1. Given data

• Radius of smaller coil: $r_1 = 1\,\text{cm} = 10^{-2}\,\text{m}$
• Radius of larger coil: $r_2 = 1000\,\text{cm} = 10\,\text{m}$
• Number of turns in smaller coil: $N_1 = 10$
• Number of turns in larger coil: $N_2 = 200$

Since the coils are concentric and coaxial, and one coil is much smaller than the other, the magnetic field produced by the larger coil may be taken as nearly uniform over the area of the smaller coil.

2. Formula for mutual inductance

For two coaxial circular coils, when one coil is much smaller than the other,

$M = \frac{\mu_0 N_1 N_2 A_{\text{small}}}{2R_{\text{large}}}$

where

• $A_{\text{small}} = \pi r_1^2$
• $R_{\text{large}} = r_2$
• $\mu_0 = 4\pi \times 10^{-7}\,\text{H/m}$

3. Area of smaller coil

$A_{\text{small}} = \pi (10^{-2})^2 = \pi \times 10^{-4}\,\text{m}^2$

4. Substitute into formula

$M = \frac{(4\pi \times 10^{-7})(10)(200)(\pi \times 10^{-4})}{2(10)}$

Now simplify step-by-step:

$M = \frac{4\pi^2 \times 10^{-7} \times 2000 \times 10^{-4}}{20}$

Since

$\frac{2000}{20} = 100$

we get

$M = 4\pi^2 \times 100 \times 10^{-11}$

$M = 4\pi^2 \times 10^{-9}$

Given $\pi^2 = 10$,

$M = 4 \times 10 \times 10^{-9} = 40 \times 10^{-9}$

$M = 4 \times 10^{-8}\,\text{H}$

5. Final integer value

Since mutual inductance is $\underline{\hspace{1cm}} \times 10^{-8}\,\text{H}$,

$\boxed{4}$

6. Comparison with stored answer

• Derived answer: $4$
• Stored correct answer: $4$

They match.