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Calculate the self inductance of a solenoid having 500 turns about a cylindrical core of 2 cm radius in which µr = 50 for 0 < rho < 0.5 cm. |
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Answer» Answer: The self-inductance is given as, L= l μ 0
N 2 A
= 50×10 −2
4π×10 −7 ×(500) 2 ×π×(2×10 −2 ) 2
=7.896×10 −4 H Thus, the self-inductance of an air SOLENOID is 7.896×10 −4 H. Explanation: please mark as brainlist |
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