InterviewSolution
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InterviewSolution
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The torque required to hold a small circular coil of 10 turns, area `1mm^(2)` and carrying a current of `((21)/(44))A` in the middle of a long solenoid of `10 ^(3)"turns"//"m"` carrying a current of 2.5A, with its axis perpendicular to the axis of the solenoid is A. `1.5xx10^(-6)Nm`B. `1.5xx10^(-8)Nm`C. `1.5xx10^(6)Nm`D. `1.5xx10^(8)Nm` |
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Answer» Correct Answer - B Here, For small circular coil, Number of turns, `N` = 10, Area , `A`=1 `"mm"^(2)` = `1xx10^(-6)" m"^(2)` Current, ` I_(1)=(21)/(44)A ` For a long solenoid, Number of turns per metre, `n=10^(3)` per m Current, `I_(2)=2.5` A Magnetic field due to a long solenoid on its axis is `B=mu_(0)nI_(2)" "...(i)` Magnetic moment of a circular coil is ` M= NAI_(1)" "...(ii)` ` vec(tau)=vec(M) xx vec(B) ` ` tau=MBsintheta=MB" "(because theta=90^(@)("Given"))` `tau=(NA I_(1))(mu_(0)nI_(2))" "("Using"(i)and (ii)) ` ` tau=10xx1xx10^(-6) xx(21)/(44 )xx4xx(22)/(7)xx10^(-7)xx10^(3)xx2.5` `=1.5xx10^(-8)" N m"` |
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