A single-layer coil (solenoid) has length $l$ and cross-section radius $R$ . A number of turns per unit length is equal to $n$ . The magnetic induction at the centre of the coil when a current $I$ flows through it is given as $B = \frac{x μ_{0} n I}{\sqrt{1 + {(\frac{2 R}{l})}^{2}}}$ . Find $x$ .

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Solution

Magnetic field at the centre of a finite solenoid , $B_{c e n t r e} = \frac{μ_{o} n I}{2} [cos θ_{1} + cos θ_{2}]$
Here, $θ_{1} = θ_{2} = θ$
$⟹$ $B_{c e n t r e} = μ_{o} n I cos θ$
Now $cos θ = \frac{l / 2}{\sqrt{R^{2} + (\frac{l}{2})^{2}}} = \frac{1}{\sqrt{1 + (\frac{2 R}{l})^{2}}}$
Thus, $B_{c e n t r e} = \frac{μ_{o} n I}{\sqrt{1 + (\frac{2 R}{l})^{2}}}$
Thus, $x = 1$

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