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A current of 2 ampere is passed in a coil of radius 0.5 m and number of turns 20.The magnetic moment of the coil isA. `0.314 Am^(2)`B. `0.14A-m^(2)`C. `314A-m^(2)`D. `31.4A-m^(2)`

Answer» Correct Answer - D
I=2A,r=0.5m,n=20,M=?
M=nIA
`nlpir^(2)`
`20xx2xx3.14xx0.25`
`=3.14Am^(2)`


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