A rod with circular cross-section area 2 cm$$^2$$ 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 the windings is $$4\pi \times 10^{-6}$$ Wb. The relative permeability of the rod is
(Given: Permeability of vacuum $$\mu_0 = 4\pi \times 10^{-7}$$ N A$$^{-2}$$)

Given: Cross-section area $$A = 2$$ cm$$^2$$ $$= 2 \times 10^{-4}$$ m$$^2$$, length $$L = 40$$ cm $$= 0.4$$ m, $$N = 400$$ turns, $$I = 0.4$$ A, total magnetic flux $$\Phi = 4\pi \times 10^{-6}$$ Wb.

The magnetic flux inside the solenoid is:

$$ \Phi = B \times A = \mu_r \mu_0 n I \times A $$

where $$n = N/L$$ is the number of turns per unit length.

$$ n = \frac{400}{0.4} = 1000 \text{ turns/m} $$

$$ \Phi = \mu_r \times 4\pi \times 10^{-7} \times 1000 \times 0.4 \times 2 \times 10^{-4} $$

$$ \Phi = \mu_r \times 4\pi \times 10^{-7} \times 400 \times 2 \times 10^{-4} $$

$$ \Phi = \mu_r \times 4\pi \times 10^{-7} \times 8 \times 10^{-2} $$

$$ \Phi = \mu_r \times 32\pi \times 10^{-9} = \mu_r \times 3.2\pi \times 10^{-8} $$

Setting this equal to the given flux:

$$ \mu_r \times 3.2\pi \times 10^{-8} = 4\pi \times 10^{-6} $$

$$ \mu_r = \frac{4 \times 10^{-6}}{3.2 \times 10^{-8}} = \frac{4}{0.032} = 125 $$