Answer:
281 turns
Explanation:
Energy stored (U) = 0.395 J
[tex]u \: = \frac{1}{2} l {i}^{2} = 0.395[/tex]
After Rearranging, we get
[tex]l(self \: inductance) = \frac{2u}{ {i}^{2} } = 5.05 \times {10}^{ - 3} [/tex]
Therefore,
[tex]l = \frac{ \beta {n}^{2}a }{2\pi r} = 5.05 \times {10}^{ - 3}\ H [/tex]
Where,
[tex] \beta (permeability \: of \: free \: space) = 4\pi \times {10}^{ - 7} [/tex]
[tex] n \: = number \: of \: turns[/tex]
Now
[tex]n \: = \sqrt{ \frac{2\pi \: r \: \times l}{ \beta \times a} } [/tex]
[tex]n = \sqrt{ \frac{2 \times 3.14 \times15.5 \times 5.05 \times {10 }^{ - 3} }{4\pi \times {10}^{ - 7} \times4.95 } } [/tex]
[tex]n = 281.1 (approx.) = 281 turns [/tex]
