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If the magnetic moment of a solenoid carrying current 2A and having area of crosssection `1.0 xx 10^(-4) m^(2) ` is 0.050 `"Am"^(2)` then calculate the number of turns in the solenoid.

Text Solution

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Given : A = `10^(-4) m^(2), I` = 2A, N= ?
m = 0.050 `"Am"^(2)`
We know that , m = NIA
i.e., `" " ((m)/(IA))= N`
i.e., `" " N = (0.05)/(2xx10^(-4)) = 0.025 xx10^(4)= 250 `
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