InterviewSolution
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InterviewSolution
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Figure shows a square loop ABCD with edge length `a`. The resistance of the wire ABC is `r` and that of ADC is `2r`. Find the magnetic field B at the centre of the loop assuming uniform wires. ` A. `(2mu_0I)/(pir)`B. `(sqrt2mu_0I)/(pir)`C. `(sqrt2mu_0I)/(3pir)`D. `(2mu_0I)/(3pir)` |
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Answer» Correct Answer - C Let `I_1` and `I_2` be the currents in arms PQS and PTS respectively. Then `I=I_1+I_2`. As resistance of arm `PQS=R` and resistance of arm `PTS=2R` Therefore, `I_1=(2I)/(3)` and `I_2=I/3` Magnetic field induction at O due to current through PQ or QS is `B_(PQ)=B_(QS)=(mu_0)/(4pi)(I_1)/(r//2)[sin pi/4+sin pi/4]` `=(mu_0)/(4pi)(2I//3)/(r//2)xx2/sqrt2=(sqrtmu_0I)/(3pir)` acting perpendicular to the plane of loop downwards. Magnetic field induction at O due to current through PT or TS is `B_(PT)=B_(TS)=(mu_0)/(4pi)(I//3)/(r//2)(sin pi/4+sin pi/4)=(sqrt2mu_0I)/(6pir)` acting perpendicular to the plane of loop upwards. The resultant magnetic field induction at O is `B=B_(PQ)+B_(QS)-B_(PT)-B_(TS)` `=(sqrt2mu_0I)/(3pir)+(sqrt2mu_0I)/(3pir)-(sqrt2mu_0I)/(6pir)-(sqrt2mu_0I)/(6pir)` `=(sqrt2mu_0I)/(3pir)` |
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