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The inductance `L` of a solenoid of length `l`, whose windings are made of material of density `D` and resistivity `rho`, is (the winding resistance is`R`)A. `(mu_(0))/(4 pil) (Rm)/(rho D)`B. `(mu_(0))/(4piR) (lm)/(rho D)`C. `(mu_(0))/(4pil) (R^(2)m)/(rhoD)`D. `(mu_(0))/(2 pi R) (lm)/(rhoD)` |
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Answer» Correct Answer - A For a solenoid, `L = mu_(0)N^(2) (A)/(l)*` if `x` is the length of the wire and `a` is the area of cross-section, then `R = (rho x)/(a)` and `m = axD` `Rm = (rhox)/(a) axD, x = sqrt((Rm)/(rho D))` Also, `x = 2 pi rN, N = (x)/(2 pir) ( :. L = (mu_(0)N^(2)A)/(l))` `:. L = mu_(0)((x)/(2pir))^(2) (pir^(2))/(l) = (mu_(0))/(4 pil) (Rm)/(rhoD)` |
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