Aditi Chauhan
Last Activity: 11 Years ago
Sol. B base H = 24 × 10^–6 T T base 1 = 0.1’
B = B base H – B base wire = 2.4 × 10^–6 – μ base 0/2π i/r = 24 * 10^-6 – 2 * 10^-7 *18/0.2 = (24 - 10) * 10^-6 = 14 * 10^-6
T = 2π√I/MB base H T base 1/T base 2 = √B/B base H
⇒ 0.1/T base 2 = √14 * 10^-6/24 * 10^-6 ⇒ (0.1/T base 2)^2 = 14/24 ⇒ 0.01* 14/24 ⇒ T base 2 = 0.076