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A long solenoid has $ 200 $ turns per $ cm $ and carries a current $ i $ . The magnetic field at its centre is $ 6.28 \times {10^{ - 2}}Wb{m^{ - 2}} $ . Another long solenoid has $ 100 $ turns per $ cm $ and it carries a current $ i/3 $ . The value of magnetic field at its centre is
A. $ 1.05 \times {10^{ - 2}}Wb{m^{ - 2}} $
B. $ 1.05 \times {10^{ - 5}}Wb{m^{ - 2}} $
C. $ 1.05 \times {10^{ - 3}}Wb{m^{ - 2}} $
D. $ 1.05 \times {10^{ - 4}}Wb{m^{ - 2}} $

Answer
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Hint: Using the Ampere’s circuital law establishes the relation for the magnetic field inside a solenoid. From that equation we can calculate the current flowing in the first solenoid. Then substitute this value of current for the second solenoid.

Formulas used:
 $ \oint {\overrightarrow B } \cdot d\overrightarrow l = {\mu _0}i $ where $ B $ is the magnetic field around any closed path, $ {\mu _0} $ is the permeability of free space and $ i $ is the current flowing through the area enclosed by the path.
 $ B = {\mu _0}ni $ where $ n $ is the number of turns of the coil per unit length.

Complete step by step answer:
A solenoid is a type of electromagnet made by wounding coils into a tightly packed helix. When current flows through the conducting coil, a magnetic field is generated inside the solenoid such that it behaves as a magnet. We can calculate the strength of the magnetic field inside the solenoid using Ampere’s law.
Ampere’s circuital law states that the line integral of the magnetic field $ \overrightarrow B $ around any closed oath is equal to $ {\mu _0} $ times the net current $ i $ flowing through the area enclosed by the path. That is,
 $ \oint {\overrightarrow B } \cdot d\overrightarrow l = {\mu _0}i $
where $ {\mu _0} $ is the permeability of free space (constant).
Using this law we can establish the relation of the magnetic field at the centre of a long solenoid which is given by the expression $ B = {\mu _0}ni $ where $ n $ is the number of turns of the coil per unit length.
According to the question we have, $ B = 6.28 \times {10^{ - 2}}Wb{m^{ - 2}} $ and $ n = 200 \times {10^2}{m^{ - 1}} $
 $ B = {\mu _0}ni $
 $ \Rightarrow $ $ i = \dfrac{B}{{{\mu _0}n}} $
 $ \Rightarrow i = \dfrac{{6.28 \times {{10}^{ - 2}}}}{{{\mu _0} \times 200 \times {{10}^2}}} $
Now for the second solenoid, we have $ n = 100 \times {10^2}{m^{ - 1}} $
Thus, substituting the value of $ i $ from the previous equation we have,
  $ {B_2} = {\mu _0} \times 100 \times {10^2} \times \dfrac{{6.28 \times {{10}^{ - 2}}}}{{{\mu _0} \times 200 \times {{10}^2} \times 3}} $
 $ \Rightarrow {B_2} = 1.05 \times {10^{ - 2}}Wb{m^{ - 2}} $
Therefore, the correct option is A.

Note
The field $ \overrightarrow B $ is independent of the length and the diameter of the solenoid and is uniform over the cross-section of the solenoid. The uniform magnetic field within a long solenoid is parallel to the solenoid axis. Also the expression used in this solution to calculate the magnetic field is only valid for a very long solenoid. For a solenoid with definite length, the formula is different.