Deepak Patra
Last Activity: 10 Years ago
Sol. γ base 1 = 40 oscillations/minute
B base H = 25 μT
m fo second magnet = 1.6 A-m^2
d = 20 cm = 0.2 m
(a) For north facing north
γ base 1 = 1/2π√MB base H/I γ base 2 = 1/2π√M(B base H – B)/I
B = μ base 0/4π m/d^3 = 10^-7 *1.6/8 * 10^-3 = 20 μT
γ base 1/γ base 2 = √B/B base H – B ⇒ 40/γ base 2 √25/5 ⇒ γ base 2 = 40/√5 = 17.88 = 18 osci/min
(b) For north pole facing south
γ base 1 = 1/2π√MB base H/I γ base 2 = 1/2π√M(B base H – B)/I
γ base 1/γ base 2 = √B/B base H – B ⇒ 40/γ base 2 = √25/45 ⇒ γ base 2 = 40/√(25/45) = 53.66 = 54 osci/min