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A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth’s horizontal magnetic field.

Radhika Batra , 10 Years ago
Grade 11
anser 1 Answers
Kevin Nash

Last Activity: 10 Years ago

Sol. M/B base H (found in the previous problem) = 3.75 *10^3 A-m^2 T^-1 θ = 37°, d = ? M/B base H = 4π/μ base 0 (d^2 - ℓ^2)^3/2 tan θ ℓ << d neglecting ℓ w.r.t.d ⇒ M/B base H = 4π/μ base 0 d^3 Tanθ⇒ 3.75 * 10^3 = 1/10^-7 * d^3 * 0.75 ⇒ d^3 = 3.75 * 10^3 * 10^-7/0.75 = 5 * 10^-4 ⇒ d = 0.079 m = 7.9 cm

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