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(B) The magnetic field $B$ at the centre of a long solenoid is given by $B = \mu_0 n I$,where $n = \frac{N}{L}$ is the number of turns per unit length

Vedclass · Accepted Answer

$2 \pi 	imes 10^{-3} \, T$

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(B) The magnetic field $B$ at the centre of a long solenoid is given by $B = \mu_0 n I$,where $n = \frac{N}{L}$ is the number of turns per unit length.Given that the wire has a radius $r = 0.5 \, mm$,the diameter of the wire is $d = 2r = 1 \, mm = 10^{-3} \, m$.Since the turns are closely wound,the number of turns per unit length is $n = \frac{1}{d} = \frac{1}{2r} = \frac{1}{10^{-3} \, m} = 10^3 \, m^{-1}$.The current $I = 5 \, A$.Substituting these values into the formula:$B = \mu_0 n I = (4 \pi 	imes 10^{-7} \, T \cdot m/A) 	imes (10^3 \, m^{-1}) 	imes (5 \, A)$$B = 20 \pi 	imes 10^{-4} \, T$$B = 2 \pi 	imes 10^{-3} \, T$.

$A$ long solenoid is fabricated by closely winding a wire of radius $0.5 \, mm$ over a cylindrical frame so that the successive turns nearly touch each other. What is the magnetic field at the centre of the solenoid if it carries a current of $5 \, A$?

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