There are $50$ $turns$ of a wire in every $cm$ length of a long solenoid. If $4\, ampere$ current is flowing in the solenoid, the approximate value of magnetic field along its axis at an internal point and at one end will be respectively
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(c) The magnetic field in the solenoid along its axis

$(i)$ At an internal point$ = {\mu _o}ni$
$ = 4\pi \times {10^{ - 7}} \times 5000 \times 4 = 25.1 \times {10^{ - 3}}\,Wb/{m^2}$
$({\rm{Here }}n = 50\,turns/cm = 5000\,turns/m)$
$(ii)$ At one end
${B_{end}} = \frac{1}{2}{B_{in}} = \frac{{{\mu _0}ni}}{2} = \frac{{25.1 \times {{10}^{ - 3}}}}{2}$$ = 12.6 \times {10^{ - 3}}\,Wb/{m^2}$

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