Aditi Chauhan
Last Activity: 10 Years ago
Sol. B base 1 = 2.5 x 10^-3 B base = 2.5
A = 4 x 10^-4 m^2 , n = 50 turns/cm 5000 turns/m
(a) B = μ base 0 ni,
⇒ 2.5 x 10^-3 = 4 π x 10^-1 x 5000 x i
⇒ i = 2.5x 10^-3/ 4π x 10^-7 x 5000 = 0.398 A ≈ 0.4 A
(b) I = B base 2 / μ base 0 – H = 2.5 / 4π x 10 ^ -7 – (B base 2 – B base 1) = 2.5 / 4π x 10^-7 – 2.497 = 1.99 x 10 ^ 6 ≈ 2 x 10^6
(c) I = M/V ⇒ ml/Al = m/A
⇒ m = IA = 2 x 10^6 x 4 x 10^-4 = 800 A-m