circular cross-section area 2cm² and length 40 cm is wound uniformly with 400 turns of an insulated wire. If a current of 0.4 A flows in the wire windings, the total magnetic flux produced inside windings is $4 π \times 10^{- 6} Wb$ . The relative permeability of the rod is
( Given : Permeability of vacuum $μ_{0} = 4 π \times 10^{- 7} N A^{- 2}$ )

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125

12.5

$\frac{32}{5}$

$\frac{5}{16}$

answer is A.

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Detailed Solution

$ϕ = NBA = NμmiA$

$\Rightarrow μ = \frac{ϕ}{NmiA}$

$\Rightarrow μ = \frac{4 π \times 10^{- 6}}{400 \times \frac{400}{0.4} \times 0.4 \times 2 \times 10^{- 4}}$

$\Rightarrow μ = \frac{4 π}{32} \times 10^{- 6}$

$\Rightarrow μ_{r} = \frac{μ}{μ_{0}} = \frac{\frac{4 π}{32} \times 10^{- 6}}{4 π \times 10^{- 7}} = \frac{5}{16}$

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