Answer 1

Given:
Magnitude of current, i = 5 A
Radius of the wire, b\[= 10 \text{ cm }= 10 \times {10}^{- 2}\]  m

 For a point at a distance a from the axis,
Current enclosed, \[i'   =   \frac{i}{\pi b^2} \times \pi a^2\]

By Ampere's circuital law,

\[\oint B . dl = \mu_0 i'\]

For the given conditions,

\[B \times 2\pi a   =    \mu_0 \frac{i}{\pi b^2} \times \pi a^2 \] 

\[ \Rightarrow B   =   \frac{\mu_0 ia}{2\pi b^2}        \ldots\left( 1 \right)\]

\[(a)\text{  a = 2 cm }= 2 \times {10}^{- 2} \] m

Again, using the circuital law, we get

 

\[B   =   \frac{4\pi \times {10}^{- 7} \times 5 \times 2 \times {10}^{- 2}}{2\pi \times {10}^{- 2}}\] 

\[ =   2 \times  {10}^{- 6}   T = 2  \mu \] T

(b) On putting  \[\text{ a = 10 cm }= 10 \times {10}^{- 2} \] m in (1), we get

B = 10 `μ T` 

(c)Using the circuital law, we get

\[\oint B . dl = \mu_0 i\]
\[B = \frac{\mu_0 i}{2\pi a} = \frac{2 \times {10}^{- 7} \times 5}{20 \times {10}^{- 2}}\]
\[ = 5 \times {10}^{- 6} T = 5 \mu \] T