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(B) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$,where $n$ is the number of turns per unit length and $I$ is

Vedclass · Accepted Answer

$6.28 	imes 10^{-2} \ 	ext{Wb/m}^2$

Vedclass · Answer

(B) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$,where $n$ is the number of turns per unit length and $I$ is the current.Given:Number of turns per $cm$,$n = 200 \ 	ext{turns/cm} = 200 	imes 10^2 \ 	ext{turns/m} = 2 	imes 10^4 \ 	ext{turns/m}$.Current,$I = 2.5 \ A$.Permeability of free space,$\mu_0 = 4\pi 	imes 10^{-7} \ 	ext{Wb/A} \cdot 	ext{m}$.Substituting the values into the formula:$B = (4\pi 	imes 10^{-7}) 	imes (2 	imes 10^4) 	imes 2.5$$B = 4\pi 	imes 10^{-7} 	imes 5 	imes 10^4$$B = 20\pi 	imes 10^{-3} \ 	ext{T}$$B = 2\pi 	imes 10^{-2} \ 	ext{T} \approx 6.28 	imes 10^{-2} \ 	ext{Wb/m}^2$.

$A$ long solenoid has $200$ turns per $cm$ and carries a current of $2.5 \ A$. The magnetic field at its centre is $(\mu_0 = 4\pi \times 10^{-7} \ \text{Wb/A} \cdot \text{m})$.

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