A closely wound solenoid 80 cm …
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Sia ? 6 years, 5 months ago
Here, l= 80 cm = 0.8 m, N = 5 {tex}\times{/tex} 400 = 2000
I = 8.0 A, D = 1.8 cm, n = no. of turns per unit length
Magnitude of magnetic field inside a solenoid near its center
{tex}n = \frac{{Total\;turn}}{{length}}{/tex}
{tex}n = \frac{{2000}}{{0.80}}{/tex}
{tex}B = {\mu _0}nI = \frac{{4\pi \times {{10}^{ - 7}} \times 2000 \times 8.0}}{{0.80}}{/tex}
{tex} = 8\pi \times {10^{ - 3}}T = 2.5 \times {10^{ - 2}}T{/tex}
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