A solenoid of length ‘l’ has N turns of wire closely spaced, each turn carrying a current i. If R is cross sectional radius of the solenoid, the magnetic induction at ‘P’(only axial component).

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$\frac{μ_{0} iN}{2 \sqrt{R^{2} + l^{2}}}$

$\frac{μ_{0} iN}{2 \sqrt{l^{2} + 4 R^{2}}}$

$\frac{μ_{0} Ni}{4 \sqrt{R^{2} + l^{2}}}$

$\frac{μ_{0} Ni}{2 R^{2} + l^{2}}$

answer is B.

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Detailed Solution

$\begin{array}{l} B = \frac{μ_{0} ni}{2} (\cos θ_{1} + \cos θ_{2}) \\ \cos θ_{1} = \frac{1}{\sqrt{l^{2} + R^{2}}} \\ n = \frac{N}{l} \\ \cos θ_{2} = \cos 90 = 0; B = \frac{μ_{o} iN}{2 \sqrt{l^{2} + R^{2}}} \end{array}$

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