Explanation Given info: The inductance of the inductor is 450 μH , the length and diameter of the inductor to be made is 12 cm and 4.0 cm respectively. Explanation: The number of turns that a solenoid needs to result the desired inductance is to be found out. Write the expression for the self-inductance of a solenoid. L = μ 0 n 2 A l Here, L is the inductance μ 0 is the permeability of free space n is the number of turns per unit length of the solenoid A is the cross sectional area of the solenoid l is the length of the solenoid. Since the number of turns per unit length is n = N l where, N is the total number of tuns, the expression for the inductance can be modified as, L = μ 0 N 2 A l = μ 0 N 2 ( 1 4 π d 2 ) l Here, d is the diameter of the solenoid Solve for N . N = L l μ 0 ( 1 4 π d 2 ) Substitute 450 μH for L , 12 cm for l , 1.2566 × 10 − 6 m ⋅ kg ⋅ s -2 ⋅ A -2 for μ 0 and 4

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ISBN: 9780134988566

Author: Wolfson, Richard

Publisher: Pearson Education,

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Chapter 27, Problem 18E

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The number of turns to be wound to make the inductor.

Solution Summary: The author explains the inductance of a solenoid, the length and diameter of which is 185.

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