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Two concentric coplanar circular loops of radii `r_(1)` and `r_(2)` carry currents of respectively `i_(1)` and `i_(2)` in opposite direction (one clockwise and the other anticlockwise). The magnetic induction at the centre of the loops is half that due to `i_(1)` alone at the centre. if `r_(2)=2r_(1)`. the value of `i_(2)//i_(1)` isA. `2`B. `1//2`C. `1//4`D. `1` |
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Answer» Correct Answer - D Magnetic field at centre due to smaller loop `B_(1)=(mu_(0))/(4pi). (2pii_(1))/(r_(1)).......(i)` Due to Bigger loop `B_(2)=(mu_(0))/(4pi). (2pii_(2))/(r_(2)).......(i)` So net magnetic field at centre `B=B_(1)-B_(2)=(mu_(0))/(4pi)xx2pi((i_(1))/(r_(1))-(i_(2))/(r_(2)))` According to question `B=1/2xxB_(1)` `(mu_(0))/(2pi).2pi((i_(1))/(r_(1))-(i_(2))/(r_(2)))=1/2xx(mu_(0))/(4pi). (2pii_(1))/(r_(1))` `(i_(1))/(r_(1))-(i_(2))/(r_(2))=(i_(1))/(2r_(1)) implies (i_(1))/(2r_(1))=(i_(2))/(r_(2)) implies (i_(2))/(r_(2))=1{r_(2)=2r_(1)}` |
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