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The needle of a dip circle shows an apparent dip of 45° in a particular position and 53° when the circle is rotated through 90°. Find the true dip.

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
Navjyot Kalra
Sol. We know B base H = μ base 0in/2r Give : B base H = 3.6 * 10^-5 T θ = 45° i = 10 mA = 10^-2 A tan θ = 1 n = ? r = 10 cm = 0.1 m n = B base H tan θ * 2r/μ base 0i = 3.6 * 10^-5 * 2 * 1 * 10^-1/4π * 10^-7 *10^-2 = 0.5732 * 10^3 = 573 turns
Last Activity: 11 Years ago
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