A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is 1.0 T. The magnetic intensity in SI units is $$\alpha \times 10^5$$. The value of $$\alpha$$ is ______.

(Free space permeability $$\mu_0 = 4\pi \times 10^{-7}$$ SI units.)

The magnetic flux density, magnetic field strength and permeability are related by
$$B = \mu\,H = \mu_{0}\,\mu_{r}\,H$$

Hence
$$H = \frac{B}{\mu_{0}\,\mu_{r}}$$

Given data:
$$B = 1.0\ \text{T},\qquad \mu_{r} = 400,\qquad \mu_{0} = 4\pi \times 10^{-7}\ \text{H m}^{-1}$$

Substituting the values:
$$H = \frac{1.0}{\bigl(4\pi \times 10^{-7}\bigr)\,(400)}$$

Simplify the denominator:
$$4\pi \times 10^{-7} \times 400 = 1600\pi \times 10^{-7} = 1.6\times10^{-4}\,\pi$$

Therefore
$$H = \frac{1}{1.6\times10^{-4}\,\pi}\ \text{A m}^{-1}$$

Working out the numerical factor:
$$\frac{1}{1.6\times10^{-4}} = \frac{1}{1.6}\times10^{4} = 0.625\times10^{4} = 6.25\times10^{3}$$

Hence
$$H = \frac{6.25\times10^{3}}{\pi}\ \text{A m}^{-1}$$

Express this in the form $$H = \alpha \times 10^{5}\ \text{A m}^{-1}$$:
$$\frac{6.25\times10^{3}}{\pi} = \frac{0.0625\times10^{5}}{\pi}$$

Thus
$$\alpha = \frac{0.0625}{\pi} = \frac{1}{16\pi}$$

Therefore, the correct choice is:
Option B which is: $$\frac{1}{16\pi}$$