An iron rod 0.2 m long, 10 mm in diameter and of permeability 1000 is placed inside a long soleniod wound with 300 turns per meter. If a current of 0.5 ampere is passed through the rod, find the magnetic moment of the rod.
Answered Feb 05, 2023
Correct Answer - `0.2325 Am^(-1)`
Correct Answer - `m=(1)/(3)piNI(R_(1)^(2)+R_(1)R_(2)+R_(2)^(2))=2.2 xx 10^(-3)A m^(2)`
Correct Answer - `[16 xx 10^(-2)T]`
Correct Answer - `[3.78 xx 10^(-6)Nm]`
Correct Answer - [5.76 T]
Correct Answer - `2xx10^(-3)WB m^(-2), 30 A "turns" m^(-1), 500`
Correct Answer - 2000 turns `m^(-1) , 7.9 xx 10^(5) "A turns" m^(-1), 397`
Correct Answer - `0.3 Wbm^(-1) 397.9 A "turns" m^(-1), 0.2995 Wbm^(-2)`
Correct Answer - `15.8 s^(-1)`
Correct Answer - `25I mu V`
Correct Answer - `0.86muH,4.3xx10^(-4)`V
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