InterviewSolution
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InterviewSolution
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Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potentail difference applied across them so that the magnetic field at their centres is the same?A. 3B. 4C. 6D. 2 |
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Answer» Correct Answer - B At the centre of coil-1 `B_(1)=(mu_(0))/(4pi)xx(2pii_(i))/(r_(1))" "......(i)` At the centre of coil-2 `B_(2)=(mu_(0))/(4pi)xx(2pii_(2))/(r_(2))" ".......(ii)` But `B_(1)=B_(2)` `:.(mu_(0))/(4pi)(2pii_(1))/(r_(1))=(mu_(0))/(4pi)(2pii_(2))/(r_(2))or(i_(1))/(r_(1))=(i_(2))/(r_(2))` As `r_(1)=2r_(2)` `:.(i_(1))/(2r_(2))=(i_(2))/(r_(2))ori_(1)=2i_(2)` Now, ratio of potential differences, `(V_(2))/V_(1)=(i_(2)xxr_(2))/(i_(1)xxr_(2))=(i_(2)xxr_(2))/(2i_(2)xx2r_(2))=(1)/(4)` `implies(V_(1))/(V_(2))=(4)/(1)` |
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