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The linear loop has an area of `5 xx 10^(-4)m^(2)` and a resistance oof `2 Omega`. The larger circular loop is fixed and has a radius of `0.1m`. Both the loops are concentric and coplanner. The smaller loop is rotated with an angular velocity `omegarads^(-1)` about its dismeter. The magnetic flux with the smaller loop is A. `2pi xx 10^(-6) "weber"`B. `pi xx 10^(-9) "weber"`C. `pi xx 10^(-9) cos omega t "weber"`D. zero

Answer» Correct Answer - C
Magnetic field due to larger loop
`=(mu_(0)1)/(2r)=(4pixx10^(-7)xx1)/(2xx0.1)T=2pixx10^(-6)T`
Now, magnetic flux linked withthe smaller loop, `phi`
`=NBA cos omega t=1xx2pixx10^(-6) xx5xx10^(-4) cos omegat`


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