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The magnetic field  B  inside a long solenoid carrying a current of 10 A, is  3.14 \times 10^{-2} \mathrm{~T}. Find the no of turns per unit length of the Solenoid.


 

Option: 1

1500 turns /m


Option: 2

3500 turns /m


Option: 3

2000 turns /m


Option: 4

2500 turns /m


Answers (1)

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For, solenoid having  n  turns, then

\text { magnetic field, } B = { \mu _{0}\,ni }

\Rightarrow 3.14 \times 10^{-2}=4 \pi \times 10^{-7} \times n \times 10 \\

\Rightarrow n=\frac{3.14 \times 10^{-2}}{4 \pi \times 10^{-7} \times 10} \\

\Rightarrow n=2500 \text { turns } / \mathrm{~m}

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