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. `

Options

A. `(2mu_0I)/(pir)`
B. `(sqrt2mu_0I)/(pir)`
C. `(sqrt2mu_0I)/(3pir)`
D. `(2mu_0I)/(3pir)`

Answer

C

Explanation

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)`