To manufacture a solenoid of length 1 ~m and inductance 1 mH , the length of thin wire required is (cross - sectional diameter of a solenoid is considerably less than the length)

Step-by-step solution

Inductance of solenoid, L = ₀ ~N ^2 ~A l where, l= length of solenoid. A = r ^2= area of solenoid Let ' x ' be length of wire required. x = circumference of solenoid no. of turns =2 rN . N = x 2 r Substituting in equation (i), aligned L & = ₀ ( x ^2 4 ^2 r ^2 ) r ^2 l L & = ₀ x ^2 4 l x ^2 & = 4 L l ₀ x & = 4 L l ₀ aligned Substituting the values, aligned & x= 4 10⁻³ 1 4 10⁻⁷ & x= 10^4 ~m & x=0.10 ~km aligned