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Question: A long solenoid has \( 200 \) turns per \( cm \) and carries a current \( i \) . The magnetic field ...

A long solenoid has 200200 turns per cmcm and carries a current ii . The magnetic field at its centre is 6.28×10−2Wbm−26.28 \times {10^{ - 2}}Wb{m^{ - 2}} . Another long solenoid has 100100 turns per cmcm and it carries a current i/3i/3 . The value of magnetic field at its centre is
A. 1.05×10−2Wbm−21.05 \times {10^{ - 2}}Wb{m^{ - 2}}
B. 1.05×10−5Wbm−21.05 \times {10^{ - 5}}Wb{m^{ - 2}}
C. 1.05×10−3Wbm−21.05 \times {10^{ - 3}}Wb{m^{ - 2}}
D. 1.05×10−4Wbm−21.05 \times {10^{ - 4}}Wb{m^{ - 2}}

Explanation

Solution

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:
∮B→⋅dl→=μ0i\oint {\overrightarrow B } \cdot d\overrightarrow l = {\mu _0}i where BB is the magnetic field around any closed path, μ0{\mu _0} is the permeability of free space and ii is the current flowing through the area enclosed by the path.
B=μ0niB = {\mu _0}ni where nn 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 B→\overrightarrow B around any closed oath is equal to μ0{\mu _0} times the net current ii flowing through the area enclosed by the path. That is,
∮B→⋅dl→=μ0i\oint {\overrightarrow B } \cdot d\overrightarrow l = {\mu _0}i
where μ0{\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=μ0niB = {\mu _0}ni where nn is the number of turns of the coil per unit length.
According to the question we have, B=6.28×10−2Wbm−2B = 6.28 \times {10^{ - 2}}Wb{m^{ - 2}} and n=200×102m−1n = 200 \times {10^2}{m^{ - 1}}
B=μ0niB = {\mu _0}ni
⇒\Rightarrow i=Bμ0ni = \dfrac{B}{{{\mu _0}n}}
⇒i=6.28×10−2μ0×200×102\Rightarrow i = \dfrac{{6.28 \times {{10}^{ - 2}}}}{{{\mu _0} \times 200 \times {{10}^2}}}
Now for the second solenoid, we have n=100×102m−1n = 100 \times {10^2}{m^{ - 1}}
Thus, substituting the value of ii from the previous equation we have,
B2=μ0×100×102×6.28×10−2μ0×200×102×3{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}}
⇒B2=1.05×10−2Wbm−2\Rightarrow {B_2} = 1.05 \times {10^{ - 2}}Wb{m^{ - 2}}
Therefore, the correct option is A.

Note
The field B→\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.